Ordinary Differential Equations and Infinite Series [2 ed.] 9780176716141, 0176716149, 9781774124994

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Ordinary Differential Equations and Infinite Series [2 ed.]
 9780176716141, 0176716149, 9781774124994

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Ordinary Differential Equations and Infinite Series

Second Edition

Sam Melkonian

School of Mathematics and Statistics Carleton University Ottawa, Canada

NELSON

NELSON

COPYRIGHT@ 2016 by Sam Melkonian. ALL RIGHTS RESERVED. Published by Nelson Education Ltd. Printed and bound in Canada 6 7 8 9 19 18 17 16 For more information contact Nelson Education Ltd., 1120 Birchmount Road, Toronto, Ontario, M1 K 5G4. Or you can visit our Internet site at nelson.com

No part of this work covered by the copyright herein may be reproduced, transcribed, or used in any form or by any means-graphic, electronic, or mechanical, including photocopying, recording, taping, Web distribution, or information storage and retrieval systems­ without the written permission of the publisher. For permission to use material from this text or product, submit all requests online at cengage.com/permissions. Further questions about permissions can be emailed to [email protected] Every effort has been made to trace ownership of all copyrighted material ·and to secure permission from copyright holders. In the event of any question arising as to the use of any material, we will be pleased to make the necessary corrections in future printings. This textbook is a Nelson custom publication. Because your instructor has chosen to produce a custom publication, you pay only for material that you will use in your course.

ISBN-13: 978-0-17-671614-1 ISBN-10: 0-17-671614-9 Consists of Original Works: Ordinary Differential Equations and Infinite Series

Sam Melkonian Cover Credit:

EtiAmmos/Shutterstock

111

\

Preface The prerequisites for this book are elementary differential and integral calculus and linear algebra. The exercises at the end of each section follow the progression of the topics in that section, and are approximately in increasing order of difficulty. The exercises at the end of certain chapters cover the material of the entire chapter, may be more difficult, may involve material from earlier chapters, or may be of a more theoretical nature. The solutions of all of the exercises are provided in Appendix C. Proofs of theorems which are mere computations are furnished in order to provide the reader with a better understanding of the subject. More abstract or otherwise difficult proofs are omitted from the main text. The proofs of selected theorems are given in Appendix B. Appendix A contains formulas and techniques from trigonometry, calculus and linear algebra which may prove useful to the reader. The second edition is identical to the first, with the exception of minor changes and corrections. Sam Melkonian April 2015

Contents Preface

I 1

111

Ordinary Differential Equations Introduction 1.1 Basic Concepts Exercises 1.1 ..

3 3 7

2 First-Order Equations . 2 1 Separable Equations ...... 2.1.1 Orthogonal Trajectories Exercises . 2 1 ....... . 2 2 Homogeneous Equations Exercises 2.2 ....... . 2 3 Linear Equations .... . 2 . 3 1 Bernoulli Equations . Exercises . 2 3 ........ 2.4 Functions of Two Variables 2.4.1 Partial Derivatives 2.4.2 T he Chain Rule . Exercises 2.4 ........ . 2 5 Exact Equations ..... 2.5.1 Integrating Factors Exercises 2.5 .. Chapter 2 Exercises ..... 3

1

9 9 1 1 1 4 1 5 8 1 8 1 2 1 2 3 2 4 2 4 26 27 8 2 3 3 36 38

Second-Order Equations . 3 1 Basic Definitions ... Exercises 3.1 ........... . 3 2 Linear Homogeneous Equations 3.2.1 Equations with Constant Coefficients . 3 . 2 2 Cauchy-Euler Equations ...... V

43 4 3 6 4 7 4 8 4 5 2

CONTENTS

Vl

Exercises 3.2 . . . . . . . . . . . . . . . . . . . . . Linear Nonhomogeneous Equations . . . . . . . . 3.3.1 The Method of Undetermined Coefficients 3.3.2 Variation of Parameters . . . . . . . . Exercises 3.3 . . . . . . . . . . . . . . . . . . . 3.4 Equations Reducible to First-Order Equations Exercises 3.4 . . Chapter 3 Exercises . . . . . . . . . . 3.3

55

57 57 67 74 76 83 84

4 Higher-Order Linear Equations 4.1 Homogeneous Equations . . . . . . . . . . . 4.1.1 Equations with Constant Coefficients 4.1.2 Cauchy-Euler Equations Exercises 4.1 . . . . . . . . . . . 4.2 Nonhomogencous Equations . . 4.2.1 Variation of Parameters 4.2.2 Undetermined Coefficients Annihilators Exercises 4.2 . . Chapter 4 Exercises .

87 88 88 98 101 103 103 111 118 119

5

121 123 123 126 135 146 168 171 186 186

II 6

Linear Systems 5.1 Homogeneous Systems . . . . . . . . . . . 5.1.1 General Theory . . . . . . . . . . . 5.1.2 Systems with Constant Coefficients Complex Eigenvalues . . . Generalized Eigenvectors . Exercises 5.1 . . . . . . . . 5.2 Nonhomogeneous Systems Exercises 5.2 . . Chapter 5 Exercises . . . . . . .

Infinite Series Sequences and Series 6.1 Sequences . . Exercises 6.1 . . . . 6.2 Series . . . . . . . . 6.2.1 The Integral Test Approximations of Series . 6.2.2 The Comparison Tests . . 6.2.3 Alternating Series . . . . . Approximations of Alternating Series

189 191 191 205 208 215 219 223 229 231

CONTENTS

Vll

6.2.4 Absolute and Conditional Convergence Exercises 6.2 . . Chapter 6 Exercises .

233 240 243

7 Taylor Series 7.1 Power Series . Exercises 7.1 . 7.2 Representations of Functions by Power Series The Binomial Series . . . . . . . . . . . Taylor Polynomials and Approximations Exercises 7.2 .

245 245 251 252 263 269 279

8

283 283 297 299 311 313

9

Fourier Series 8.1 Fourier Series of Exercises 8.1 . . 8.2 Fourier Series of Exercises 8.2 . . Chapter 8 Exercises .

Periodic Functions . . . . . . . . . . . Functions on F inite Intervals . . . . . . . . . . . . . . .

Series Solutions of Differential Equations 9.1 Solutions About Ordinary Points Exercises 9 .1 . . . . 9.2 Periodic Solutions . Exercises 9.2 . . . .

317 317 326 328 332

Appendices A Formulas and Techniques A.I Trigonometric Identities A.2 Integration Techniques A.3 Matrix Inversion . . . .

B Proofs of Selec ted Theorems B.1 B.2 B.3 B.4 B.5 B.6

The The The The The The

Comparison Test . . . . Limit Comparison Test Alternating Series Test Ratio Test . . . . . Root Test . . . . . . Binomial Theorem .

335 335 335 340 341 341 342 343 343 345 346

viii

CONTENTS

C Solutions C.1 Chapter 1 Solutions . C.2 Chapter 2 Solutions . C.3 Chapter 3 Solutions . C.4 Chapter 4 Solutions . C.5 Chapter 5 Solutions . C.6 Chapter 6 Solutions . C.7 Chapter 7 Solutions . C.8 Chapter 8 Solutions . C.9 Chapter 9 Solutions .

347 347 348 360 377 390 408 419 426

Index

445

439

Part I Ordinary Differential Equations

1

...

Chapter 1 Introduction 1.1

Basic Concepts

Definition 1. 1. An ordinary differential_ equation of order n � l is a relation which contains the derivative of order n of an unknown function y, called the dependent variable, and may also contain lower-order derivatives of y as well as given functions of an independent variable x or t. Example 1. 1. y'

+ y = x2 + ex is a first-order ordinary differential equation.

Example 1.2. y111 = 2y2

+ y' + 1 is a third-order ordinary differential equation.

The term ordinary refers to differential equations in which the unknown function y is a function of a single variable. In contrast, differential equations which contain unknown functions of more than one variable are called partial differential equations. This book is concerned only with ordinary differential equations, to which we shall refer simply as differential equations, or equations for short. Definition 1.2. A solution of a differential equation is any differentiable function f such that, if y is replaced by J(x), y' by f'(x), y" by J"(x), etc., then the equation is satisfied for all x in an interval.

....

Example 1.3. The function f (x) = e 2x is a solution of the equation y' - 2y = 0 because f' (x) = 2e 2x and, hence, f' ( x) - 2f ( x) = 2e 2x - 2e 2x = 0 for all x, the relevant interval here being the entire real line IR = (-oo, oo). 3

4

CHAPTER 1. INTRODUCTION

Exampl� 1.4. The first-order equation y' +x and reqmres no special techniques:

Y +x=0 ⇒ y I

/

= -x

= 0 can be solved simply by integration

⇒ y

=

J

x2

-- + c

-xdx =

2

x where c is an arbitrary constant of integration. Thus, f(x) = -2 for every value of· c. In particular, 2

f(x)

= - x , J(x) = - x + 1 2 2 2

2

are solutions, corresponding to c are displayed in Figure 1.1.

=

0, c

=1

and f(x)

and c

=

+ c is a

solution

= -2 - 2 x2

-2, respectively. Their graphs

y

X

Figure 1.1: Particular solutions of y' + x

= 0 and directions.

Definition 1.3. The solution of a first-order equation which contains one arbitrary constant is called the general solution. It is a one-parameter family of solutions, i.e., a set of solutions in which every solution corresponds to a particular value of the parameter c. A solution in which c is assigned a particular value is called a particular solution. In Example 1.4, y y

2

= -�

2

+ c is the general solution, and

= -x ,

2

y=-

x2

2

+1

and y

=-

2 -2

x2

1.1. BASIC CONCEPTS

5

are particular solutions. At every point (x, y) on any one of these curves, the slope of the tangent line is :�

= -x, i.e., y' + x = 0,

which is the differential equation. Thus,

the differential equation defines a direction at every point (x, y) in the plane (where it is defined), and the graph of every solution is a curve which follows these directions at every point on the curve. A few directions are indicated in Figure 1.1. Definition 1.4. The set of directions defined by a first-order differential equation of the form y' = g(x, y) is called a direction field, and the graph of a solution is called a solution curve. In Definition 1.4, g(.rc, y) denotes an expression which may include both x and y. For example, g( x, y)

= x 2 + y2 or g( x, y) = '!!_.

discussed at greater length in Section 2.4.

X

Such functions of two variables will be

Definition 1.5. An initial-value problem for a first-order equation consists of a first­ order equation, together with an initial condition, which specifies the value of y at a single point x and, thereby, determines a particular solution. For the equation y' + x = 0 in Example 1.4, the initial condition y(0) = 1 forces c = 1. Hence, the solution of the initial-value problem y'

+ x = 0,

The initial condition y(4) problem y'

=

y(0) = 1, is y

-1 forces

+ X = 0,

y(4)

=-

x2

2 + 1.

c = 7.

Hence, the solution of the initial-value

= -1,

is y

=-

x2

2 + 7.

Example 1.5. Let y(t) > 0 denote the amount of a substance present at time t, and suppose that the rate at which y changes with time is proportional to the amount present at time t. This means that dy = ky(t), dt which is a first-order ordinary differential equation, where k is the proportionality constant. This equation determines, for example, the (approximate) number y(t) of bacteria present in a culture at time t. Since y is an increasing function oft, y' > 0 and, hence, k > 0. This equation also determines the amount y(t) of a radioactive substance present at time t. Since the substance decays with time, y' < 0 and, hence, k < 0. If a quantity y(t) is constant in time, then y' = 0 and, hence, k = 0.

6

CHAPTER 1. INTRODUCTION

If we attempt to solve the equation as in Example 1.4, we obtain y'

*

= ky

y=

J

kydt,

and the integral on the right cannot be evaluated because y is not known as a function of t. It is therefore necessary to perform certain algebraic manipulations before an integration can be performed. Express the equation in the equivalent form � y ' y . respect to t to obtam y'

= ky *

By the substitution rule, J

=k *

ty'

1 , . dt = -:;/

J

=k

and integrate both sides with

ty' dt =

Jy

1 dy dt = dt

Jy 1

J

kdt.

dy,

and the integral on the right can be evaluated without knowledge of y as a function of t. Thus, we obtain

j

tdy =

Evaluating both integrals, we obtain

j

kdt.

where c1 and c2 are arbitrary constants of integration and the combination c2 - c1 may be denoted by the single arbitrary constant c3. Taking the exponential of both sides and employing the fact that e10 (a) = a for any a > 0, we obtain where c = ±ec3 is an arbitrary constant since c3 is arbitrary. At t = 0, we obtain y(0) = c. Thus, the solution may be expressed as y(t) = y(0)e kt _ The quantity y grows exponentially if k > 0, decays exponentially if k < 0, and has the constant value y(0) if k = 0. Example 1.6. The equation Example 1.5 with k = 2.

dy dt

= 2y

has the general solution y(t)

Example 1. 7. The initial-value problem y' = -3y, y(0) y(t) = 2e- 3t by Example 1.5 with k = -3 and y(0) = 2.

=

y(0)e 2t by

2, has the solution

1.1. BASIC CONCEPTS

7

= 0 and solve the initial-value

Example 1.8. Find the general solution of y' - xy problem given the initial condition y(0) = -2.

=

If y(x) 0 (i.e., y = 0 for all x), then y'(x) y(x) ¢. 0, then, proceeding as in Example 1.5, y' - xy = 0



y' = xy



tv'

=x ⇒

= 0, and the equation is satisfied. If

j tv'

dx

j

=

x dx

jt



dy

=

j

x dx

2

x c x2 /2 ⇒ ln lvl = ⇒ y = ±ecex2 /2 = Kex2 /2 + C ⇒ lvl = e e 2

=

is the general solution, where K = 0 corresponds to the solution y 0. y(0) = -2 ⇒ 2 K = -2 ⇒ y = -2ex / 2 is the solution of the initial-value problem. The method employed in solving the first-order equations in the present chapter ca,11 be applied to all equations of a particular type, and will be discussed in greater generality in Chapter 2, as well as the methods required to solve other types of first-order equation. Second-order equations will be discussed in Chapter 3, where the concepts general solution, particular solution and initial-value problem will be suitably generalized.

Exercises 1.1 1. Find the general solution of the equation y' + 2x 1 2. Find the general solution of the equation -y' X

3. Find the general solution of the equation

= 4cos(2x).

!� =

4. Find the general solution of the equation (

= xe x .

dy dx)

2x cos(x2 ). 2

= 4x2e 2x .

1 5. Find the general solution of the equation -- - y' COS (X ) 6. Solve the initial-value problem

dy

dt

7. Solve the initial-value problem y'

= ln(t),

2

= 2 sin(x).

l > 0, y(l)

= 2.

= 4cos 2(x), y (�) =

. . . 8. Fm d the genera1 so1ut10n of the equation y' =

i

x+3 x _x_6 2

8

CHAPTER 1. INTRODUCTION y 9. Find the general solution of the equation d x d 10. Solve the initial-value problem

!� =

y 11. Solve the initial-value problem d

dt

3y, y(O)

=

12e 2x sin(3x).

= 2.

= -2y, y(O) = 3.

12. Solve the initial-value problem y' + cos(x)y = 0, y(1r) = -3. 13. Solve the initial-value problem 3y2 y'

=

1, y(O) = 2.

Chapter 2 First-Order Equations 2.1

Separable Equations

Definition 2.1. A first-order differential equation is separable if it can be placed in the form dy = (x) g(y) dx J , possibly after some algebraic manipulation. The essential characteristic of a separable equation is that the quantity g(y) which dy multiplies is independent of x, whereas the right-hand side f(x) of the equation is dx independent of y. Either or both of these functions may, of course, be constant. The reader may readily verify that the equations in Examples 1.4, 1.5 and 1.8 in Chapter 1 are separable. The method employed in their solution is applicable to any separable equation. Thus, employing the substitution rule, we obtain g(y) :�

= f(x) ⇒

J

g(y) :� dx

=

J

J(x) dx



J

g(y) dy

=

J

J(x) dx,

and evaluation of the last two integrals gives a relation between x and y containing one arbitrary constant. (As seen in the examples in Chapter 1, a difference c2 - c 1 of two arbitrary constants may be replaced by a single one.) It may or may not be possible to solve the relation for y explicitly as a function of x. Nevertheless, it constitutes, at least in principle, the general solution. Example 2.1. The equation 2d dy form 2ydx

J 2y �� dx

= 3x =

2

j

.

Then

3x 2 dx



j

dy

X

=

2y dy

3x 2 is separable because it can be placed in the y

=

j

3x 2 dx

9



y2

= x3 + c ⇒

y

= ±Jx3 + c

10

CHAPTER 2. FIRST-ORDER EQUATIONS

is the general solution. If the initial condition y(O) = 2 is imposed, then it requires that 2 = ±Jc, which gives c = 4 and we must take the plus sign. The solution of the initial-value problem is then y = ✓x3 + 4. If, instead, the initial condition is y(O) = -2, then,again, c = 4, but we must take the minus sign, and y = - ✓x3 + 4. Example 2.2. The equation (x2 in the form 2yy' =

j

2yy' dx =

j

-!--. Then +4 X

: x2 4

dx =>

j

+ 4)y' = .!__ is separable because it can be placed 2y

2y dy =

j

employing the substitution rule with u = x2

J

x x2

Hence, y =

dx =

+4

±

✓�

1

2

ln(x2

J

2x

1

� dx = 2

J

x2

:

4

dx => y2 = � ln(x2

du dx

+ 4) + c,

+ 4, - = 2x,to obtain

1 du 1 = dx ; dx 2

J

1

; du =

1

2 ln(u) + c.

+ 4) + c.

Example 2.3. Consider the initial-value problem 2

2xyy' = e- Y ln(x), x > 0,

y(l)

= 2.

2 , ln(x) The equation can be placed in the form 2yeY y = --,and is separable. Thus,

j

2yeY y' dx = 2

j

ln x) � dx =>

j

Employing the substitution rule with u = y2 , and v = ln(x),

X

2

2ye Y dy =

j

ln x) . � dx

du = 2y, in the integral on the left, dy

-

dv = ! ,in the integral on the right, we obtain dx x

=> y2 = ln { �[ln(x)] 2

+ c} => y = ± In { �[ln(x)] 2 + c}

is the general solution, and

y(l)

= 2 =>

±� = 2 => ln(c) = 4 => c = e4 => y =

In { �[ln(x)] 2

+ e4 }

is the solution of the initial-value problem.

.

\

.

11

2.1. SEPARABLE EQUATIONS

2.1.1

Orthogonal Trajectories

A circle of radius r > 0 centred at the origin can be described by the equation x2 + y2 = r 2 . The set ( or family or collection) of all such circles constitutes a one­ parameter family of circles, one circle for each value of the parameter r. If we let J(x, y) = x 2 + y 2 , then the family may be described by the equation f(x, y) = r2 or, alternatively, g(x, y, r) = x 2 + y 2 - r2 = 0. A straight line through the origin with slope m can be described by the equation y = mx. The set of all such lines constitutes a one-parameter family of lines, one line for each value of the parameter m. With f(x) = x, the family may be described by y = mf(x) or, alternatively, g(x, y, m) = mx - y = 0. A few circles centred at the origin and a few lines through the origin are displayed in Figure 2.1. y

X

Figure 2.1: Circles centred at the origin and lines through the origin. More generally, a relation between x and y which includes a parameter (constant) defines a set of curves in the plane, one curve (or more) corresponding to each value of the parameter.

Definition 2.2. A one-parameter family of curves is a set of curves defined by an equation of the form g(x, y, c) = 0, where c is the parameter. The relation g(x, y, c) = 0 my take more particular forms, such as y = cf(x), y = J(x, c), or J(x, y) = c. In the first case, g(x, y, c) = y - cf(x) = 0, in the second, g(x, y, c) = y - f(x, c) = 0, and, in the third, g(x, y, c) = f(x, y) - c = 0.

12

CHAPTER 2. FIRST-ORDER EQUATIONS

Notice that, in Figure 2.1, every line intersects every circle orthogonally, i.e., at right angles. More precisely, at every point of intersection, the line tangent to a circle at that point is orthogonal (perpendicular) to the line through the origin. Definition 2.3. A curve is an orthogonal trajectory of a one-parameter family of curves if it intersects every member of the family at right angles. Thus, every straight line through the origin is an orthogonal trajectory of the family of circles centred at the origin, and every circle centred at the origin is an orthogonal trajectory of the family of lines through the origin. Given a one-parameter family of curves, it is often necessary to determine its orthogonal trajectories. To this end, we note that if a line L has slope m, then a 1 line perpendicular to L must have slope --. This fact enables us to determine the m orthogonal trajectories of a given family of curves with the aid of differential equations. X

Example 2.4. The general solution of the first-order equation y' = -- is a one­ y parameter family of curves. Find the general solution and its orthogonal trajectories. The equation is separable, and YY

I

=

2

-x

2

y X + c1 ⇒ ⇒ 2 = -2

y2

=

-x 2



+ 2c1

x2

+ y2 = 2c1 = c,

which defines a one-parameter family of circles centred at the origin for c > 0. Its orthogonal trajectories must satisfy the equation y' = '}!_, which is also separable. If X

y ¢. 0, then

�y' y

=� X



ln IYI

= ln lxl + C



c lvl = e ixl

⇒ y = ±ecx = K x,

=

which defines a one-parameter family of lines through the origin, in agreement with the result inferred from Figure 2.1. The constant function y 0 (the x-axis) satisfies the equation and corresponds to K = 0. The vertical line x = 0 (the y-axis) is an orthogonal trajectory, but is excluded from the solution y = K x. It can be included by regarding x as a function of y, expressing the equation y' = '}!_ as X

dx 1 1 x x, =-=-=-=dy

and noting that x

= 0 is a solution.

dy dx

y'

y'

2.1. SEPARABLE EQUATIONS

13

In Example 2.4, the one-parameter family of curves was obtained by solving a first-order equation. If no equation is given, then an equation must be determined for which the given family is the general solution. Thus, given a one-parameter family of curves, g(x, y, c) = 0, express the relation as f(x, y) = c and differentiate with respect to x, regarding y as a function of x, to eliminate c. Or, eliminate c from the two equations g(x, y, c) = 0 and its derivative with respect to x, regarding y as a function of x. Example 2.5. Find the orthogonal trajectories of the one-parameter family of parabolas y = x 2 + c. Differentiation with respect to x gives y' = 2x. The orthogonal trajectories must 1 1 therefore satisfy y' = --, the general solution of which is y = -- ln lxl + k, and 2x 2 defines the orthogonal trajectories (Figure 2.2). y

Figure 2.2: The parabolas y

)

= x2 + c and their orthogonal trajectories.

Example 2.6. Find the orthogonal trajectories of the one-parameter family of parabolas y = kx2 . x2 y' - 2xy . as Y = We may express the equat10n k and then differentiate to obtain ---X2 x4 2y .1.e., xyI - 2 y = Q, or yI = -. Alternatively, X

y

= kx2 ⇒ :2 = k

and y'

= 2kx ⇒ y' = 2x ( :2 ) =

2

:,

= O'

14

CHAPTER 2. FIRST-ORDER EQUATIONS

as before.

= _ _.:._,

The orthogonal trajectories must therefore satisfy y'

2y

separable. Thus, y'

= -;

y

==> 2yy'

= -x

==> y2

2

= -� + c

=

==> y

±g,

2

or �

which is

+ y2 = c,

which defines a one-parameter family of ellipses for c > 0. A few members of each family are displayed in Figure 2.3

X

Figure 2.3: The parabolas y

= kx 2 and their orthogonal trajectories.

Exercises 2 .1 1. Solve the initial-value problem y' = 3x 2y2, y(0) = _ •





I

2. Fmd the general solution of the equation y = 3. Solve the initial-value problem (x 2

.

+ 1) .

dy dx

4. Find the general solution of the equation

= 4x2

1

2

°

Jx+I . y3

2x + 1 y(0) 3y2 ,

y'

+ 3x- 2

= 3. X

- 5y

4

- 4y3 + 1-

, ln(x) . .. 5. Solve the m1tial-value problem xy - -- = 0, x > 0, y(l) = -3. y 6. Find the general solution of the equation xy' = (x + l)e Y .

---- ---

I

15

2.2. HOMOGENEOUS EQUATIONS 7. Solve the initial-value problem y

'

=

xy + x , y(2)

= -2.

xy-y

8. Solve the initial-value problem cos(x) ln(y)

9. Find the general solution of the equation y'

1�

= ysin(x), Y > 0, y(O)

= e 2. .

x2

.

= - and its orthogonal traJectones. y

=

10. Find the general solution of the equation y' trajectories.

2x(y2

+ 1) and its orthogonal

11. Find the orthogonal trajectories of the one-parameter family of curves y

= x3 +c.

12. Find the orthogonal trajectories of the one-parameter family of curves y

= kx 4 .

13. Find the orthogonal trajectories of the one-parameter family of curves y

= ecx.

2.2

Homogeneous Equations

Definition 2.4. A first-order equation is homogeneous if it can be placed in the form y' = f possibly after some algebraic manipulation.

(�),

For example, the equations y'

=

y' =

'!!_, X

and y'

X

+ X

y

= l + '!!._ X

x y l y x2 + y 2 =---=-+-=-+xy

y

x

y/x

x

/ X2 + Y Y 1s . not, smce . x and y may appear only are homogeneous, whereas y = -=x+as the combination

X

'!!._. X

X

A homogeneous equation is solved by making the substitution (also called a change of variable) u = '!!_, where u is a new function of x. The homogeneous equation for the X unknown function y is then transformed into a separable equation for the unknown function u: Employing the product rule, we obtain u=

� ⇒

y = xu



y'

= u + xu' ⇒

⇒ xu' = J (u) - u ⇒

u + xu' = y' 1

f(u) -

U

u' -

=f 1

X

,

(;-) = f (u)

CHAPTER 2. FIRST-ORDER EQUATIONS

16

which has the form g(u)u' = h(x), and is separable. After u is found, y is obtained by y = XU. ,

Example 2.7. The equation

' I

2 + 2 / X 1 y Y y=--=-+xy y/x x

is homogeneous,

as

shown above. The substitution u = '!!... gives

,-L--

---

X

1 y l + XU,=y,= - + - = - + U. : U yjx X U

1 Then xu' = - ,which is separable. Thus, u xu'

1

U



1 I uu = X





1 -u2=ln x l l + c1 2

u = ±J2 ln lxl + c





u2=21n lxl + 2c1 = 21n lxl + c

y=xu = ±xJ2 ln lxl + c ..

Example 2.8. The equation

xy' = yn l (y) -yln(x), x > 0, y > 0,

can be placed in the form

IX (I), X

y'='!!... [ln(y) - ln(x)]= ln X

and is homogeneous. Then u = '}!_ gives u + xu' = u ln( u) X

, 1 1 u =] unu-l x [l ( )



J

l [ ] du= unu-l l ( )

⇒ xu' = u ln( u) - u ⇒

l J

- dx=ln lxl + c1. x

In order to evaluate the integral on the left, make the substitution v= ln(u) - 1. l dv Then - =-,and we obtain u du J

l ] du= unu [l ( ) - l

J

l dv - -du= v du

J

l - dv=ln v l l = ln I ln(u) - 11v

\

17

2.2. HOMOGENEOUS EQUATIONS Example 2.9. Consider the initial-value problem

x2 y' = x2

+ xy + y2 , y(l) =

The equation can be placed in the form y I '/1,

y = X



=}

y = XU

f1

=} U + XU

1 - du=

+ u2

J

I

= yI

l.

2 y ;; , = I + ;; + (Y)

= l + U + U2

=} XU

_!-_ dx ⇒ tan-1(u) = ln lxl X

I

+c

. h omogeneous. and 1s

= 1+ U 2

1

=} --- U 2 1

+u

1

/

X

⇒ u = tan(ln lxl + c).

= x tan(ln lxl + c) is the general_ solution. y(l) = 1 ⇒ tan(c) = and the solution of the initial-value problem is y = x tan (in lxl + �). Hence, y

1⇒ c

7f

= ,

4

Example 2.10. The equation

can be placed in the form y'

=}

X

U

J J

⇒ Integration by parts gives

e- y/x and is homogeneous. y1 X

y

= - + --,

+ XUI = y I =

ueu du =

U

J�

e-U

+u

dx = ln lxl

+ c.

ueu du= ueu - eu . Hence, ueu - eu

= ln lxl + c, i.e.,

It is not possible to solve this relation for y explicitly in terms of x. Nevertheless, , this is the general solution, with y defined implicitly by the above relation. Definition 2.5. A solution of a first-order equation defined by a relation between x and y is called an implicit solution. If it contains one arbitrary constant, then it is the general solution and has the form g(x, y, c) = 0. In Example 2.10, the solution may be placed in the form

g(x, y, c)

(

= '!/__ ey/x - ey/x - ln lxl - c = 0. X

18

CHAPTER 2. FIRST-ORDER EQUATIONS

Exercises 2. 2 1. Solve the initial-value problem y' =

x+ X

Y, y(l) = 2.

2. Solve the initial-value problem xyy' = y2 3. Solve the initial-value problem y' = V:, vY

-

x2 , y(-1)

+ '!!_, x > 0,

y > 0, y(l) = 9.

X

4. Find the general solution of the equation y'

=

'}!_

X

= 2.

+� x x2 y + ,/xY + X 3

5. Solve the initial-value problem y +x sec(�) -xy' = 0, y(l) =



2x + 3Y 6. Find the general solution of the equation y' = 3x - 2y 7. Find the general solution of the equation y' =

x3 +x2 y +xy2 + y3 X 3 + X 2 y +xy2

8. Solve the initial-value problem yy' = 6x + y, y(l) = 4.

2.3

Linear Equations

Definition 2.6. A first-order equation is linear if it has the form

a(x)y' + b(x)y = c(x), where a(x) ¢ 0, b(x) and c(x) may be any functions of x whatsoever (or constants), but are independent of y and y'. An equation which is not linear is called nonlinear. For example, the equations ex y' + ln(x)y = x3 + 1, y' = 6 and y' + x2 y linear, but yy' = l, y' + sin(y) = 0 and y' + y2 = x are nonlinear.

=

In order to solve a linear equation, divide the equation by a(x) and let P(x) = . c(x) to place the equation in the standard form and Q(x) = ax ( ) J

0 are ) b(x ) ax (

y' + P(x)y = Q(x).

Multiply the standard form of the equatiun by a function I(x), to be determined, to obtain I(x)y' + I(x)P(x)y = I(x)Q(x).

19

2.3. LINEAR EQUATIONS

The function J(x) is chosen so that the left-hand side I(x)y'+ I(x)P(x)ybecomes [J(x)y]'. Employing the product rule, this requirement becomes

I (x)y'+ I(x)P(x)y= [J(x)y]'= I(x)y'+ I'(x)y, = I' (x) .

which gives l(x)P(x)y= I '(x)y, or I(x)P(x) employing the substitution rule, we obtain

j} :�dx =JP(x)dx

J



from which we obtain

j

}dI= P(x)dx



Then 1!·

'(x) (x)

= P(x)

and,

j

lnlIx()I = P(x)dx ,

J(x) = ef P(x )dx ,

calle c =

+c=

1 -[ln(x)] 2 2

+ c'

X

3 - ==> y = -[ln(x)] 2 - 3x. 2

Example 2.13. The equation xexy' + ex y = 4xe-x is linear. In standard form, it is 1 y' + - y = 4e-2x, with the integrating factor X

I(x)

and I(x)

= x will do .

J

and y

=

4x e-2x dx e-2x

-2e-2x - -X

= ±x,

J

= -2xe-2x -

The equation then becomes · , (xy)' = 4xe-2x xyI + y = 4xe-2x, 1.e.

Integration by parts then gives xy =

l l = lxl e' n x

= ef ½dx =

= -2xe-2x +

2e-2x dx

e-2x + c,

C

+ -. X

Example 2.14. Consider the initial-value problem cos(x)y' + sin(x)y

=

The equation is linear, with the standard integrating factor is J(x)

and I ( x)

=

= eftan( x )dx =

= form y' + tan(x)y =

2sin(x)cos3(x), y(1r)

eln lsec(x)I

=

- 2. 2sin(x)cos2 (x). The

1 I sec() x I = ±--, cos(x)

will do. The equation then becomes __J_ COS() X 1 , sin(x) . y = 2sm(x)cos(x), - y + 2 () COS(X COS X -)

l.C.'

v]'

= 2sin(x)cos(x) = sin(2x), [cos\x) integration of which gives l 1 l )cos(2x) + ccos(x). - - y = sin(2x) dx = -- cos(2x) + c ==> y = --cos(x 2 2 COS(X ) 1 5 1 5 Then y(1r) = - 2==> -2 = - c ==> c = ==> y = - cos(x)cos(2x) + cos(x).

J

2

2

2

2

21

2.3. LINEAR EQUATIONS

2.3.1

Bernoulli Equations

Definition 2. 7. A first-order equation is a Bernoulli equation if it has the form a(x)y' + b(x)y = c(x)ya , wherea(x) ¢. 0, b(x) and c(x) are functions of x (or constants), but are independent of y and y', and a is a real number. If a= 0, then the equation isa(x)y' + b(x)y = c(x), which is linear, and if a= 1, then the equation isa(x)y' + b(x)y = c(x)y, i.e.,a(x)y' + [b(x) - c(x)]y = 0, which is also linear. For a -/= 0, 1, a Bernoulli equation can be transformed into a linear equation by means of the substitution (change of variable) u = y 1 -a. Then

and, empioying the chain rule, 1 1-a

---1 I --U 1 0 U

YI =

1

=

1 1-a

_g_ I --U1-0U

and the equation becomes

1

et

I

--a(x)u 1 -au 1-a

+ b(x)u� = c(x)uT=o. I

a

-Ct

Multiplication by u� then gives 1

� a(x)u' + b(x)u = c(x), a

which is linear. Once the linear equation is solved for u(x), the solution y(x) of the Bernoulli equation is obtained by y = u �. Example 2.15. The equation xy' + 2y = y2 is a Bernoulli equation with a = 2. With 1 - a= -1, u = y- 1 ⇒ y = u- 1 ⇒ y' = -u- 2 u1 , y2 = u- 2, and the equation becomes -xu' + 2u which is linear, with the integrating factor

=

2 1, or u' - -u =

x

1

--

x'

22 Then

CHAPTER 2. FIRST-ORDER EQUATIONS

l , 2 1 u - u= 2 3 x x x3



' 1 ( 2 u) x

⇒ u=- 1 +cx2 ⇒ 2

1

y =u

-1

=

is the general solution.

Example 2.16. The equation 2y' - y

With 1 - a

=

-2, u

equation becomes

=

y

-2

⇒y=

=

u- 1/2 i.e.'

1



x3

I_u=j-I_dx= --+c x2 x3 2x2

1 2

1

+

cx2 -

2 1 + kx2

exy3 is a Bernoulli equation with a



y'

- 31 2

= -tu

u',

3 y

=

u- 3 1 2 ,

=

3.

and the

X -uI -u= e, or u' +u= -ex ,

which is linear, with the integrating factor I(x)

= ef 1 dx = ex .

Then

is the general solution. Note thatu112 = ±Ju. In fact, it is evident from the Bernoulli equation that if y is a solution, then so is -y, because

Example 2.17. Consider the initial-value problem y'

+ 2x2 y=

2x2

vY,

y

> 0,

The equation is a Bernoulli equation with a= -

and the equation becomes i.e.'

y (O) =4.

1

2.

With 1 - a=

I 3 2,

2.3. LINEAR EQUATIONS

23

which is linear, with the integrating factor J(x) 3

e x u'

+ 3x2 ex3 u = 3x2 ex3 ⇒

(

3

'

ex u)

= ef3x

= 3x2 ex3 ⇒

⇒ u = 1 + ce-x ⇒ y = u 3

2

dx

= ex3 .

3

ex u =

J

Then 2

3x ex dx = ex 3

3

+c

= (1 + ce-x3 ) 2 /3

2 /3

is the general solution, and y(0) = 4 ⇒ (1 + c)2 1 3 = 4 ⇒ c 213 3 is the solution of the initial-value problem. y = (1 + 7e- x )

43/2

-

1

7

⇒ \

Exercises 2. 3 1. Solve the initial-value problem y' + y = e- x , y(0) = -2. 2. Find the general solution of the equation y' - 3y 3. Solve the initial-value problem ex y'

= 1.

+ 2ex y -1 = 0, y(0) = -3.

4. Find the general solution of the equation y' + 2xy = 2x. 5. Solve the initial-value problem xy' + y =

1

-, X

x > 0, y(l)

= 2.

6. Find the general solution of the equation x 3 y' - 2y + 1 = 0. 7. Find the general solution of the equation x2 y' 8. Solve the initial-value problem cos(x)y'

+ xy =

1 , x ln(x)

> 0.

+ sin(x)y = cos4 (x), y(1r) = . 7f

9. Find the general solution of the equation xy' + 2y

= e-

2

x

.

x3 lO. Find the general solution of the equation xy' - y = _ __ x+l e-x X 2 + 2x + 2 22x , x > 0. 12. Find the general solution of the equation x ln(x)y' + y = X -1 1 13. Solve the initial-value problem 3y' - y = 2, y(0) = 2. y 11. Find the general solution of the equation y' + y

=

14. Find the general solution of the equation y' + 4y 15. Solve the initial-value problem 2x2 y' - 2xy

= 2e- x y'Y, y > 0.

= -y3 sin(x),

16. Find the general solution of the equation xyy' + y 2 17. Find the general solution of the equation x ln(x )y'

-

xex

y (�)

= 1r.

= 0.

+ y = 2yy/y, x > l, y > 0.

24

CHAPTER 2. FIRST-ORDER EQUATIONS

2 .4

Functions of Two Variables

The study of a type of first-order equation called exact, to be defined in Section 2.5, requires that certain basic properties of functions of two variables be established first.

Definition 2.8. A function f of two variables assigns to any pair (x, y) in its domain a real number denoted byf ( x, y), where the domain of f is either specified or taken to be all points (x, y) in the plane where f ( x, y) is defined. Several such functions have already been encountered in the preceding sections. For example,

f(x, y)

= x2 + xy - y3,

g(x, y)

=-+1 X

y

and h(x, y)

=

Jx - y

are functions of two variables. In the.above, the domain off is the entire plane IR2, that of g is the set of all points (x, y) with y i= 0, and that of his the set of all points (x, y) with x - y 2'. 0, i.e., x 2'. y. It is important to note that, in the definition of a function of two variables, y is not a function of .1:, and that x and y may be assigned values independently. In other words, x and y are independent variables. For example,

J(x, y)

= y - x2 + 1

⇒ J(l, 2)

= 2,

J(2, 1)

=

-2, and f(0, 0)

=

1.

It is only a relation, such as f(x, y) = 0 or f(x, y) = c, which defines y as a function of x. For example, f(x, y) = y - x 2 + 1 = 0 ⇒ y = x 2 - 1.

2.4.1

Partial Derivatives

The (first-order) derivative of a functionf of a single variable x is the function f' or df defined by dx f J(x + h) - J(x). J'(x) = d (x) = lim 1i-o h dx

f(x), then dz = J'(x) is the rate of change of z with respect to x. dx In contrast to functions of one variable, there are different ways in which functions of two variables can be differentiated. If z

=

25

2.4. FUNCTIONS OF TWO VARIABLES

Definition 2.9. Let f be a function of two variables x and y. The partial derivative

off with respect to x is the function j� or �� of two variables defined by f(x+h,y)-J(x,y) _of x, y -l· ( ) - Im ! X ( x, y) . h->O h uX !:}

The partial derivative off with respect to y is the function J� or �� of two variables defined by

a f(x,y) _- 1.Im J(x,y + h)-J(x,y)

fy (x,y ) _ If z

z ox

!:}

uy

= J(x,y), then o =

y held constant, and �: x held constant.

=

h

.

fx (x,y) is the rate of change of z with respect to x with

j�(x, y) is the rate of change of z with respect toy with

Example 2.18. Let f(x,y)

fx (x,y) = 4x3y2

h->O

= x4y2 + 2x- 3y+ e x - ln(y) - sin(y) + cos(x) + 2.

+ 2 + e x - sin(x)

Then

and fy (x,y) = 2x4y - 3 - � - cos(y). y

In the computation of fx , y is regarded as a constant, and all expressions which are 'independent of x, such as y, ln(y) and sin(y) have derivative with respect to x equal to 0. In the computation of fy , xis regarded as a constant, and all expressions which are independent of y, such as x, e x and cos(x) have derivative with respect toy equal to 0. A function J of two variables has four second-order partial derivatives, namely, the first-order partial derivatives of the two functions fx and fy • The four possibilities are as follows:

fxx

O af = ox ox ox f � fx = �� = uy uy ux f fy ox· = ox oy =

a Jx =

a

aa

a

aa

a

= �� =

1 � uy ·y

a af

uy uy

3 f. h . h respect to x, . 1 denvat1ve . . of fx wit 1s t e partia ox2 a2 f . � .1s tl1e partia . 1 denvative . . of 1·x with respect toy, uyux o2 f . . . . i denvative . of fy with respect to x, 1s the partia oxoy a2 f . . . i denvative . . of fy with respect toy. 1s the partia 2 uy 2

!:}

26

CHAPTER 2. FIRST-ORDER EQUATIONS

= x3 y4 + e 2x ln(y) + x2

Example 2.19. Let f(x,y)

-

y3

-

3. Then

fxx (x,y)

3x2 y4 + 2e2x ln(y) + 2x, e2x 4x3 y3 + - - 3y2 , y 4 6xy + 4e2x ln(y) + 2,

fxy (x,y)

12x2 y3

+ --,

12x2 y3

+ -�-,

12x3 y 2

-

fx (x,y) fy (x,y)

2e 2x y ?e 2x

y e 2x y2

-

6y.

Note that, in this example, fxy = fyx · Although this is not true in general, the conditions under which it is true are given by the following: Theorem 2.1. If fxy and fyx are continuous in an open region R of the (x,y)-plane (i.e., R excludes its boundary), then f�y (x,y) = fyx (x,y) for all (x,y) in R. Continuity of a function of two variables is defined in a manner analogous to the definition for a function of a single variable: Definition 2.10. A function f of two variables is continuous at a point (a, b) if lim

(x,y)-+(a,b)

f(x,y) = f(a, b).

f is continuous in a region R if f is continuous at every point (a, b) in R.

2.4.2

The Chain Rule

The chain rule for functions of a single variable states that, if z = f ( x) and x = g ( t), then the composition h(t) = f(g(t)) has derivative

h'(t) = .

.

.

!

h(t) = J'(g(t))g'(i). �

In other words, 1f z 1s a funct10n of x and xis a function oft, then dt

��

= --. dx dt

Several versions of the chain rule exist for functions of two (or more) variables. The one of interest in the present study is the following:

2.4. FUNCTIONS OF TWO VARIABLES

27

Theorem 2.2. (The Chain Rule) If f is a function of two variables and x and y are functions of a single variable t, then the composition h(t) = f(x(t), y(l)) has derivative

h'(l) =

d � J(x(t), y(t)) = fx(x(t), y(t)) dl dl

� + fy (x(l), y(l)) dt .

(2.1)

In other words,if z = f (x, y) and x and y are functions of t, then

dz dt

8 f dx 8x d t

8 f dy 8y dl

-=--+--. Note that, in the special case where z = f(x) is independent of y, �� = 0 and dz df dx d z dx . . 8f df - =-,and the above formula reduces to - = -- = -- , 1.e.,the cham rule dl d x dl dx dl 8x dx for functions of one variable. Example 2.20. Let f(x, y) = x2 + y3, x(t) = t3 3 h(t) = J(x(t), y(l)) = (t

-

-

6l + 1 and y(t) =et +3t. Then

6t +1)2 +(e t +3t)3

is a function of t, and h' (t)

8 ! dy 8 ! dx + = 2x(t)(3t2 - 6) +3[y(t)] 2(e l +3) 8y d t 8X dt 2 t 2(t3 - 6l +1)(3t2 - 6) +3(e +3t) (i +3).

In Theorem 2.2, consider the special case where x(t) = l. In that case,

and Equation (2.1) reduces to

dx =1, dt

or, employing the fact that x =t,

d dy J(x, y(x)) = fx(x, y(x)) + fy (x, y(x)) dx , dx which is the version of the chain rule required in the study of exact equations.

Exercises 2.4 1. Let f(x, y) = x3 (a) Determine

-

y2 + x2 y2 + 1.

J(l, 0), f(3, 1), f(l, 3) and f(2, -1).

(2.2)

28

CHAPTER 2. FIRST-ORDER EQUATIONS (b) Find fx (x,y) and fy (x,y), and the values of these functions at the points (1, 0), (3, 1), (1, 3) and (2, -1). (c) Find fxx (x,y), fxy (x,y), fyx (x,y) and fyy (x,y), and the values of these functions at the points (1, 0), (3, 1), (1, 3) and (2, -1). 2. Let J(x,y)

= x2 ln(y) - y3 ex + x3 - y2 + 1.

3. Let f(x,y) = x4 d f ( X (t) ' y ( t)). dt 4. Let f(x,y)

=

x2

+ 3x2 y2 + y4, x(t) = -

t2

+ 2t

cos(y) + x sin(y), x(t)

Determine : f(x(t),y(t)). t 5. Let f(x,y) = sin(xy) d f ( X (t)' y (t)). dt 6. Let f(x,y)

+

el

y!XY, x(t)

Find fx , fy , fxx , fxy , fyx and fyy · and y(t)

8. Let f(x,y) =

X2

+y

2

+ xe- y

and y(x)

- 2t.

Determine

2t 2

+ 3t -

1 and y(t) =et + 2t.

+

l and y(t)

ln(t). Determine

- x2 + y3 ln(x) and y(x) = sin(x). l

t3

=

= x4 + 3x2 y2 + y4 and y(x) = x3 + 6x2 - x.

7. Let f(x,y) = eY

=

= sin(2x).

Determine

J(x,y(x)). dx

..!!:.._

Determine :J(x,y(x)). d

d Determine xf(x,y(x)). d

= x4 - y4 - l. The relation J(x,y) = 0 defines y as a function of x. dy Determine - m . terms of x and y.

9. Let f(x,y)

dx

= x3 + y2 - eY .

The relation f(x,y) dy x. D eterminc - m . terms of x and y. dx

10. Let f(x,y)

2.5

= 0 defines y as a function of

Exact Equations

Consider a first-order equation expressed in the form

P(x,y) + Q(x,y)

dy = 0, dx

(2.3)

where P and Q are functions of two variables. Suppose that there exists a function J of two variables such that fx fy = Q. Then Equation (2.3) takes the form fx

+ fy

dy = dx 0

=

P and

29

2.5. EXACT EQUATIONS which, by Equation (2.2), is

!I:_ J(x,y(x)) = 0. dx

Integration with respect to x then

yields f(x,y(x)) = c, an arbitrary constant. Thus, the general solution of Equation (2.3) is J(x,y) = c. The problem, then, is the determination of the function f, called a potential function. Definition 2.11. An equation of the form P(x,y) + Q(x,y) ;�

= 0,

with P and Q

continuous in an open region R of the (x, y)-plane, is exact in R if there exists a potential function J defined for all (x, y) in R. Example 2.21. Consider the equation y + xy' = 0. Evidently, J(x,y) = xy is a potential function because fx = y = P and fy = x = Q. Hence, the equation is exact in IR 2 , with the general solution f(x,y) = c, i.e., xy = c. Theurern 2.3. Suppose that P and Q have continuous first-order partial derivatives in an open region R of the plane. If the equation P(x,y) + Q(x,y)y' = 0 is exact in R, then Py (x,y) = Qx (x,y) for all (x,y) in R.

Proof. Since the equation is exact, there exists a potential function f. Then fx = P fxy = Py and fyx = Qx , Since Py and Qx are continuous in R, fxy and and fy = Q are continuous in R, hence equal by Theorem 2.1. Thus, Py = Qx in R. fyx

*

Theorem 2.3 is equivalent to the following: Theorem 2.4. If Py =I= Qx in R, then the equation P(x,y) + Q(x,y)y' exact in R.

=

0 is not

Proof. Suppose that Py =I= Qx in R. If the equation is exact, then Py = Qx by Theorem 2.3, which contradicts Py =I= Qx . Hence, the equation is not exact. Example 2.22. Consider the equation x 2

+ y2 + xy

P(x,y) = x 2 + y2 , Q(x,y) = xy, Py not exact, by Theorem 2.4.

Qx

= 2y,

=

dy dx

= 0.

y, Py =I= Qx, hence, the equation is

It is not true, in general, that if Py = Qx, then the equation P(x, y)+Q(x,y)y' = 0 is exact. In order to state a converse of Theorem 2.3, i.e., a theorem which gives the conditions under which exactness follows from Py = Qx , the following definition is required.

30

CHAPTER 2. FIRST-ORDER EQUATIONS

Definition 2.12. A region R in the (x, y)-plane is connected if it consists of a single "piece." It is simply connected if it is connected and has no "holes" in it. Definition 2.12 is, of necessity, merely informal, since the mathematically rigorous definitions of connected and simply connected sets are beyond the scope of this book. For example, the region R enclosed by a non-self-intersecting closed curve (e.g., a circle) is simply connected. The entire plane is simply connected. However, the plane with the origin excluded is not simply connected due to the "hole" at the origin. Theorem 2.5. Suppose that P and Q have continuous first-order partial derivatives in an open, simply connected region R of the plane. If Py = Qx in R, then the equation P(x, y) + Q(x, y)y' = 0 is exact in R. Note that, if the region R is not simply connected, then the equation may not be exact, even if Py = Qx Example 2.23. Consider the equation x 2

+ y2 + 2xyy' = 0.

2 P(x, y) = x + y 2 , Q(x, y) = 2xy, Py = 2y, Qx = 2y, Px = 2x and Qy = 2x are continuous in the plane, which is simply connected, and Py = Qx . Hence, the equation is exact by Theorem 2.5. A potential function f exists, and satisfies the two conditions fx = P and fy = Q.

fx = P



f(x, y) =

J

P(x, y) dx

=

J

x2

3

+ y2 dx = � + xy2 + g(y),

where g(y) is an arbitrary function of y. Note that, in the above integration with respect to x, y is regarded as a constant, and g(y) has the role of the arbitrary d "constant" of integration, being the most general expression such that g(y) = . 0 dx Then fy = Q ⇒

2xy + g'(y) = 2xy



g'(y) = 0



g(y)

and the general solution of the equation is

f ( x, y) =

x3

2 3 + xy + c1 = c2,

or

= c1

=?

f ( X, Y) =

x3

3

+ xy2 + C1,

2.5. EXACT EQUATIONS

31

Example 2.24. Consider the equation 4x3

-

2x + 3y + (3x + 2y)y' = 0.

P(x,y) = 4x3 - 2x + 3y, Q(x,y) = 3x + 2y, all first-order partial derivatives of both P and Q are continuous in the plane, which is simply connected, and Py = 3 = Qx . Hence, the equation is exact by Theorem 2.5. A potential function f exists, and satisfies the two conditions fx = P and fy = Q. fx

=

P

=>

f(x,y)

=

J

P(x,y)dx

=

f

2

4x3 -2x+3ydx=x4 -x +3xy+g(y),

where g(y) is an arbitrary function of y. Then fy = Q 3x + g'(y)

= 3x + 2y ⇒



J(x,y)

= 2y ⇒

g'(y)

⇒ g(y)

= y2 + c1

= x4 - x2 + 3xy + y2 + c1,

and the general solution of the equation is I

'f(x,y)

=

x4 -x2 +3xy+y2 +c 1

=

c2, or x4 -x2 +3xy+y2

Note that, if we begin with the condition fy will be the same. Thus, fy

=

Q =>

J(x,y)

=

J

= Q instead of

Q(x,y)dy=

f

=

fx

=

k.

P, then the result

2 3x+2ydy= 3xy+y +g(x),

where g(x) is an arbitrary function of x, being the most general expression such that d g(x) = 0. In the above integration with respect toy, x is regarded as a constant. dy Then fx = P => 3 3y+g'(x)=4x -2x+3y

=>



f(x,y)

g'(x)=4x3 -2x



g(x)=x4 -x2 +c 1

= 3xy + y2 + x4 - x 2 + c 1,

as before. Example 2.25. Consider the equation x2

+ y2 + xyy' = 0.

As shown in Example 2.22, this equation is not exact because Py =I= Qx. Thus, any attempt to determine a function f such that fx = P and fy = Q must fail. Proceeding as in the last two examples, fx and fy

=

P =>

=Q⇒

f(x,y)

=

J

P(x,y)dx

2xy + g'(y)= xy

=



J

x 2 + y2 dx

3

= � + xy2 + g(y),

g'(y)= -xy,

32

CHAPTER 2. FIRST-ORDER EQUATIONS

which is a contradiction because the left-hand side is independent of x, whereas the right-hand side is not, making their equality impossible. It follows that a function f(x,y) with fx = P and fy = Q does not exist. Example 2.26. Consider the initial-value problem x ye + sin(y) +[e x +xcos(y) + 1]:� = 0, y(0)= -3.

P(x,y) = ye x + sin(y), Q(x,y)=ex +xcos(y) + 1, all first-order partial derivatives of both P and Q are continuous in the plane, and Py = e x +cos(y) = Qx - Hence, the equation is exact by Theorem 2.5. fx = P ⇒ f(x,y) = and fy = Q

J

P(x,y)dx =

J

x ye x + sin(y) dx = ye + x sin(y) + g(y),



ex +xcos(y) + g'(y)=e x +xcos(y) + 1





g'(y) = 1

⇒ g(y) = y + c1

f(x,y)=ye x +xsin(y)+y +c1,

and the general solution is ye x + x sin(y) + y = k. Setting x = 0 and y = -3 then gives k = -6, and the solution of the initial-value problem is ye x + x sin(y) + y = -6. Example 2.27. Consider the initial-value problem

2 yln(x)+2x-[x- xln(x)+3y ]y'=0, x>0, y(1)=2.

P(x,y) = y ln(x) +2x, Q(x,y) = -x+x ln(x)-3y2 , all first-order partial derivatives of both P and Qare continuous in the right half-plane x > 0, which is simply connected, and Py = ln(x) = Qx - Hence, the equation is exact by Theorem 2.5. fy

=

Q



and fx = P ⇒

J(x,y)

J Q(x,y)dy = f-x +xln(x)- 3y dy 2

-xy + xy ln(x) - y3 + g(x),

-y + y ln(x) + y + g'(x) = y ln(x) + 2x





g'(x) = 2x



3 2 J(x,y) = -xy + xy ln(x) - y + x + c 1,

g(x) = x2 + c 1

and the general solution is xy ln(x) - xy + x2 - y3 = k. Setting x = 1 and y = 2 then gives k = -9, and the solution of the initial-value problem is

xyln(x)- xy +x 2 - y3 = -9.

33

2.5. EXACT EQUATIONS

2.5.1

Integrating Factors

If the equation P(x, y) + Q(x, y)y' = 0 is not exact because Py =/- Qx , then it can be made exact by multiplication by an integrating factor I(x, y), chosen such that the resulting, equivalent, equation

T(x, y)P(x, y) + I(x, y)Q(x, y)y' = 0 is exact. The exactness of Equation (2.4) requires that (IP)y the product rule, this condition becomes

(2.4)

= (IQ)x -

Employing

This partial differential equation for the unknown function I ( x, y) is more difficult to solve than the given ordinary differential equation, and simplifying assumptions must be made, which may or may not succeed. Seek an integrating factor J(x) which is independent of y. Then Iy and the condition (2.5) becomes

= 0, Ix

=

I'(x) I(x)

I'(x), (2.6)

If the right-hand side of Equation (2.6) is independent of y, then I(x) is determined by Equation (2.6). Otherwise, Equation (2.6) is a contradiction because the left-hand side is independent of y. In that case, I ( x) does not exist. If I(x) does not exist, then seek an integrating factor I(y) which is independent of x. Then Ix = 0, Iy = J'(y), and the condition (2.5) becomes

I'(y) I(y)

(2.7)

If the right-hand side of Equation (2.7) is independent of x, then I(y) is determined by Equation (2.7). Otherwise, Equation (2.7) is a contradiction because the left-hand side is independent of x. In that case, I (y) does not exist. Example 2.28. Consider the equation x 2

+ y 2 + xyy' = 0.

As shown in Example 2.22, this equation is not exact because Py =/- Qx - Since

Py - Qx Q

2y -y

1

xy

X

is independent of y, I ( x) exists and is determined by

J'(x) I(x)

1 X

(

CHAPTER 2. FIRST-ORDER EQUATIONS

34

⇒ ln III= ln /xi ⇒

I(x) = ±x,

where arbitrary constants of integration are unnecessary and have been suppressed, because we seek only one solution I(x), and not all possible solutions. Since both ±x are integrating factors, we select the simpler one, I(x) = x. The equation then becomes x

and is exact. Then



fx = P

+ xy2 + x2 yy' = 0,

3

f(x,y) =

J

and fy = Q ⇒



2

2

x y+g'(y)=x y

x



I

g (y)=0

l

1

+ xy2 dx = 4 x4 + 2 x2 y2 + g(y),

3

1

1

f(x,y)= x + x y +c1,



g(y)=c 1

4

. 1 . and the general solution 1s x 4

4

2

1 2 2 k 2 2 4 y = 4k = k14 + 2 x y = �, or x + 2x

Example 2.29. Consider the equation jx + y

3

+ 2xy2 y' = 0,

X

2 2

> 0.

3 2 2 2 P(x,y) = ..jx + y , Q(x,y) = 2xy , Py (x,y) = 3y , Qx (x,y) = 2y , Py =/= Qx, hence,

the equation is not exact. Since

2 2y2 = � Py - Qx - 3y

2xy2

Q

is independent of y, I ( x) exists and is determined by I'(x) = � I(x) 2x



ln II(x)I =

! ln(x) = ln (x 2

1 12

)

2x



I(x) = ±x 112 = ±Jx.

With I(x) = x 112, the equation becomes x

and is exact. Then fx = p



+ x 1f2y3 + 2x3f2y2 y' = o,

J(x,y) = J x + x

2 y dx = 1x

1f 2 3

+ �x3f2y3 + g(y),

and fy = Q ⇒ 2x3 f 2 y2

+ g'(y) = 2x3f2y2 ⇒

. . 1 2 and the general solut10n 1s x

2

+

2

g(y) = c1 3/2 3

3x

y



J(x,y) =

1

2

2

x

+

2

3/2 3

3x

= k�, or 3x2 + 4x3/ 2 Y 3 _ - k 1-

y + c1,

35

2.5. EXACT EQUATIONS Example 2.30. Consider the equation X

Py= 1, Q x

= x + 1 + y,

x2

)

+y+ (

+ X + XY y1 = 0.

2

Py # Q x , hence, the equation is not exact. Since

-x-y x2

+ X + xy

is not independent of y, I(x) does not exist. Since

Qx

-

p

Py

=

x+ y x+y

=

is independent of x, J(y) exists and is determined by ThP equation then becomes

l

��?

= 1, which gives J(y)

= eY .

2

(x+y)e Y + (� +x+xy)eY y'=O, and is exact. Then

and fu

= Q ==> )e Y (�x2 +xy+x)e Y +g'(y)= (�x2 +xy+x 1 f (x, y)

==>

and the general solution is (�x2

==>

g(y)=c1

= ( � x2 + xy)eY + c1,

+ xy)e Y = k, or ( x2 + 2xy) e Y = k 1.

Example 2.31. Consider the initial-value problem 4xy +

(x 2 + v'Y) y' = 0,

y > 0, y(l)

= 4.

Py = 4x, Q x = 2x, Py -I Qx , hence, the equation is not exact. An integrating factor I (y) exists and is determined by

J'(y) J(y)

-2x 4xy

1

2y

==>

1

ln II(y)I=-- ln(y) ==> I(y) =y- 112. 2

36

CHAPTER 2. FIRST-ORDER EQUATIONS

The equation then becomes

and is exact. Then fx and fv

= 4xy112 ⇒

= J 4xy112 dx = 2x2y 112 + g(y),

f(x, y)

=Q⇒ x 2 y-l/ 2 + g'(y)

= x 2y -l/2 + 1 ⇒



g'(y)

=

1



g(y)

= y + C1

J(x, y) = 2x2y 112 + y + c1,

and the general solution is 2x 2 y 112 + y = k. Setting x = 1 and y = 4 gives k = 8, and the solution of the initial-value problem is 2x 2 y112 + y = 8.

Exercises 2. 5 1. Find the general solution of 3x2

+ y + 1 - (3y2 - x + l)y' = 0.

2. Find the general solution of xy + x + 1 +

3. Find the general solution of 3x2y2 - 2x + (2x3 y + 3y2)y' 4. Solve the initial-value problem y(0) = 1.

y

3

- 3x2y2

5. Solve the initial-value problem y' =

y(l) = 4.

.

9. Find the general solution of ex+y

= 0.

+ 4x + (3xy2 - 2x 3

+ y + ( � + x) 2

7. Find the general solution of (6xy + eY)y' + xex + 3y2 .

0.

3

y

- 4y3 ) y'

=

0,

3x 2y2 - y - 2xy3 , y(2) = 1. 3x 2y2 - 2x 3 y+x

6. Solve the initial-value problem 21

2+ Y2

8. Fmd the general solution of 2xe x

!� =

( �x 2 + y + 1)

��

= 0, x > 0,

y

> 0,

= 0. d

+ 4x3 + ( 2yex2+Y2 - 2y dy = 0. x )

+ x + y + cos(x) + (ex+y + x - y) y' = 0.

x y y - 2x + ( 2 10. Solve the initial-value problem 2 + 2y) dd 2 2 x x+y+l x+y+l y(0) = 0.

= 0,

37

2.5. EXACT EQUATIONS 11. Find the general solution of

dy = . 2y - 3 2x + 2 -------O + 1 + -------2 2 2 2 x + y + 2x - 3y + 3 x + y + 2x - 3y + 3 dx

12. Consider the equation

✓x x+ y 2

2

+1+ (

(a) Show that Py = Qx in the region R

✓x y+ y 2

=

(b) Solve the equation.

2

+ 1)

dy = 0. dx

{(x,y): (x,y) =/= (0,0)}.

(c) Is the equation exact in R?

13. Consider the equation

X2

y

+y

X X2

2

+y 2

yI = O.

(a) Show that Py = Qx in the region R = {(x, y): (x, y) =/= (0, 0)}.

(b) Solve the equation.

(c) Is the equation exact in R?

14. Solve the initial-value problem x + 6y2 + 4xyy' = 0, y(2) = 0, and express y in terms of x. 15. Find the general solution of x3

+ 2y2 + xyy' = 0.

16. Find the general solution of e- x - cos(y) + sin(y )y' 17. Find the general solution of xex

2

= 0.

+ xy2 + yy' = 0.

18. Find the general solution of ft+ y2 19. Find the general solution of x + y2

+ 4xyy' = 0, x > 0.

+ 3xyy' = 0.

20. Solve the initial-value problem 4xy + (6x 2

+ y)y' = 0, y(l) = 1.

21. Find the general solution of sin(x) + [e-Y - cos(x)] y' = 0. 22. Find the general solution of y + (1 + 2x + xy)y' 23. Find the general solution of

= 0.

dy (y) = ;r� , y > 0. dx x n y + 2x

\

38

CHAPTER 2. FIRST-ORDER EQUATIONS

Chapter 2 Exercises 1. Consider the equation xy + y2

x2 y'

-

= O.

(a) Solve it as a homogeneous equation. (b) Solve it as a Bernoulli equation. 2. Consider the equation

y2 X2

-

y'

= 0.

(a) Solve it as a separable equation. (b) Solve it as a homogeneous equation. (c) Solve it as a Bernoulli equation. (d) Find an integrating factor which makes the equation exact and solve it. (e) Find the orthogonal trajectories of the one-parameter family of curves defined by the general solution. 3. Consider the equation x + y - xy' = 0. (a) Solve it as a homogeneous equation. (b) Solve it as a linear equation. (c) Find an integrating factor which makes the equation exact and solve it. ( d) Find the orthogonal trajectories of the one-parameter family of curves defined by the general solution. 4. Consider the equation 3xy' + 2y

= 0.

(a) Solve it as a separable equation. (b) Solve it as a homogeneous equation. (c) Solve it as a linear equation. (d) Find an integrating factor which makes the equation exact and solve it. (e) Find the orthogonal trajectories of the one-parameter family of curves defined by the general solution. 5. Find the orthogonal trajectories of the one-parameter family of curves defined by the equation 2x - 2e-x - y2 = k. 6. Find the orthogonal trajectories of the one-parameter family of curves defined by the equation y = 2x ln Jxl + ex. 7. The relation f(x, y) = 0 defines y as a function of x, and x as a function of y. dy dx dy dx Determine - and - in terms of Jx and fy, and express - in terms of -. dy dX dX dy

CHAPTER 2 EXERCISES

39

8. Consider the equation 3x - 2y + (y - 2x)y' = 0. (a) Solve it as an exact equation and express yin terms of (b) Solve it as a homogeneous equation.

x.

9. Consider the equation xex - x ln(x) + y + xy' = 0, x > 0. (a) Solve it as an exact equation and express yin terms of x. (b) Solve it as a linear equation. 10.

(a) Show that any separable equation can be expressed as an exact equation. (b) Solve the equation y equation.

+ x ln(x) ln(y)y' = 0, x > 0, y > 0, as a separable

(c) Solve the separable equation in part (b) as an exact equation. 11. Consider an equation of the form y' = J(ax +by+ c), b-=/- 0. (a) Show that u( x) = ax + by + c transforms the equation into one which is separable. (b) Solve y' = (x (c) Solve y' =

+ y + 1)2. 1 , x + y > 2. + y- 2

Jx (d) Solve y' = (x + y) ln(x + y) - 1, x + y > 0. 12. Consider an equation of the form a(y) + [b(y)x - c(y)]y' = 0. (a) Show that an integrating factor I (y) exists. (b) Regard y as the independent variable and x as the unknown function of y. Show that the resulting equation for x(y) is linear, employing the fact

1

that y' = - by the chain rule (see Exercise 7).

x'

13. Consider the equation y + [(2y2

+ l)x - 2y]y' = 0.

(a) Find an integrating factor which makes the equation exact and solve it. (b) Proceed as in Exercise 12(b) and solve the linear equation. 14. Consider the equation y2

+ (3xy + 2)y' = 0.

(a) Find an integrating factor which makes the equation exact and solve it. (b) Proceed as in Exercise 12(b) and solve the linear equation. 15. Solve the equation 1 + (xy - x- 3 y3 )y' = 0 by transforming it into an equation for x as a function of y (see Exercise 12(b)).

40

CHAPTER 2. FIRST-ORDER EQUATIONS

16. Consider the equation y'

= 2-jy=1".

(a) Find the general solution. (b) By inspection of the equation, find a solution which is not a particular solution, i.e., it does not correspond to any particular value of the arbitrary constant c. Such a solution is called a singular solution, and may occur only as the solution of a nonlinear equation. (c) The envelope of a one-parameter family of curves is a curve C with the property that, at every point of C, it is tangent to the graph of a member of the family. The envelope of the one-parameter family of curves defined by the general solution, if it exists, is a singular solution. However, a singular solution is not necessarily an envelope. Confirm that the graph of the singular solution found in part (b) is the envelope of the one-parameter family of curves defined by the general solution found in part (a). 17. Given the general solution J(x, y, c) = 0 of a nonlinear equation, where f is a (single-valued) function with continuous first-order partial derivatives, if a singular solution exists, then it is obtained by the elimination of c from the two equations f(x, y, c)

(a)

= 0 and of Be (x, y, c) = 0.

Find the general solution of the equation (y') 2

1

y2

= �­ y

(b) By inspection of the equation, find two singular solutions. (c) Express the general solution as f (x, y, c) = 0 and determine the singular solutions by elimination of c from J(x, y, c) = 0 and fc (x, y, c) = 0.

(d) Confirm that each singular solution is an envelope of the general solution.

18. Consider the equation

(y') 2 + 1 = 1. (y-xy')2

(a) Show that y = mx arbitrary constant.

±

✓m2 + 1

is the general solution, where m is an

(b) Express the general solution in part (a) as f(x, y, m) = 0 and determine x2 as a function of m 2 from the equations f(x, y, m) = 0 and fm (x, y, m) = 0. (c) Employ the result of part (b) to determine y2 as a function of m2. (d) Employ the results of parts (b) and (c) to eliminate m, thereby obtaining a relation g(x, y) = 0 between x and y, which defines a singular solution.

(e) The general solution y = mx± ✓m 2 + l of part (a) defines a one-parameter family of straight lines of slope m and y-intercept ± ✓m2 + 1. Identify the graph of the singular solution obtained in part (d) as a well-known geometrical object, and confirm that it is the envelope of the family of straight lines.

CHAPTER 2 EXERCISES

41

19. Peano 's existence theorem states that, if f is continuous on a rectangular region R with sides parallel to the axes, and if (x0 , y0) is an interior point of R, then the initial-value problem y' = f(x, y), y(x0) = y0, has a solution. (a) Find the general solution of the equation y' of x.

=

'#._

X

and express y as a function

(b) Show that the initial-value problem with the initial condition y(0) no solution, and give the reason.

=

1 has

20. Picard 's existence and uniqueness theorem states that, if f and fy are continuous on a rectangular region R with sides parallel to the axes, and if (x0, y0) is an interior point of R, then the initial-value problem y' = J(x, y), y(x0) = y0, has a unique solution. (a) Find the general solution of the equation y'

= 3xy 1 1 3.

(b) Solve the initial-value problem with the initial condition y(0) (c) Find a singular solution (which is not an envelope).

= 0.

(d) Show that the solution of the initial-value problem is not unique, and give the reason.

L

Chapter 3 Second-Order Equations 3.1

Basic Definitions

Definition 3.1. The general solution of a second-order equation is the solution which contains two arbitrary constants which cannot be combined. It is a two-parameter family of solutions, i.e., a set of solutions such that every solution corresponds to particular values of the arbitrary constants. A solution in which the constants are assigned particular values is called a particular solution. Example 3.1. The second-order equation y" = x can be solved by two integrations: yII =

X



yI = 1 x 2 2

+ C1

is the general solution since it contains two arbitrary constants c1 and c2 which cannot be replaced by a single arbitrary constant. Particular solutions are y=

1 3 1 x , y = x3

6

6

1 3

+ 2x + 1, and y = x - 4x - 3,

6

corresponding to c1 = c2 = 0, c1 = 2, c2 = 1, and c1 = -4, c2 = -3, respectively.

Definition 3.2. An initial-value problem for a second-order equation consists of the equation, together with two initial conditions, which specify the values of y and y' at a single point x and, thereby, determine a particular solution. In Example 3.1, if the initial conditions y(0)

1 3 + C1X + C2 and y 0 = 2 ⇒ ()

y = 6x

C2

= 2 and y'(0) = 3 are imposed, then

12 + C1, = 2 ⇒ y = 61x3 + C1X + 2 ⇒ yI = 2x

1 and y' (0) = 3 ⇒ c 1 = 3 ⇒ y = -x 3 6

43

+ 3x + 2

44

f

CHAPTER 3. SECOND-ORDER EQUATIONS

is the solution of the initial-value problem. In contrast to first-order equations, only a few types of second-order equation can be solved in terms of elementary functions. Definition 3.3. A second-order equation is linear if it has the form a(x)y" + b(x)y' + c(x)y = g(x),

!�:; ,

where a(x) :/= 0, b(x), c(x) and g(x) are functions of x (or constants). Otherwise, the equation is nonlinear. Dividing the equation by a(x) and letting p(x) q(x) =

c(x) and J(x) a(x)

=

g(x) , the equation may be placed in the standard form a(x) y"

=

=

+ p(x)y' + q(x)y =

f(x).

If f(x) 0, then the equation y" +p(x)y' +q(x)y = 0 (or a(x)y" +b(x)y' +c(x)y = 0) is called homogeneous. Otherwise, it is nonhomogeneous. For example, the equation

is linear and homogeneous, with the standard form y" + e-xy' - 2e-xy The equation

2y"

= 0.

+ 5xy' + y = sin(x)

is linear and nonhomogeneous, with the standard form 5x , l y11 + -y + -y

2

2

1

. (X ) . = - Slll

2

The equations yy" + y

= x2 ,

y"

+ yy' = 0

and y" + y'

+ eY = x

are nonlinear. Definition 3.4. Two functions y1 and y2 are linearly dependent on an interval I if there exist constants c1 and c2, not both zero, such that C1Y1 + C2Y2 = 0 on I (i.e., c1 y1 ( x) + c2y2 ( x) = 0 for all x in I). Two functions y1 and Y2 are linearly indepen­ dent on I if they are not linearly dependent on I, i.e., if c1 y1 + C2Y2 = 0 on I, then C1 = C2 = 0.

3.1. BASIC DEFINITIONS

45

It follows by Definition 3.4 that two functions are linearly dependent on I if and only if one of the functions is a constant multiple of the other on I: Suppose that C1Y1 + C2 Y2 = 0. -C1 -C2 If Ci-/=- 0, then Yi = -y2 , and if C2 -/=- 0, then Y2 = -yi. � � = Conversely, if Yi = cy2 , then 1 · y1 - cy2 = 0 (with ci 1 -/=- 0 and c2 = -c), and if y2 = cy1, then cy1 - 1 · Y2 = 0 (with c1 = c and c2 = -1 -/=- 0). Two functions are linearly independent on J if and only if neither one is a constant multiple of the other on J. Example 3.2. y 1 = x and y2 = ex are linearly independent on IR because neither one is a constant multiple of the other. If c1x + c2 ex = 0 for all x, then x = 0 ⇒ c2 = 0 ⇒ cix = 0 for all x, and x = 1 ⇒ c1 = 0. Example 3.3. y1 = ex and y2 = e 2x are linearly independent on IR because neither one is a constant multiple of the other. If ciex + c2 e 2x = 0 for all x, then x = 0 ⇒ Ci+ c2 = 0 =} c2 = -c1 ⇒ c1(ex - e 2x) = 0 for all x, and x = 1 ⇒ ci(e - e 2 ) = 0 =} Ci = 0 =} C2 = 0. More generally, Yi = ri -/=- r2.

e

rix

and y2 =

e

rz x

are linearly independent on IR if and only if

Example 3.4. Yi = e 3x and Y2 = 2e 3x are linearly dependent on IR because y2 = 2yi for all x. Equivalently, 2yi + (-l)y2 = 0 for all x, with Ci = 2 and c2 = -1. Example 3.5. Yi = cos(x) and y2 = sin(x) are linearly independent on IR because neither one + c2 sin(x) = 0 for all x, . is a constant multiple of the other. If c1 cos(x) � . then x = 0 ⇒ ci = 0 ⇒ c2 sm(x) = 0 for all x, and x = ⇒ C2 = 0.

2

The terms "dependent" and "independent" are often employed to mean "linearly dependent" and "linearly independent," respectively. If an interval J is not specified, then it is taken to be any interval on which y 1 and y2 are defined. The definitions of linear dependence and independence can be extended to any number of functions: Definition 3.5. The functions Yi, Y2 , · · ·, Yn are linearly dependent on an interval T if there exist constants ci, C2 , · · · , en, not all zero, such that ciyi + c2y2 + · · · + CnYn = 0 on I. The functions Yi, Y2, · · ·, Yn are linearly independent on I if they are not linearly dependent on I, i.e., if CiYi + C2 Y2 + · · ·+ CnYn = 0 on I, then C1 = C2 = · · · = Cn = 0.

46

CHAPTER 3. SECOND-ORDER EQUATIONS

Exercises 3 .1 1. Find the general solution of the equation y" + 6x2

+ sin(x) = 2.

2. Find the general solution of the equation xy" = ln(x), x > 0.

3. Solve the initial-value problem y" - 4 cos2 (x) = 2, y(0) = 0, y'(0) = 1. 4. Solve the initial-value problem y"

5.

= xex , y(0) =

l, y'(0)

= 1.

Determine whether the given equation is linear and homogeneous, linear and nonhomogeneous, or nonlinear.

(a)

=0 2y" + ln(x)y = sin(x)

(c)

y"+y'+(x2 +y2)=0

(b)

xy" - y' + 2y

= y3 (e) x2 y" + xy' = 2y (f) y" + y' + y + X = 0 (g) y" + yy' + y = 0

(d)

y" + xy'

6. Show that the functions y1 and only if r1 = r2.

= er i x and y2 = er2 x

are linearly dependent on JR if

7. Show that the functions y 1 = x0 and y2 = xf3 are linearly dependent on I if and only if a = /3, where I is any interval where both functions are defined. 8. Show that the functions y 1 = xr and y2 = esx are linearly dependent on I if and only if r = 0 and s = 0, where I is any interval where both functions are defined. 9. Show that the functions y 1, Y2, · ··, Yn are linearly dependent if and only if (at least) one of the functions is a linear combination of the others. 10. Are the functions Y1

= x, Y2 = ex and y3 = e- x linearly independent?

11. Are the functions y1

= x, y2 = ex

and y3

= 2x linearly independent?

12. Under what condition(s) is the single function y linearly independent?

47

3.2. LINEAR HOMOGENEOUS EQUATIONS

3.2

Linear Homogeneous Equations

Before methods of solution of linear, second-order equations can be discussed, certain fundamental results must be established. Theorem 3.1. If y1 and y2 arc any two solutions of a linear, homogeneous equation y" + p(x )y' + q(x)y = 0, then any linear combination y = C1Y1 + C2Y2 is a solution. Proof. y" + p(x)y'

(ciY1 + C2Y2) 11 + p(x)(ciyi + c2y2)' + q(x)(ciYt + c2y2) ci[y� + p(x)y� + q(x)yi] + c2[y� + p(x)y; + q(x)y2] 0.

+ q(x)y

Theorem 3.2. The equation y" + p(x)y' + q(x)y = 0 has precisely two linearly independent solutions y 1 and Y2, and the general solution is y = CiY1 + c2y2, where ci and c2 are arbitrary constants. Example 3.6. Let y1 = cos(x) and Y2 = sin(x). Then y� = -y1 ⇒ y� + Yi = 0, and YI = -y2 ⇒ YI + Y2 = 0. Hence, Y1 and Y2 are solutions of y" + y = 0. Since y1 and y2 are linearly independent on JR, the general solution of y" + y = 0 is y = c1 cos(x) + c2 sin(x). Note that, if we take Yi = cos(x) and Y2 = 2 cos(x), then, although both Y1 and Y2 are solutions of y" + y = 0, they are not linearly independent since y2 = 2yi. Then y = C1Yi + C2Y2 = ci cos(x) + 2c2 cos(x) = (c1 + 2c2) cos(x) = kcos(x) is not the general solution. Thus, it is essential that Yi and y2 be linearly independent solutions in order for C1Y1 + C2 Y2 to be the general solution. It is often possible to obtain one solution Yi of an equation easily. The method of reduction of order, discussed below, is a method of determining a second, independent solution y2 (i.e., a solution y2 such that Yi and y2 are independent). Reduction of Order Suppose that Yi is a solution of y" + p(x)y' + q(x)y

i.e., y� +p(x)y� +q(x)y 1 by the product rule, Y2

= uyi

= 0,

= 0, and let y2 = u(x)yi, where u is to be determined. ==>-

Y2 I

= u Yi + UYi, I

I

Y2

fl

=u

ff

Y1

' + 2' u Yi + UY1 , ff

Then,

48

CHAPTER 3. SECOND-ORDER EQUATIONS

and Y2 is a solution if and only if y� + p(x)y� [u"y1

+ q(x)y2 = 0,

i.e.,

+ 2u'y� + uy�] + p(x)[u'y1 + uy�] + q(x)[uy1] = b.

Rearranging terms, we obtain u[y� + p(x)y�

Since yJ + p(x)y�

+ q(x)y1] + u"y1 + u'[2y� + p(x)y1] = 0.

+ q(x)y1 = 0,

Equation (3.1) reduces to

u"y1

(3.1)

+ u'[2y� + p(x)y1] = 0,

which is a first-order, separable equation for the unknown function v = u'. Thus, v'y1 =}

+ v[2y� + p(x)y1] = 0

lnlvl

=}

= -2 ln lY1I- J p(x)dx

v'

-

V

=}

y�

2 = -- p(x)

Yi

V

=

1 e- f p(x)dx -

Yi

(3.2)

where arbitrary constants of integration and± signs have been omitted, since we seek 1 only one function u(x), and not all such functions. Since 2ef p(x) dx =/= 0 (provided Y1 that p is continuous), u(x) is not a constant and, hence, y1 and Y2 = uy1 are linearly independent.

3.2.1

Equations with Constant Coefficients

Definition 3.6. A linear, homogeneous equation with constant coefficients has the general form ay" + by' + cy = 0, where a, b and c are constants.

Seek solutions in the form y = e rx , where r is a constant to be determined. Then and y

=

e rx is a solution if and only if ar2 e rx ar2

+ bre rx + ce rx = 0,

+ br + c = 0,

i.e.,

(3.3)

called the indicial equation. Employing the quadratic formula, the solutions of the indicial equation (3.3) are given by -b ± Jb - 4ac r=-----2

2a

(3.4)

49

3.2. LINEAR HOMOGENEOUS EQUATIONS

There are three possibilities, namely, b2 - 4ac > 0, b2 - 4ac = 0, and b2 - 4ac < 0, and each must be analysed separately. Case (i) If b2 - 4ac > 0, then

-b - Jb 2 - 4ac -b + Jb2 - 4ac and r2 = -----2a 2a r x r2 x are two distinct, real roots, and y 1 = e i and y2 = e are two independent solutions. The general solution is then Y = C1er1 x + C2er2 x _ r1

=

Example 3. 7. Consider the constant-coefficient equation 3y"

+ 7y' + 2y = 0.

Setting y = erx gives the indicial equation 3r 2 + 7r + 2 = 0, with the solutions 1 -7 ± J49 - 24 . , 1.e., r1 = - and r2 = -2. Since the roots of the quadratic are r= 6 3 real and distinct, y 1 = r½x and y2 = e-2x are two independent solutions, and the general solution is y = c1e-½ x + c2 e-2x _ Example 3.8. Consider the initial-value problem

y" - y' - 6y = 0, y(0) = 0, y'(0) = 10. Setting y = erx gives the indicial equation r2 - r - 6 = (r - 3)(r + 2) = 0, with the solutions r 1 = 3 and r2 = -2. Since the roots of the quadratic are real and distinct, y 1 = e3x and Y2 = e- 2x are two independent solutions, and the general solution is y = c1e3x + c2 e-2x . Then y(O) = 0 =} C1 + C2 = 0 =} y = c1(e3x - e-2x ) and y'(0) problem.

10

=}

c1

=

2

=}

y

=

=}

y' = c1(3e3x

+ 2e-2x ),

2(e3x - e-2x ) is the solution of the initial-value

-b Case (ii) If b2 - 4ac = 0, then r = - is the only root, and one solution is y1 = erx _ 2a In order to determine a second, independent solution, employ reduction of order. The standard form of the equation ay" +by'+cy = 0 is y" + �y' + 0 } lxl= { -X,

X

0 by replacing x by lxl. Example 3.13. Consider the Euler equation 2x 2 y" + xy'-3y For x > 0, y

= xr ⇒

2r(r-l)+r-3=0 ⇒ r1 =



= 0.

2r2 -r-3=0



r=



✓ 1 + 24 4

3

Since r1 and r2 are real and distinct, the general solution for 2 and r2 = -1. 3 2 1 1

any xi- 0 is y=c1lxl / + c2 ixlby c3x- 1, where C3= ±c2 .

.

Since lxl= ±x, the term c2ixl- may be replaced

54

CHAPTER 3. SECOND-ORDER EQUATIONS

Example 3.14. Consider the initial-value problem X

For x > 0, y

2 /I

y

+ xyI -y = 0

r1

= 0,

y'(l)

=

-4.

= xr ⇒ r ( r - l) + r - l



y(l)

= 1 and r2 = -1.

..

= r2 - 1 = (r - l) (r + l) = 0

The general solution for any x =I= 0 is

Then y(l) = 0 ⇒ c3 + C4 = 0 ⇒ y = c3(x - x-1 ), y'(x) = c3 (1 + x-2), and y'(l) ⇒ c3 = -2 ⇒ y = -2(x - x- 1) is the solution of the initial-value problem.

=

-4

Example 3.15. Consider the Euler equation 4x2 y" + l6xy' + 9y For x > 0, y = xr

= 0.



4r(r - 1) + 16r + 9

= 4r2 + 12r + 9 = (2r + 3) 2 = 0

2 is the only root. The general solution, for any x =I= 0, is y = lxl-3/ (c1 + c2 ln lxl). Note that lxi- 3/2 =I= ±x- 3 1 2, and x -3/2 is not real for x < 0. Hence, the absolute values must be retained. Example 3.16. Consider the initial-value problem x2y" - 5xy' + 9y For x > 0, y

= xr

= 0,

y(l)

= 0,

y'(l)

= 2.



r(r - 1) - 5r + 9

= r 2 - 6r + 9 = (r - 3) 2 = 0 ⇒ r = 3

is the only root. The general solution for any x =I= 0 is

Then y(l) = 0 ⇒ C3 = 0 ⇒ y = c4x3 ln lxl, y' = c4 (3x2 ln lxl + x2), and y'(l) c4 = 2 ⇒ y = 2x3 ln lxl is the solution of the initial-value problem.

=2⇒

55

3.2. LINEAR HOMOGENEOUS EQUATIONS Example 3.17. Consider the Euler equation

3x2 y" - 2xy' + 4y For x > 0, y

= 0.

⇒ 3r(r - 1) - 2r + 4 = 3 r2 - 5r + 4 = 0 5 ± J25 - 48 5 J23 . ⇒ r=-----=-±-i 6 6 6

= xr



y - lxl'i' [c, cos ( � ln lxl)

+ c, sin ( � ln lxl)]

is the general solution for any x =/= 0. Example 3.18. Consider the initial-value problem

x2 y" - 5xy' + 13 y = 0, For x > 0, y = xr

⇒ r(r - 1) - 5r + 13 = r

=



= 0,

y'(l)

6r + 13

=0

y(l) r2

-

3 - 52 j 6 = 3 ± 2i 2

= 6.





= lxl3 [c1 cos(2ln lxl) + c2 sin(2ln lxl)] = x3 [c3 cos(2ln lxl) + C4 sin(2ln lxl)] is the general solution for any x =/= 0. Then y(l) = 0 ⇒ c3 = 0 ⇒ y

and y'(l)

=6⇒

c4

= 3.

Hence, the solution of the initial-value problem is y = 3x3 sin(2ln lxl)-

Exercises 3.2 1 . Find the general solution of the equation y" - y' - 2y 2. Solve the initial-value problem y" determine y(l) and y'(l).

+ y' - 6y =

3. Solve the initial-value problem y" + 5y' + 4y

= 0.

0, y(0)

= 0,

y(O)

0 , y'(0)

=

= 0,

4. Find the general solution of the equation 4y" + y' - 3y

=

y'(O)

=

10, and

-6.

= 0.

5. Find two independent solutions of the equation y" - 3y' + y = 0. 6 . Solve the initial-value problem y" - y y(l) and y(-1).

=

0, y(0)

=

1, y'(0)

= 2,

and determine

56

CHAPTER 3. SECOND-ORDER EQUATIONS 7. Find two independent solutions of the equation 9y" - 24y' 8. Solve the initial-value problem y" - lOy' + 25y

= 0,

+ l6y = 0.

= 1,

y(0)

y'(0)

= 1.

9. Find the general solution of the equation 4y" - 4y' + y = 0. 10. Employ two different methods to solve the equation y"

= 0.

11. Find the general solution of the equation y" + y' + y = 0. 12. Solve the initial-value problem y" - 2y' + 2y

= 0,

y(0)

= 0,

13. Find the general solution of the equation 2y" - 4y' + 3y

y'(0)

= 1.

= 0.

14. Find the general solution of the equation y" + 4y = 0. 15. Consider the equation y" + y = 0. (a) Find the general solution. (b) Solve the initial-value problem, given the initial conditions y(0) y'(0) = 0.

1,

(c) Solve the initial-value problem, given the initial conditions y(0) y'(0) = 1.

0,

16. Find the general solution of the equation x 2y" + 2xy' - 6y

= 0.

17. Find the general solution of the equation 6x2 y" + 7xy' - 2y = 0. 18. Solve the initial-value problem x2 y" - 7xy' determine y(-1) and y(e).

+ l6y = 0, y(l) = 0, y'(l) = 2, and

19. Find the general solution of the equation 9x2 y" - 3xy' + 4y 20. Employ three different methods to solve the equation x2 y"

= 0.

+ xy' = 0.

21. Find the general solution of the equation -2x 2 y" + 2xy' - 5y 22. Solve the initial-value problem x2 y" - 3xy' + 13y 23. Find the general solution of the equation x2 y"

= 0.

= 0, y(l) = 0, y'(l) = 6.

+ xy' + 4y = 0.

57

3.3. LINEAR NONHOMOGENEOUS EQUATIONS

3.3

Linear Nonhomogeneous Equations

Theorem 3.3. The general solution of the nonhomogeneous equation

a(x)y" + b(x)y' + c(x)y = g(x)

is y = YP + Yh, where yP is any particular solution, and Yh solution of the associated homogeneous equation

3 ( .7)

= C1Y1 + C2Y2 is the general

a(x)y" + b(x)y' + c(x)y = 0,

(3.8)

with y 1 and y2 any two linearly independent solutions.

Proof. y = YP + Yh is a solution of Equation 3 ( .7) because a(x)y" + b(x)y' + c(x)y

a(x)(yp + Yh)" + b(x)(yp + Y1i)' + c(x)(yp + Yh) = [a(x)yi + b(x)y� + c(x)UE] + [a(x)y� + b(x)y� + c(x)yh] = g(x) + 0 = g(x) .

Since y contains two arbitrary constants, it is the general solution of Equation (3.7). The solutions y 1 and y2 of Equation (3.8) are called the homogeneous solutions, ( .7) is called the nonhomogeneous term. and g(x) in Equation 3 In view of Theorem 3.3, given a nonhomogeneous equation, it is necessary to find a particular solution yP in order to obtain the general solution, the determination of two independent solutions of the associated homogeneous equation already having been discussed in Section 3.2.

3.3.1

The Method of Undetermined Coefficients

Consider a nonhomogeneous equation

ay" +by'+ cy = g(x),

(3.9)

where a, band care constants, i.e., the associated homogeneous equation

ay" +by'+cy has constant coefficients. I\

=0

(3.10)

If g(x) consists of certain, particular forms, then the form of a particular solution , containing unknown constants, of Equation 3 ( .9) can be deduced. The unknown yp constants in YP are then determined by forcing YP to be a solution of Equation (3.9). This is called the method of undetermined coefficients. If g(x) = C ef3x , where C and /3 are constants, then YP is not a solution of Equation (3.10).

=

Aef3x , provided that ef3x

58

CHAPTER 3. SECOND-ORDER EQUATIONS

Example 3.19. Consider the nonhomogeneous equation I x y II -y2y= 3e.

The associated homogeneous equation is y II-yI - 2y= 0 and y = erx =? r 2-r -2 = (r -2)(r + 1) = 0:::} y1 = e2x, Y2 solution of the homogeneous equation is Yh = c1 e2x + c 2 e- x .

= e-x,

and the general

Since g(x) = 3ex and ex is not a solution of the homogeneous equation, YP = Aex . Then y'p = Aex yp11= Aex and )

)

3 - ex and y = YP + Yh 2 the nonhomogeneous equation. Thus, YP

=

3

= -2 ex + c1e2x + c2 e-x

is the general solution of

Example 3.20. Consider the initial-value problem

y11 + y

= 4ex ,

y(O)= l, y'(O)

= 2.

The associated homogeneous equation is yll + y= 0, and y = erx ⇒ r2 + l = 0 ⇒ r = ±i ⇒ y1 = cos(x), Y2 = sin(x), and the general solution of the homogeneous equation is Y1t = c1 cos(x) + c 2 sin(x). Since g(x) = ex , which is not a solution of the homogeneous equation, YP = Aex . - 2ex , - 2 . H ence, YPThen Yp'- Aex , Yp11 - Aex, and Yp11 + YP - 4ex 1"f and on1y 1"f Aand the general solution of the nonhomogeneous equation is Then y(O) = 1 ⇒ c 1 =-1 , and y'(O) solution of the initial-value problem.

=

2

⇒ c2 = 0,

and y

=

2ex - cos(x) is the

If ef3x is a solution of Equation (3.10), then so is Aef3x and, hence, it cannot be a solution of Equation (3.9). In that case, YP = Axef3x, provided that xef3x is not a solution of Equation (3.10).

59

3.3. LINEAR NONHOMOGENEOUS EQUATIONS Example 3.21. Consider the nonhomogeneous equation y"

- 4y

= 8e2x .

The associated homogeneous equation is y"

- 4y

= 0,

2 2 and y = erx ⇒ r2 - 4 = 0 ⇒ r = ±2 ⇒ Yi = e x, y2 = e- x, and the general solution of the homogeneous equation is Yh = c1e 2x + c2e- 2x .

Since g(x) = 8e 2x and e 2x is a solution of the homogeneous equation but xe 2x is not, YP = Axe2x . Then y; = A(e 2x + 2xe 2x ) = Ae 2x (l + 2x), yi = Ae 2x (4 + 4x), and yi

- 4yp = 8e 2x if and only if Ae 2x (4 + 4x) - 4Axe2x = 8e 2x, i.e., A= 2.

Thus, YP = 2xe2x and y = YP + Yh the nonhomogeneous equation.

= 2xe2x + c1 e2x + c2e-2x

is the general solution of

If both ef3x and xef3x are solutions of Equation (3.10), then yP = x2 ef3x . In this case, x2 ef3x cannot be a solution of Equation (3.10) because a linear, second-order equation has precisely two independent solutions (Theorem 3.2). Example 3.22. Consider the nonhomogeneous equation y"

- 6y '

+ 9y = 4e3x .

The associated homogeneous equation is y"

- 6y ' + 9y = 0,

and y = erx ⇒ r 2 - 6r + 9 = (r - 3) 2 = 0 ::::} Yi = e 3x, Y2 = xe3x, and the general solution of the homogeneous equation is Yh = c1 e3x + c2 xe3x . Since g(x) = 4e 3x and both e 3x and xe 3x are solutions of the homogeneous equation, YP = Ax2e 3x . Then y; = Ae3x (2x + 3x2 ), yi = Ae3x (2 + 12x + 9x2 ), and YpII - 6YpI + 9YP

= 4e3x

if and only if Ac3x [(2 + 12x + 9x2 )

-

6(2:r; + 3x 2 )

Thus, YP = 2x2 e3x and y = YP + Yh of the nonhomogeneous equation.

+ 9x2 ] = 4e3x ,

1.e., A= 2.

= 2x2 e3x + cie3x + c2 xc3x is the general solution

60

CHAPTER 3. SECOND-ORDER EQUATIONS The case where g(x) = Ce13x can be summarized by the following:

Theorem 3.4. A particular solution YP of the nonhomogeneous equation

ay" +by'+ cy where a, b, c, C and

= Cef°3x ,

/3 arc constants, is given by

l. YP = Ae13x , provided that e13x is not a solution of the associated homogeneous equation. 2. YP = Axe13x, if ef3x is a solution of the associated homogeneous equation and xe13x is not. 3. YP = Ax2e/3x , if both ef3x and xef3x are solutions of the associated homogeneous equation. If g(x) = Ccos(f3x) + D sin(/Jx), where C, D and YP

/3 are constants, then

= Acos(f3x) + B sin(/Jx),

provided that cos(/Jx) and sin(/Jx) are not solutions of Equation (3.10 ). Example 3.23. Consider the nonhomogeneous equation y" + 4y' + 4y

= 6sin(3x).

The associated homogeneous equation is y" + 4y' + 4y

= 0,

and y = erx ⇒ r 2 + 4r + 4 = (r + 2)2 = 0 ⇒ Y1 = e- 2x , Y2 = xe- 2x, and the general solution of the homogeneous equation is Yh = c 1 e- 2x + c2 xe- 2x. Since g(x) = 6sin(3x) and sin(3x) and cos(3x) are not solutions of the associated homogeneous equation, YP = Acos(3x) + Bsin(3x). Then

y� = -3A sin(3x) + 3B cos(3x), vi = -9A cos(3x) - 9B sin(3x), and yi + 4y; + 4yp

= 6 sin ( 3x)

if and only if

[-9A cos(3x) - 9B sin(3x)] + 4[-3A sin(3x) + 3B cos(3x)] +4[A cos(3x) + Bsin(3x)] = 6sin(3x),

61

3.3. LINEAR NONHOMOGENEOUS EQUATIONS 1.e.'

(12B - 5A) cos(3x) - (12A + 5B) sin(3x)

= 6 sin(3x),

which holds if and only if A and B satisfy the system { ���: the solution A

=-

72

169

and B

72

=-

30

169

. Thus, yP

)

=-

30 . (

)

72

169

�2/} : !6 } , with

cos(3x) -

30 .

169

sm(3x) and

+ c 1 e-2x + c2 xe-2x 169 is the general solution of the nonhomogeneous equation. Note that YP includes both a cosine term and a sine term, even though g(x) includes no cosine term. Y

= YP + Yh = - cos ( 3x 169

sm 3x

-

If cos((Jx) and sin((Jx) are solutions of Equation (3.10), then YP

= x[Acos((3x) + Bsin((3x)].

In this case, xcos((3x) and x sin((3x) cannot be solutions of Equation (3.10) because a linear, second-order equation has precisely two independent solutions (Theorem 3.2). Example 3.24. Consider the nonhomogeneous equation

y" + y

= 2cos(x).

The associated homogeneous equation is

= 0, y 1 = cos(x),

y" + y

y2 = sin(x), and the general and y = erx =} r 2 + 1 = 0 =} r = ±i =} solution of the homogeneous equation is Yh = c 1 cos(x) + c2 sin(x). Since g(x)

= 2cos(x) and cos(x), sin(x) satisfy the associated homogeneous equation, YP = x[Acos(x) + Bsin(x)].

Then y; y:

and Yi+ YP

= =

[Acos(x) + Bsin(x)] + x[-Asin(x) + Bcos(x)],

2[-Asin(x) + B cos(x)]

+ x[-A cos(x)

- B sin(x)],

= 2cos(x) if and only if 2[-Asin(x) + Bcos(x)] = 2cos(x),

i.e., A= 0 and B = l. Thus, YP = xsin(x) and Y=

YP

+ Yh = X sin(X) + C1 cos(X) + C2 sin(X)

is the general solution of the nonhomogeneous equation.

62

CHAPTER 3. SECOND-ORDER EQUATIONS The case where g(x) = C cos(/3x) + D sin(/3x) can be summarized by the following:

Theorem 3.5. A particular solution YP of the nonhomogeneous equation

ay" +by'+ c:y

= Ccos(f3x) + Dsin(/3x),

where C, D and /3 are constants, is given by l. YP = A cos(/3x) + B sin(/Jx), provided that cos(/3x) and sin(/3x) are not solutions of the associated homogeneous equation. 2. YP = x[Acos(/3x) + Bsin(/3x)], whenever cos(/3x) and sin(/3x) are solutions of the associated homogeneous equation. If g(x) is a polynomial of degree n 2: 0, i.e.,

g(x) =Co+C1x + C2 x2 + · · · + Cnxn , then YP is a polynomial of degree n, YP =Ao+A1x + A2 x2 +

· · · + Anxn ,

provided that no term in YP satisfies Equation (3.10). Example 3.25. Consider the nonhomogeneous equation

y" - Sy' + 6y = 3 + 12x. The associated homogeneous equation is

y" - Sy'+ 6y = 0, and y = erx ⇒ r 2 - Sr+ 6 = (r - 2)(r - 3) = 0 ⇒ Y1 = e 2x , Y2 = e 3x , and the general solution of the homogeneous equation is Yh = c1e 2x + c2e 3x . Since g(x) = 3 + 12x and no term in a first-degree polynomial is a solution of the homogeneous equation, YP =A+Bx. Then y; = B, yi = 0, and

y; - sy; + 6yp

= 3 + 12x 13

6.

if and only if - SB+ 6(A +Bx)= 3 + 12x, 13

.

+ 2x, and the general solut10n of the 6 . . 13 x e 2x + c2e3x . nonhomogeneous equat10n 1s y = YP + Yh = 6 + 2 + c1 which gives B

---

= 2 and A=

Hence, YP

=

63

3.3. LINEAR NONHOMOGENEOUS EQUATIONS

If any term in Ao+A 1 x + A 2x2+· · ·+An xn is a solution of Equation (3.10), then

provided that no term in YP is a solution of Equation (3.10). Example 3.26. Consider the nonhomogeneous equation y"-2y' = 2-3x. The associated homogeneous equation is y"-2y' = 0, and y= erx ⇒ r 2 -2r= r(r - 2) = 0 ⇒ y1 = 1, y2 = e 2x, and the general solution of the homogeneous equation is Yh = c1 + c2 e 2x. Since g(x) = 2 -3x and a constant is a solution of the homogeneous equation, YP i= A+Bx, but YP = x(A +Bx)=Ax+ Bx2. Then y; =A+2Bx, y�=2B, and

y:-2y�=2-3x if and only if 2B-2(A+2Bx)=2-3x, which gives B=

l

and A=

-1-

Hence, YP =

-lx lx2 +

,

and the general solution

of the nonhomogeneous equation is y = YP +Yh =-ix+ �x2 +c1 +c2 e2x . If both Ao +A1x + A 2 x2 + · · · + A n xn and Aox+ A1x2 + A 2 x3 + · · · + An xn+l contain terms which are solutions of Equation (3.10), then

The latter cannot contain any terms which satisfy Equation (3.10), because the only polynomial which can satisfy a homogeneous equation with constant coefficients has degree zero or one, and every term in the above polynomial has degree at least two. Example 3.27. Consider the nonhomogeneous equation y" = 24-36x + 12x2. The associated homogeneous equation is

y" = 0,

64

CHAPTER 3. SECOND-ORDER EQUATIONS

and y = erx ⇒ r2 = 0 ⇒ Y1 equation is Yh = C1 + C2X.

=

1, Y2

= x, and the general solution of the homogeneous

Since g(x) = 24 - 36x + 12x2 and both a constant and x satisfy the homogeneous equation, YP i- A+Bx+Cx2 and YP i- x(A +Bx+Cx2 ), but YP Then y� y�

= x2 (A+Bx+Cx2 ) = Ax2 + Bx3 + Cx4.

= 2Ax + 3Bx2 + 4Cx3,

= 24 - 36x + 12x2

yi

= 2A + 6Bx + 12Cx2 ,

and

if and only if 2A + 6Bx + 12Cx2

which gives A = 12, B = -6 and C = 1, i.e., YP solution of the nonhomogcneous equation is then

=

= 24 - 36x + 12x2 ,

12x2 - 6x3

+ x4 . The general

Of course, this equation can be solved much more easily simply by two integrations: y"

= 24 - 36x + 12x2 ⇒

y'

= 24x - 18x2 + 4x3 + k1

as above. The case where g(x) = C0 following:

+ C1x + C2 x2 + · · · + Cn xn can

be summarized by the

Theorem 3.6. A particular solution of the nonhomogeneous equation

ay"

+ by'+cy = Co + C1x + C2 x2 + · · · + Cn xn

is given by l. YP = Ao + A1x + A2 x2 + · · · + An xn , provided that no term in YP is a solution of the associated homogeneous equation. 2. YP = x(Ao + A1x + A 2 x2 + · · · + An xn ) = Aox + A1x2 + A2 x3 + · · · + An xn+l if Ao+A 1 x + A2 x2 + · · · + An xn contains a term which satisfies the associated homogeneous equation and x(Ao + A1 x + A2 x2 + · · · + A n xn ) does not. 3. YP = x2 (Ao + A1x + A2 x2 + · · · + An xn ) = Aox2 + A1x3 + A2x4 + · if both Ao+ A1x + A2x2 + · · · + An xn and x(Ao + A1x + A2 x2 + contain terms which satisfy the associated homogeneous equation.

· · + An xn+2 · · · + An xn )

65

3.3. LINEAR NONHOMOGENEOUS EQUATIONS Theorem 3. 7. If YPi and yp2 are particular solutions of ay" + by' + cy respectively, then YP

= 91(x)

= YPi + yp2

and ay" + by' + cy

= 92(x),

is a particular solution of

ay" +by'+ cy

= 91 (x) + 92(x).

More generally, if YP; is a particular solution of ay" +by'+cy

= YP i + yp2 + · · · + YP is a particular solution of ay" +by'+cy = 91(x) + 92(x) + · · · + 9n(x).

for 1 � i � n, then YP

Proof. For n

= 2,

II

ayP

b

+

= 9i(x)

n

Yp + cyp I

a(YP1 + YP2 ) + b(YP1 + YP2 ) (ay:1 + by� 1 + cyp1 ) + (ay:2 91(x) + 92(x). 11

1

+ c(YP1 + YP2 ) + by�2 + cyp2)

The general case is proved similarly. Example 3.28. Consider the nonhomogencous equation y" + 4y

= 4x2 - 5ex .

The associated homogeneous equation is

= 0, and y = erx ⇒ r2 + 4 = 0 ⇒ r = ±2i ⇒ y1 = cos(2x), y2 = sin(2x), and the general solution of the homogeneous equation is Yh = c 1 cos(2x) + c 2 sin(2x). y" + 4y

= 4x2, YP i =A+ Bx+Cx2. Then y� 1 = B + 2Cx, yi1 = 2C, and 2 2 y:1 + 4yp 1 = 4x2 if and only if 2C + 4(A +Bx+Cx ) = 4x ,

For y" + 4y

which gives C For y" + 4y

=

1, B

= -5ex '

which gives D

1

= 0, and A = - .

2

Hence, Yp 1

1

= - + x2 . 2

x x = Dex . Then yP2 ' = De ' y"P2 = De ' and y:2 + 4yp2 = -5ex if and only if Dex +4Dex = -5ex ,

= Y

yP2

-1. Hence, yp2

=

-ex , and YP

= YPi + yp2 = -� + x2 - ex , and

= YP + Yh = - 1 + x2 - ex +c1 cos(2x) + c2 sin(2x) 2

is the general solution of the nonhomogeneous equation.

66

CHAPTER 3. SECOND-ORDER EQUATIONS If g(x) consists of products such as

P(x)ef3x , P(x)[Ccos(,x) + Dsin(,x)], e-8x [Ccos(,x) + Dsin(,x)],

or

P(x)elh [Ccos(,x) + Dsin(,x)],

where P is a polynomial and C, D, corresponding forms or

/3 and I are constants,

then YP must have the

x x Q(x)ef3 , Q(x) cos(,x) + R(x) sin(,x), ef3 [A cos(1x) + Bsin(,x)],

ef3x [Q(x) cos(,x) + R(x) sin(,x)],

multiplied by x or x2 if necessary, in order that no term in YP be a solution of the associated homogeneous equation, where Q and R are polynomials of the same degree as that of P. Example 3.29. Consider the nonhomogeneous equation y"

- 2y' + 2y = ex cos(x).

The associated homogeneous equation is y"

- 2y'

+ 2y = 0,

and y = erx ⇒ r2 - 2r + 2 = 0::::} r = 1 ± i::::} y1 = ex cos(x), Y2 = ex sin(x), and the general solution of the homogeneous equation is Yh = c1ex cos(x) + c2 ex sin(x). Since g(x) = ex cos(x) and ex cos(x) is a solution of the associated homogeneous equation, YP = xex [A cos(x) + Esin(x)]. Then y�

= ex [A cos(x) + Bsin(x)] + xex [(A + B) cos(x) + (B - A) sin(x)],

= ex [2(A + B) cos(x) + 2(B - A) sin(x)] + xex [2B cos(x) - 2A sin(x)], and yi - 2y� + 2y = ex cos(x) if and only if y�

p

ex [2Bcos(x)-2A sin(x)] = ex cos(x),

which gives A= 0 and B = t- Hence, YP = txex sin(x), and y

= YP + Yh = �xex sin(x) + c1ex cos(x) + c2 ex sin(x)

is the general solution of the nonhomogeneous equation.

A more efficient method for the determination of particular solutions in examples such as Example 3.29 will be discussed next.

3.3. LINEAR NONHOMOGENEOUS EQUATIONS 3.3.2

67

Variation of Parameters

Consider a nonhomogeneous equation y

11

+ p(x)y' + q(x)y = J(x)

(3.11)

in standard form. The associated homogeneous equation is y

11

+ p(x)y' + q(x)y = 0.

(3.12)

Let y1 and y2 be any two linearly independent solutions of Equation (3.12). Seek a particular solution of Equation (3.11) in the form (3.13) where u 1 (x) and u2 (x) are functions to be determined by forcing YP to be a solution of Equation (3.11). Then, by the product rule, (3.14) The requirement that YP be a solution of Equation (3.11) imposes one constraint upon the two functions u1 and u2. We may therefore impose a second constraint, chosen in order to simplify the procedure. Thus, impose the condition (3.15) Then Equation (3.14) reduces to

from which we obtain Hence, y� + p(x)y; [u�y�

Yp = U1Y1 + U1Y1 II

I

I

II

+ U2Y2 + U2Y2· I

I

11

+ q(x)yp = f(x) if and only if

+ U1Y� + u;y� + U2Y�] + p(x)[u1y� + U2Y�] + q(x)[u1y1 + U2Y2]

= f(x).

Rearranging terms, the latter condition becomes

Since Y1 and y2 are solutions of Equation (3.12), y�

+ p(x)y� + q(x)y1

= 0 and y; + p(x)y�

+ q(x)y2 = 0,

and condition (3.16) reduces to (3.17)

68

CHAPTER 3. SECOND-ORDER EQUATIONS

Combined with condition (3.15), u� and u; must satisfy the system YiU'i + Y2U2 y� u� + y� u; I

= =

0 f ( x)

(3.18)

of algebraic equations. Employing Cramer's rule, we obtain I

-

0

,

Y2

J(x) y;

u, � I �; �; I

I

The quantity

yif(x)

Yi Y2 y� y�

I

t I = YiY2t - Y1Y2

is denoted by W[y1, Y2](x) or simply by W(x) and called the Wronskian of y1 and y2. Thus, ui

=J

-yd(x) dx, u2 W(x)

=J

yif(x) dx, and YP W(x)

= uiyi + U2Y2-

This method for the determination of a particular solution of Equation (3.11) is called variation of parameters or variation of constants. Note that the above formulas for u i and u2 are defined provided that W(x) This is guaranteed by the following:

i= 0.

Theorem 3.8. If p and q are continuous on an interval I, and Yi and y2 are solutions of y" + p(x)y' + q(x)y = 0, then either W 0 on I or W(x) i= 0 for any x in J. The solutions Yi and y2 are linearly independent on I if and only if W(x) i= 0 on I. (See Chapter 3 Exercises, Exercise 15, on page 85.)

=

Example 3.30. Consider the nonhomogeneous equation 2x y - 3yI+ 2y=e. ff

The associated homogeneous equation is

y" - 3y'

+ 2y = 0,

Yi Y2 I I Yi Y 2

ex e 2x = 3x = e x 2 2x e . e

x 2x , and the and y = e rx ⇒ r2 - 3r + 2 = (r - l)(r - 2) = 0 ⇒ Yi = e , Y2 = e 2x x general solution of the homogeneous equation is Yh = Ci e + c2 e . The equation is in standard form with f(x) = e 2x, and the Wronskian of Yi and Y2 is

w (X) =

I

I I

I

3.3. LINEAR NONHOMOGENEOUS EQUATIONS Hence,

2x.e2x - y 2](x)d () X =J-e 3 x dX =!-exdX J WX e x e . e2x d Y1 J ( x)d d X = J 1 X = X, J () X =J x 3 e Wx

69

=-e,x

xe2x - e2x + C1ex + C2e2x xe2x + c1ex + (c 2- l)ex2 Xe2x + C1ex + C3e2x

is the general solution of the nonhomogeneous equation.

Note that the term -e2x = -y2 has been absorbed into the term c2e2x in the general solution above. Thus, any term in YP which satisfies the homogeneous equation may be removed from YP · In fact, a particular solution of a nonhomogeneous equation is not unique because, if YP is a particular solution, then so is YP + Yi, where Yi is any solution of the associated homogeneous equation. For this reason, arbitrary constants of integration k 1 and k2 need not be added to u 1 and u 2, since this would merely add the term k1Y1 + k2 Y2 to yp, and these may be removed. Example 3.31. Consider the initial-value problem 4y" -

y

= 4e½

x

,

y (0)

= 0,

y'

(0) = 3.

The associated homogeneous equation, 4y "- y = 0, has the two independent solutions x x Y1 = e½ and Y2 = e-½, with the Wronskian W(x) = -1. The standard form of the . . yII - 1 y = e2lx . Hence, · x) = e2I x , non l1omogcncous equat10n 1s f( 4

I 1 I l and YP = xe2x - e2 x , or YP = xe2x, since -e2x = -y1 satisfies the homogeneous equation and may be removed from YP · The general solution of the nonhomogeneous l l I . . equat10n 1s y = YP + Yh = xe2x + c1e2 x + c2e-2 x . Then y( 0 ) = 0 ⇒ c2 = -c1 ⇒

and y'(O) = 3

⇒ c1 = 2.

Hence, the solution of the initial-value problem is

70

CHAPTER 3. SECOND-ORDER EQUATIONS

Example 3.32. Consider the nonhomogeneous equation

y" - 2y ' + 2y

= ex cos(x),

which has been solved in Example 3.29 by the method of undetermined coefficients. Here, we shall solve it by variation of parameters. Two independent solutions of the associated homogeneous equation were found to be y1 = ex cos(x) and y2 = ex sin(x). With f(x) = ex cos(x) and

W(x)

I

=

I=I

Y1 Y2 y � y�

u1

-

j

u2

j

ex cos(x) ex sin(x) x x e cos(x) - e sin(x) e sin(x)+ ex cos(x) x

sin(x) cos(x) dx

cos 2(x) dx = �

j

= -�

j

sin(2x) dx

=

I = e2x '

i cos(2x),

l + cos(2x) dx = }x+ 1 sin(2x),

and }cos(2x)[ex cos(x)]+ [ix+}sin(2x)] [ex sin(x)]

YP

}xex sin(x)+ }ex [cos(2x) cos(x)+ sin(2x) sin(x)]

Since

1

=

1

satisfies the homogeneous equation, it suffices to take as a 1 particular solution YP = xex sin(x), in agreement with the result of Example 3.29. 2

4

ex cos(x)

1 x 1 xe sin(x)+ :::tx cos(x). 2

4 y1

Example 3.33. Consider the nonhomogeneous equation 11

y

2ln(x)

+ 6y, + 9y = xe3x

,

X

> 0.

The associated homogeneous equation has the two independent solutions y 1 21 x ) and and y2 = xe-3x , with f(x) = n( e3x x

I

Then, by integration by parts and by substitution,

=e -6x

xe-3x 2ln(x) J -6dx = -2 J ln(x) dx = -2x ln(x)+ 2x, e- x xe3x e-3x 2ln(x) ln(x) J 6 dx = 2f dx = [ln(x)] 2, e- x xe3 x X -

= e- 3x

71

3.3. LINEAR NONHOMOGENEOUS EQUATIONS and

since 2xe- 3x = 2y2 . Thus, y

= YP

+ Yh = xe-3x [ln(x)] 2 - 2xe-3x ln(x) + c1e-3x + c2 xe- 3x

is the general solution of the nonhomogcneous equation. Example 3.34. Consider the nonhomogencous equation x2y" - 6xy'

+ 12y = x5 ex .

The associated homogeneous equation is the Euler equation x2y" - 6xy'

+ 12y = 0

and, for x > 0, y = xr ⇒ r 2 - 7r + 12 = (r - 3)(r - 4) = 0 ⇒ Y1 = x3 and Y2 = x4. Since lxl3 = ±x3 and lxl4 = x4 , these solutions are defined for all real x, with the Wronskian I x3 x4 6 W (x) = 3x2 4x3 = x . I •

The nonhomogcneous equation in standard form 1s

f(x)

=x

y

II

e . By integration by parts,

3 x

and YP = (ex - xex)x3 + exx4 = x3 ex . Hence, y = YP general solution of the nonhomogeneous equation.

6 I 12 X - -y + 2 y = x3 e X X

Thus,

+ Yh = x3ex + c1x3 + c2x4 is the

Example 3.35. Consider the nonhomogeneous equation x2y" - 3xy'

+ 3y = 4x5 ex

2

.

The associated homogeneous equation is the Euler equation x2y" - 3xy' + 3y = 0 and, for x > 0, y = x r ⇒ r2 - 4r + 3 = (r - l)(r - 3) = 0 ⇒ Y1 = x and Y2 = x3 . Since lxl = ±x and lxl3 = ±x3 , these solutions are defined for all real x, with the Wronskian x x3 3 W (x) = 1 3x2 = 2x .

I

I

72

CHAPTER 3. SECOND-ORDER EQUATIONS 3

The standard form of the nonhomogeneous equation is y " - � y' +

f (x

)

= 4x3 e x2 .

Thus, by integration by parts,

J

-

3 3 x x . 4x e dx --2x3 2

- x 2e x

J

X .

2

+

J

4x e 2x3

3 x2

2 xe

x2

dx =

X2

X

=-

J

J

= 4x3 ex , with 2

2 x2 · 2 xe x dx

dx = - x 2ex2 2 xe

y

+ ex , 2

dx = ex2

x2

and YP = xe x\1 - x2) + x3 ex = xex . The general solution of the nonhomogeneous 2 equation is then y = YP + Yh = xe x + c 1x + c2x3. 2

2

Example 3.36. Consider the nonhomogeneous equation x

2

y" - 3xy ' + 5y

= x2,

x

> 0.

The associated homogeneous equation is the Euler equation 2 x y"

and y

y2

=

xr ⇒ r 2

-

4r + 5

=

- 3xy ' + 5y

0 ⇒ r

= x2 sin[ln(x)], with the Wronskian ( )W x

I

= 0,

4±\/-4 = 2 ± i = --2

⇒ y1

= x2 cos[ln(x)]

x2 sin[ln(x)] x2 cos[ln(x)] 2xcos[ln(x)] - xsin[ln(x)] 2xsin[ln x)] + xcos[ln(x)] (

-

I-

x

3

and

·

5 The nonhomogeneous equation in standard form is y" - �y' + 2 = 1, and f(x) = 1. X X 1 du By integration by substitution with u = ln(x) and - = -,

-! J

-! J

dX

X

sin[ln(x)] x2 sin[ln(x)] d = dx = cos[ln(x)], x 3 x X 2 cos[ln(x)] x cos[ln(x)] dx = dx = sin[ln(x)], 3 x X

and YP

=x

2

cos2[ln(x)] y

+ x2 sin2[ln(x)] = x2. Hence,

= YP + Yh = x2 {1 + c1 cos[ln(x)] + c2 sin[ln(x)]}

is the general solution of the nonhomogeneous equation.

73

3.3. LINEAR NONHOMOGENEOUS EQUATIONS Example 3.37. Consider the initial-value problem

x 2 y" - xy' + y = 27x4 ln(x), x > 0, y(l)

= 3, y'(l) = -3.

The associated homogeneous equation is the Euler equation

x 2 y" - xy' + y and y = xr Wronskian



r 2 - 2r

+

1

(r - 1)2

=

I

x = W( )

=

0

= 0,



Y1

x x ln(x) 1 ln(x) + 1

=x

I = x.

and Y2

= x ln(x),

with the

1 1 The nonhomogeneous equation in standard form is y" - -y' + 2y = 27x2 ln(x), and X X f(x) = 27x 2 ln(x). By integration by parts, -

u1

J

x ln(x). 7x 2 ln(x) ! dx

J

=-

J

27x 2 [ln(x)] 2 dx

18x 2 ln(x) dx

-9x3 [ln(x)]2

+

-9x3 [ln(x)]2

+ 6x3 ln(x) -

J

6x2 dx

-9x 3 [ln(x)]2 + 6x 3 ln(x) - 2x 3, x · 27x 2 ln(x) dx = 27x 2 ln(x) dx J

U2

x

9x3 ln(x) -

J

J

9x2 dx

9x3 ln(x) - 3x3,

and YP

2 3 3 3 3 3 { -9x [ln(x)] + 6x ln(x) - 2x }x + [9x ln(x) - 3x ]x ln(x) 3x4 ln(x) - 2x4 .

Hence, the general solution of the nonhomogeneous equation is Then y(l) = 3 ⇒ c1 c2 = -3. Hence,

= 5, y' =

12x 3 ln(x) - 5x 3 + 5 + c2 [ln(x) + 1], and y'(l)

y = 3x4 ln(x) - 2x4 + 5x - 3x ln(x)

is the solution of the initial-value problem.

= -3 ⇒

74

CHAPTER 3. SECOND-ORDER EQUATIONS

Example 3.38. Consider the nonhomogeneous equation 4x2 y" + 4xy' - y = x. The associated homogeneous equation is the Euler equation 4x2 y" + 4xy' - y = 0 and, for x > 0, y = xr ⇒ 4r(r - 1) + 4r - 1 and y2 = x- 1/2, with the Wronskian

= 4r2

-

1

= O ⇒ r = ±½ ⇒

The nonhomogeneous equation in standard form is y " + � y ' - -; X � 1 x J( ) = x. Then 4 -

J

and

J

y1

= x 1 12

=� and, hence, �

J J

x-1/2 . 4� dx = ! x-1/2 dx = !x1/2 ' -x- 1 4 2 1 f 2 · .l_ x 1 1 x1/2 dx = --x3/2 ' ___4�x dx = --x- 1 4 6

1 1 1/2 . 1/2 _ -1 3/2 x . x-1/2 = -x. x x 2 6 3 Since YP is a particular solution of the nonhomogeneous equation for any real x, the general solution for any x =I= 0 is YP

=

The absolute values are required only in Yh·

Exercises 3.3 In Exercises 1-17, solve the equation or the initial-value problem by the method of undetermined coefficients. 1. y"

+ 4y' = -sex

2. y" - y = 6e2x, y(O) = 2, y'(O) = 4 3. y" - 5y' + 6y

= 4e-2x

4. y" + 3y' - 4y = -lOex

3.3. LINEAR NONHOMOGENEOUS EQUATIONS 5. y" + 6y' + 9y

=

75

6e- 3x

6. 6y"+y' - 2y=65sin(x) 7. y"+2y' +y=50cos(3x), y(0) = 0, y'(0) = 2 8. y"+y=4sin(x) 9. y"+4y = 8 cos(2x)+12sin(2x) 10. y"+4y' +4y = 4sin(x) - 3 cos(x) 11. y" - y = 4 12. y" - y' - 12y=3+12x 13. y" + 2y' + 3y 14. y" - 9y

=

=

-9x

2 - 3x + x2

15. y"+y' =1+x, y(0)=1, y'(0) = 1 16. y" - y'

=

4x3

17. y" - y=-2x+4ex +2sin(x), y(0)=0, y'(0)=9 In Exercises 18-37, solve the equation or the initial-value problem by variation of parameters . 18.

y"+y'

19.

y"

- 2y=9ex

- 4y' +4y=6xe2x

20 . y"+y' = 1, y(0)=2, y'(0)=4 21. y"+9y=6sin(3x) 22. y"+y = sec(x) 23. y"+6y' +13y =l6e- 3x cos(2x) 24.

y

II

- 4y' +13y =

9e 2x COS(3X )

25. y" - 2y' +y = ex ln ( x), x > 0 26. y"+3y' +2y =sin(ex ) 27. x2 y" - 6y=10x5

76

CHAPTER 3. SECOND-ORDER EQUATIONS

28. x2 y" - 4xy' + 6y 29. x2y"

= x 4 ex

+ 5xy' + 3y = 4xex

30. x2 y" - 2xy' + 2y 31. x2 y" + xy' - y

2

, y(l)

= e,

y'(l)

=e

= x3 cos(x)

= x2 sin(x)

32. x2 y" + 3xy' + y = 33. x2y" - 3xy' + 4y

ln(x) x> 0 X

- ,-

= x2 ln(x),

34. x2y" - 3xy' + l3y

=

9x 2

35. x2y" + 7xy' + l3y

=

3

36. x2y" - 5xy' + l3y

=

x

cos2

x>0

4 x>0 [2ln(x)]

8x3 sin[2ln(x)], x>0

37. 4x2 y" + 4xy' - y = x2

3.4

Equations Reducible to First-Order Equations

Consider a second-order equation in which the dependent variable y does not appear explicitly. For example, y'y" + x2 = 0. Let y'(x) = z(x). Then y" = z', and the equation becomes zz' + x2 = 0, which is of the first order. If the first-order equation can be solved for z, then y is obtained by y

=

j

y' dx

=

j

z dx.

The general solution z of the first-order equation includes one arbitrary constant, and its integral y contains a second arbitrary constant, and is therefore the general solution of the second-order equation. Example 3.39. Consider the equation

l y" --=-+x. X y' + l Since y does not appear explicitly, let z(x) = y'(x) to obtain the first-order equation

z' l --=-+x z+ l X

77

3.4. EQUATIONS REDUCIBLE TO FIRST-ORDER EQUATIONS which is separable. Then

1 J--dz=J!+xdx Z + 1 X



z + 1= k1xe x

212



is the general solution.

y=

J



1 2 ln lz+ll=lnlxl+- x +c1 2

zdx =

J

k1xe x

212

-

1 dx = k1ex ! 2 2

- X

+ C2

Example 3.40. Consider the equation

x./xy" =./xy' + 2x,/yi, x > 0.

Since y does not appear explicitly, let z(x) =y'(x) to obtain the first-order equation xvx z' = VX z

+

2xyz,

z which is homogeneous (also a Bernoulli equation). Then u = X

z -= u+ 2--./u u+xu' =z'= -+ 2 � X



X

u= [ln(x)+c1] 2

and, by integration by parts,





1 1 ' = Y r,;u X

2y

z = x[ln(x) + c1] 2



⇒ VU= ln(x)+c1

⇒ y= J zdx

is the general solution.

Example 3.41. Consider the equation

-1 y" =-----2 - 1 (x + y' + 1)

Since y does not appear explicitly, let z(x) =y'(x) to obtain the first-order equation -1 (x + z +

z'=-----1,

1) 2

78

CHAPTER 3. SECOND-ORDER EQUATIONS

which has the form z' = f(x + z + 1). By Exercise 11 in the Chapter 2 Exercises on page 39, let u = x + z + l to obtain 1

uI = 1 +zI = -­ u2

⇒ ⇒

y

=

J

z dx

u

=

= (k1

J

( k1

- 3x) 1 13



z

= (k1

- 3x) 1/3 - x - l dx

- 3x) 1 13

-

x-1

= - 1 ( k1 - 3x)4/3 - 21 x2 - x + c2

4

is the general solution.

Example 3.42. Consider the equation

y'y"

= x.

Since y does not appear explicitly, let z(x) zz'

= y'(x)

to obtain the first-order equation

= x,

which is separable. Then

The value of the last integral depends upon the sign of k. If k

= 0,

then

If k > 0, let k = µ2 and make the trigonometric substitution

x

= µ tan(t),

-

1r

2

to obtain

y=±

� vt )

=1

=>

/ ty dy v )

=

J

l dx

= x + c,

yielding a second arbitrary constant and, therefore, the general solution of the second­ order equation. Example 3.43. Consider the equation y" = (y') 2 ( 2y -

t)

Since x does not appear explicitly, let v(y) = y'(x) to obtain the first-order equation vv'

= v2 ( 2y -

which is separable. One solution is v

=>

J

=>

-y2

2

ye-Y dy

=

J

t)

= 0 and gives y' = 0, i.e., y = c 1.

k 1 dx =>

= ln(k3x + k4)

=>

-1e-y

y=

= k1x + k2

± ✓- ln(k3x + k4)

is the general solution. Example 3.44. Consider the equation y" + (1

2

+ 2y)(y') 3 = 0.

If v #- 0, then

81

3.4. EQUATIONS REDUCIBLE TO FIRST-ORDER EQUATIONS Since x does not appear explicitly, let v(y) which is separable. v'

= y'(x) to obtain the first-order equation vv' + (1 + 2y)v3 = 0, One solution is v = 0 and gives y' = 0, i.e., y = c1. If v i- 0, then

-- = 1 + 2y v2



J y + y2

is the general solution.

=}

!V = y + y2 + C2

+ c 2 dy = J 1 dx

=}

Example 3.45. Consider the equation

yy" + 3y2(y')3

Since x does not appear explicitly, let v(y)

yvv' + 3y2 v3

One solution is v

= 0 and gives y' = 0,

+ v2 = 0.

= -3y2u-2,

which is linear. Its standard form is

with the integrating factor J(y) J(y) = y- 1, the equation becomes

ef

=3 !u' - -2:.._u y y2

is the general solution.

= c 1.

If vi- 0, then

= v- 1 ⇒ v = u- 1 ⇒

v'

=

-u- 2 u', and

or - yu' + u = -3y2,

= 3y,

-t dy

=

e- In IYI 1

(-y u)

1.e.,

'* j 3y2 + c2y dy = j 1 dx

+ 3Y3 + C2Y = X + C3

= y'(x) to obtain the first-order equation

i.e., y

1 uI - -u y

1

=1

+ (y')2 = 0.

is a Bernoulli equation with a= 2. Hence, u the equation for v transforms into

Hence,

1 2 2,Y

yv' + v = -3y2v2

-yu- 2 u' + u- 1

(y + y2 + C2)y'

=}



y3

1

±y- 1 and, with = 3.

+ �c2y2 = x + c3

82

CHAPTER 3. SECOND-ORDER EQUATIONS

Example 3.46. Consider the equation

y" + (y') 2 = y'.

(3.19)

Since x does not appear explicitly, let v (y) = y' (x) to obtain the first-order equation

vv' + v2 = v, which is separable. One solution is v = 0 and gives y' = 0, i.e., y = c1 . If v =/- 0, then

v' + v = l =>

=>

J



-v' - = 1 => - ln\ 1-v \ = y + c2 1-v

_ dy =

1- e Y

J

y' ---= 1 1-ke-Y

l dx

=>

j __!!____ dy = x + eY - k

c3

is the general solution.

Note that y also does not appear in Equation (3.19). It may therefore also be solved by letting z(x) = y'(x). The equation then becomes

z' + z 2

= z,

which is separable. Then

1 _z_'_ = 1 => (- + �) z' = 1 => -ln\1-z\ +lnlz\ = x+c1 2 z l - z z- z kex z z x => z = + C1 => _ _ = ke -l =x => ln 11- z 1- z kex + l

is the general solution, and is equivalent to the solution obtained earlier because

3.4. EQUATIONS REDUCIBLE TO FIRST-ORDER EQUATIONS

Exercises 3. 4 Find the general solution of the given equation.

2 y" 1. --=-+3x2 X y' -2 2. X2 yll = (y')2 3. y" - x(y')3 = 0 4. y'y" + X = 0 5. xy" = y'ln(y') - y'ln(x), x > 0

6. y" + 2y' = 2e-x .,/iJ 7. x4 + (y') 2 - xy'y" = 0 8. y" = (2x - y' + 3)2 + 2 9. y" + (2 - 6y + 12y2 )(y')3 = 0 10. y" = (y')2 tan(y) 11. yy" = y'(y' -1) ln(y' - 1)

12. sin(y)y" + cos(y)(y')2 = (y')3 13. y" = 2yy' 14. y" -(y')2 = 1 15. yT+y'y" = 1 16. eY' y" = 1 17. cos(y')y" = 1 18. y" = y' ✓1 - (y') 2

83

84

CHAPTER 3. SECOND-ORDER EQUATIONS

Chapter 3 Exercises 1. Consider the equation x2 y" . . so1ut10n 1s Y1 order.

=

+ xy' + (x2 - } ) y = 0,

x > 0. Given that one

cos(x) , find a second, independent solution y2 by reduction of VX

2. Consider the equation y" + xy' + y = 0. Given that one solution is y1 = e-x / 2 , find a second, independent solution y2 by reduction of order. Note that y2 includes an integral which cannot be evaluated in terms of elementary functions. 2

3. Consider the equation 2x2 y" + (2x2 + x)y' - y = 0, x > 0. Given that one solution is y1 = x-1/ 2 e-x, find a second, independent solution y2 by reduction of order. Note that Y2 includes an integral which cannot be evaluated in terms of elementary functions. 4. Consider the equation y" + y

= cos(x).

(a) Solve the equation by the method of undetermined coefficients. (b) Solve the equation by variation of parameters. 5. Consider the equation y" - y' - 2y

= 8x2 .

(a) Solve the equation by the method of undetermined coefficients. (b) Solve the equation by variation of parameters. 6. Consider the equation y" - y = cos(x) - 5 sin(2x). (a) Solve the equation by the method of undetermined coefficients. (b) Solve the equation by variation of parameters. 7. Consider the equation y" - 2y' - 8y = 24(x2

+ l)e4x .

(a) Solve the equation by the method of undetermined coefficients. (b) Solve the equation by variation of parameters. 8. Consider the equation a(x - x0) 2y" + b(x - x0)y' + cy are constants.

= 0,

where a, b, c and

(a) Show that the change of variables t = x - x0 and y(x) the equation into an Euler equation for z(t).

= z(t)

(b) Solve the equation 2(x - 3) 2 y" + 3(x - 3)y' + y = 0 for x-=/= 3. 9. Consider the Euler equation ax2 y" + bxy' + cy = 0, x > 0.

Xo

transforms

85

CHAPTER 3 EXERCISES

(a) Show that the change of variables t = ln(x) and y(x) = z(t) transforms the Euler equation for y(x) into an equation for z ( t) with constant coefficients.

(b) Show that the indicial equation for z = e rl is identical to the one for y = xr.

(c) Express the three cases of solution z = c 1 e rit + c2er2l , z = erl (c1 + c2t) and z = e°'t [cL cos(,6l) + c2 sin(,6t)] in terms of x to obtain the three cases of solution for y(x). 10. Consider the equation x2 y" - xy' + y

=

� , x > 0. ln x)

(a) Solve the equation by variation of parameters.

(b) Let l = ln(x) and y(x) = z(l) to transform the equation for y(x) into an equation for z(l) (see Exercise 9). (c) Solve the equation obtained in part (b) for z(t). (d) Obtain y(x) from z(t) and compare with the result of part (a). 11. Consider the equation 4x2 y" + 4xy' - y

= x2 lxl 1 1 2.

> 0. (b) Solve the equation for x < 0. (You will need to employ the fact that lxl = -x for x < 0, and that �(-x)°' = -a(-x)°'-1, by the chain rule.) (a) Solve the equation for x

d (c) Combine the solutions in parts (a) and (b) into a solution which is valid for any x i= 0.

12. Consider the equation y" + y'

= 2x.

(a) Solve the equation by the method of undetermined coefficients. (b) Solve the equation by variation of parameters. (c) Solve the equation by letting z(x) 13. Consider the equation x2y" + xy'

= l,

= y' ( x).

x > 0.

(a) Solve the equation by variation of parameters. (b) Solve the equation by letting z(x) 14. Consider the equation 3y'y"

=

= y'(x).

l.

(a) Solve the equation by letting z(x)

= y'(x).

(b) Solve the equation by letting v(y) = y'(x). 15. Suppose that the functions p and q are continuous on an interval I and let y 1 and Y2 be solutions of the equation y" + p(x)y' + q(x)y = 0.

86

CHAPTER 3. SECOND-ORDER EQUATIONS (a) Show that the Wronskian W(x) = W[y1, y2](x) = Y1Y; - y�y2 satisfies the first-order equation W'(x) + p(x)W(x) = 0. (Hint: Multiply the equation y� + p(x)y; + q(x)Y2 = 0 by Y1, y� + p(x)y� + q(x)Y1 = 0 by Y2, and subtract the latter from the former.) (b) Solve the first-order equation in part (a) for W(x). (c) Deduce from the result of part (b) that either W(x) for any x in I.

= 0 on I, or W(x) =/= 0

(d) Prove that y 1 and y2 are linearly dependent on I if and only if W (x) on I.

=0

(e) Deduce from the result of part (d) that y 1 and y2 are linearly independent on I if and only if W(x) =/= 0 on I. 16. Consider the equation x 2y" - xy' = 0 on the interval

I=

(-l, 1).

(a) Find two linearly independent solutions y 1 and Y2 • (b) Compute the Wronskian W(x). (c) Show that there is a point x0 in I with W(x0)

= 0.

(d) W hy does the result of part (c) not contradict the result of part (e) in Exercise 15?

Chapter 4 Higher-Order Linear Equations The definitions and certain of the methods of solution discussed in Chapter 3 can be extended to equations of any order n 2'. 2.

Definition 4.1. The general solution of an equation of order n 2'. 2 is the solution which contains n arbitrary constants which cannot be combined. It is an n-parameter family of solutions, i.e., a set of solutions such that every solution corresponds to particular values of the arbitrary constants. A solution in which the constants arc assigned particular values is called a particular solution. Definition 4.2. An initial-value problem for an equation of order n 2'. 2 consists of the equation, together with n initial conditions, which specify the values of I

II

y, y' y' . . . ' y(n-1) at a single point x and, thereby, determine a particular solution.

Definition 4.3. An equation of order n 2'. 2 is linear if it has the form

Otherwise, it is nonlinear. The standard form of Equation (4.1) is 11 1 y(n) + Pn-1(x)y(n-l) + · · · + P2(x)y + P1(x)y + Po(x)y = f(x).

The equation is called homogeneous if g(x) nonhomogeneous. 87

= 0 (or f(x) = 0).

(4.2)

Otherwise, it is

88

4.1

CHAPTER 4. HIGHER-ORDER LINEAR EQUATIONS

Homogeneous Equations

Theorem 4.1. If y 1, Y2, · · ·, Yn are solutions of the linear, homogeneous equation

then any linear combination

is also a solution. Theorem 4.2. The homogeneous equation (4.3) has precisely n linearly independent solutions Y1, Y2, · · · , Yn , and the general solution is

where c1, c2, · · ·, 4.1.1

Cn

are arbitrary constants.

Equations with Constant Coefficients

Consider the equation an y( n )

+ an -lY( n -l) + ... + a2y" + a1y' + aoy = 0,

(4.4)

where a,, a2, · · ·, an are constants. As in the case n Then

=

2 discussed in Chapter 3, seek solutions in the form y

= e rx.

and substitution into Equation (4.4) yields

or, upon division by erx # 0, the indicial equation P(r)

= an ,n + an -lrn-l + ... + a2r2 +air+ ao = 0.

(4.5)

In the case n = 2 discussed in Chapter 3, there are three possibilities for the roots of the polynomial P in Equation (4.5), namely, two distinct real roots, one real repeated root, or two complex-conjugate roots. In the cases n 2: 3, there are more possibilities, such as repeated complex roots, or real roots which are repeated more than once, but the forms of the solutions are analogous to the ones for n = 2, and are given by Theorems 4.3-4.5.

89

4.1. HOMOGENEOUS EQUATIONS Theorem 4.3. If P has n distinct real roots ri, 1 :S: i :S: n, i.e.,

where r i f= rj if if= j, then n independent solutions of Equation (1.4) are given by and the general solution is y

= C1Y1 + c2 y2 + · · · + CnYn·

Example 4.1. Consider the equation

4 1/1 4 /I I+ 0 y - y -y y= . P(r) = 4r3

-

4r2

-

r + 1 = 0.

Since r = 1 is a root, r -1 must be a factor. Long division of r -1 into 4r3 - 4r2 - r + 1 gives the other factor 4r2 - 1. Alternatively, 4r3 3

4r

-

-

4r2 2

4r

-

-

r+1 r+1

(r - 1)(4r2 + ar - 1) 4r3 + (a - 4)r2 - (a+ l)r + 1 (r - 1)(4r2 - 1),



a= 0



as above. Thus, P(r)

= 4r3 - 4r2 - r + 1 = (r - 1)(2r - 1)(2r + 1) = 0

is the general solution. Example 4.2. Consider the initial-value problem y"' - 4y" + y'

+ 6y = 0, y(0) = 0, y'(0) = 0, y"(0) = 12. P(r)

=

r3

-

4r2

+ r + 6 = 0.

Since r = -1 is a root, r+1 must be a factor. Long division of r+1 into r3 -4r2+r+6 gives the other factor r2 - 5r + 6. Alternatively,

r3

-

4r2

+r+6

r3

-

4r2

+ r+ 6

(r+ l)(r2 + ar +6) r 3 +(a+ l)r2 +(a+ 6)r+ 6 ⇒ a= -5 ⇒ (r + l)(r2 - 5r + 6) = (r + l)(r - 2)(r - 3)

=0

90

CHAPTER 4. HIGHER-ORDER LINEAR EQUATIONS

is the general solution · Then y' = -c1 e- x +2c2 e2x +3c3 e3x , y" = c1 e- x +4c2 e2x +9c3 e3x , and the initial conditions y(0) = 0, y'(0) = 0 and y"(0) = 12 require that

The sum of the first two equations gives 3c2 + 4c3 = 0, and the sum of the last two gives 6c2 + 12c3 = 12. The solution of the system { is c2

= -4 and c3 = 3.

3c2 + 4c3 6c2 + 12c3

= 1,

Then c1

y=

-

=

0 } 12

and

e- x -

4e2x + 3e3x

is the solution of the initial-value problem. Example 4.3. Consider the equation

y( 4) P(r)

-

5y"' + 5y" + 5y' - 6y

= o.

= r4 - 5r3 + 5r 2 + 5r - 6 = 0.

Since r = 1 and r = -1 are roots, r - 1 and r + 1 must be factors. Long division of (r - l)(r + 1) = r2 - 1 into r4 - 5r3 + 5r2 + Sr - 6 gives the other factor r2 - 5r + 6. Alternatively,

r4 4

r

-

-

5r3 5r

3

+ 5r2 + 5r - 6 + 5r + 5r - 6 2

(r2 - l)(r 2 + ar + 6) r4 + ar3 + 5r2 - ar - 6 2

(r - l)(r - 5r + 6) (r - l)(r + l)(r - 2)(r - 3) 2

⇒ r1 = 1, r2 = -1, r = 2 and r4 = 3, and y1 = e

and

3

is the general solution.



x

, Y2

= e- x ,

y3

a

=

-5



=0 = e2x

and

Y4

= e x, 3

4.1. HOMOGENEOUS EQUATIONS

91

Definition 4.4. Let P be a polynomial and suppose that P(r)= Q(r)(r - ror, where Q is a polynomial with Q(r0) =I- 0 and m 2: 1 is an integer. Then r0 is a root of P(r) and m is called the multiplicity of the root r0. For example, if P(r) = r3 - 3r + 2 = (r + 2)(r - 1)2 , then r0 = -2 is a root of multiplicity 1 and r1 = 1 is a root of multiplicity 2.

If P(r) = r4 - r3 - 3r2 + 5r - 2 = (r2 + r - 2)(r - 1)2 , then the multiplicity of the root r0 = 1 is not 2 because Q(r) = r2 + r - 2 and Q(r0) = Q(l) = 0. Since P(r)= (r+2)(r-1)3 and Q(r)= r+2 with Q(l) = 3 =/- 0, the multiplicity of r0 = 1 is m= 3. Thus, the multiplicity of a root r0 of a polynomial P is the largest power of r - r0 which occurs in the factorization of P.

Theorem 4.4. Let r0 be a root of the indicial equation (4.5) obtained by letting y = erx in Equation (4.4). If r0 has multiplicity m 2: 1, i.e., P(r) = Q(r)(r - r0)m with Q(r0) =I- 0, then m linearly independent solutions of Equation (4.4) are given by

Proof. Define the linear operator L by L[y] = any (n) + an JY (n-l) + · · · + a2y" + a1y' + aoy, i.e., L assigns to a function y the expression on the left-hand side of Equation (4.4). Thus, y is a solution of Equation (4.4) if and only if L[y] = 0. Then (an 1.n + an lrn-l + · · · + a2 r2 + a1 r + ao)e TX P(r)e TX = Q(r)(r - roreTX = 0 when r= ro, if m 2: 1, L [!!,_erx ] = !!,_L eTX ] = !!,_ Q(r)(r - rore rx l dr dr [ dr [ (r - ror!!,_ [ Q(r)erxl + m(r - ror- 1 [Q(r)erx l = 0 dr when r= r0, if m 2: 2, L [.!._erx] = .!._L[e rx l = .!._[Q(r)(r - rore rx l dr2 dr 2 dr2 d2 .1 d rx rx (r - ror-[Q(r)e l + 2m(r - ror- -[Q(r)e l 2 dr dr +m(m - l)(r - ror-2 [ Q(r)e TX ] = 0 when r = ro, if m 2: 3,

92

CHAPTER 4. HIGHER-ORDER LINEAR EQUATIONS

and, in general, L[xk e rx ] = 0 when r = ro, if m > k. Thus, for k = 0, l, · · · , m - 1, xk e rox is a solution. Note that the order of differentiation with respect to r and x has been reversed. This is permissible by Theorem 2.1 in Section 2.4.1 since the partial derivatives of all orders of e rx are continuous everywhere. By Theorem 4.4, every real root of multiplicity m yields m independent solutions of Equation (4.4). The solutions which correspond to the complex roots of P(r) are discussed in Theorem 4.5. Example 4.4. Consider the equation 111

y

P(r)

-

4 " + 5 ' - 2y y y

= 0.

= r3 - 4r2 + 5r - 2 = 0.

Since r = l is a root, r - l is a factor. Long division of r - l into r3 gives the other factor r2 - 3r + 2. Hence, r3

-

+ 5r - 2 = ( r - l)(r2 - 3r + 2) = ( r -

4r2



l) 2 (r - 2)

-

4r2

+ 5r - 2

=0

r 1 = 1 is a root of multiplicity 2 and r2 = 2 is a root of multiplicity 1. Three linearly independent solutions are therefore given by Y1 = ex , Y2 = xex and y3 = e 2x , and is the general solution. Example 4.5. Consider the initial-value problem 111

y

-

3y" + 3y' - y

y = e rx ⇒

P(r)

= 0,

y(0)

=

1, y'(0)

= 2,

y"(0)

= 9.

= r3 - 3r2 + 3r - 1 = 0.

Since r = l is a root, r - l is a factor. Long division of r - l into r3 gives the other factor r 2 - 2r + 1. Hence, r3

-

3r2 + 3r - 1 = (r - l)(r2

-

2r + 1)

= (r - 1)3 = 0

-

3r2 + 3r - 1

⇒ r1 = 1 is a root of multiplicity 3, and three linearly independent solutions are given by Y1 = ex , Y2 = xex and y3 = x2 ex , and

4.1. HOMOGENEOUS EQUATIONS

93

is the general solution. Then y'

y"

C1ex + c2 (e x + xex) + c3 (2xex

+ x2ex)

(c1 + c2)ex + (c2 + 2c3)xex + c3 x2ex , (c1 + c2)ex + (c2 + 2c3)(ex + xex ) + c3(2xex + x2ex ) (c1 + 2c2 + 2c3)ex + (c2 + 4c3 )xex + C3X2ex ,

= c1 = 1, y '(0) = c1 + c2 = 2 ⇒ c2 = 1, y"(0) = c1 + 2c2 + 2c3 = 9 ⇒ C3 = 3 ⇒ 2 x x x y = e + xe + 3x e is the solution of the initial-value problem.

y (0)

Example 4.6. Consider the equation y

( 4)

P(r)

-

4y "' + 3y " + 4y ' - 4y = 0.

= r4

-

4r3 + 3r2 + 4r - 4

= 0.

Since r = 1 and r = -1 arc roots, r - 1 and r + 1 are factors. Long division of (r - l)(r + 1) = r2 - 1 into r4 - 4r3 + 3r2 + 4r - 4 gives the other factor r2 - 4r + 4. Hence,

= (r2 - l)(r2 - 4r + 4) = (r - l)(r + l)(r - 2)2 = 0 ⇒ r 1 = 1 and r2 = -1 are roots of multiplicity 1, and r3 = 2 is a root of multiplicity 2. Four independent solutions are therefore given by y 1 = ex , y2 = e-x , y3 = e2x and y4 = xe2x , and r4

-

4r3 + 3r2 + 4r - 4

is the general solution. Example 4. 7. Consider the equation 4 ( y ) - 2y "' - 3y " + 4y ' + 4y

P(r) = r4

-

2r 3 - 3r2

= 0.

+ 4r + 4 = 0.

Since r = - l and r = 2 are roots, r + l and r - 2 are factors. Long division of (r + l)(r - 2) = r2 - r - 2 into r4 - 2r3 - 3r2 + 4r + 4 gives the other factor r2 - r -2. Hence, r4 - 2r3 - 3r2 + 4r + 4 = (r2 - r - 2)2 = (r + 1)2 (r - 2)2 = 0

⇒ r1 = -1 and r2 = 2 are roots of multiplicity 2. Four independent solutions are therefore given by Yi = e-x , Y2 = xe-x , y3 = e2x and y4 = xe2x , and Y = C1e-x + C2Xe-x + C3 e 2x + C4Xe2x is the general solution.

94

CHAPTER 4. HIGHER-ORDER LINEAR EQUATIONS

Example 4.8. Consider the equation

y (4 )

-

5y'" + 6y" + 4y' - 8y = 0.

P(r) = r4

-

5r3 + 6r2 + 4r -8 = 0.

Since r = -1 and r = 2 are roots, r + 1 and r - 2 are factors. Long division of (r + l)(r - 2) = r2 - r -2 into r4 -5r3 +6r2 + 4r -8 gives the other factor r2 -4r +4. Hence,

r4

-

5r3

+ 6r2 + 4r - 8 = (r + l)(r -2)(r2 - 4r + 4) = (r + l)(r -2)3 = 0

=> r1 = -1 is a root of multiplicity 1 and r2 = 2 is a root of multiplicity 3. Four independent solutions are therefore given by y1 = e- x , y2 = e 2x , y3 = xe2x and Y4 = x 2 e 2x, and is the general solution. Example 4.9. Consider the equation y ( 4)

P(r)

-

8y111 + 24y" - 32y' + 16y

= 0.

= r4 - 8r3 + 24r2 - 32r + 16 = 0.

Since r = 2 is a root, r-2 is a factor. Long division of r-2 into r4 -8r3 +24r2 -32r+16 gives the other factor r3 - 6r2 + 12r -8. Since r = 2 is a root of the cubic, r - 2 is a factor. Long division of r -2 into r3 - 6r2 + 12r -8 gives r2 - 4r + 4. Hence,

r4

-

8r3 + 24r2

-

32r + 16

= (r - 2)2 (r2 - 4r + 4) = (r - 2)4 = 0

=> r1 = 2 is a root of multiplicity 4. Four independent solutions are therefore given by Y1 = e2x' Y2 = xe2x' y3 = x 2e2x and Y4 = x3 e2x' and is the general solution. Note that, since no root of P(r) other than r1 = 2 could be found, it may be beneficial to determine whether or not r1 = 2 is also a root of P'(r).

P'(r)

= 4r3 - 24r2 + 48r - 32,

and r1 = 2 is a root. Since r1 = 2 is a root of both P(r) and P'(r), (r - 2)2 must be a factor of P(r), and long division needs to be performed only once in order to determine the other factor r2 - 4r + 4.

4.1. HOMOGENEOUS EQUATIONS

95

In general, if P(r) is a polynomial of degree n 2: 1 and 1 ::::; k :::; n, then (r - r0)k is a factor of P(r) if and only if r0 is a root of P(r), P'(r), ..., and p(k-ll(r). Theorem 4.5. Let r0 =a+ i/3 and r0 = a - i/3 be two complex-conjugate roots of the indicial equation (4.5) obtained by letting y = e rx in Equation (4.4). Then two real, linearly independent solutions of Equation (4.4) are given by Y1

= e °'x cos(/Jx)

and Y2

= e °'x sin(/Jx).

More generally, if r0 and r0 have multiplicity m 2: 1, i.e.,

then 2m real, linearly independent solutions of Equation (4.4) are given by e °'x cos(f]x), xe°'x cos(f3x), x 2 e °'x cos(f3x), e °'x sin(/Jx), xe °'x sin(/Jx), x 2 e °'x sin(/Jx), The remaining n - 2m independent solutions of Equation (4.4) are determined by the roots of the polynomial Q of degree n - 2m.

Proof. By the proof of Theorem 4.4, the complex-conjugate functions xk e rox and xk erox , 0 :::; k :::; m - 1, are solutions. As in the case n = 2 derived in Section 3.2.1, from every one of the m pairs of complex-conjugate solutions Zk = xk e ro x and k r x Zk = x e o , 0 :::; k :::; m - 1, two real solutions Zk

+ Zk 2

= xk e°'x cos(/3x)

and

Zk - Zk 2i

= xk e°'x sin(/Jx)

are obtained, for a total of 2m real, independent solutions. By the first part of this theorem and/or Theorem 4.4, an additional n - 2m real, independent solutions are determined by the roots of Q. Example 4.10. Consider the equation y111

P(r)

-

8y"

+ 25y' - 26y = 0.

= r3 - 8r2 + 25r - 26 = 0.

Since r = 2 is a root, ,,. - 2 is a factor. Long division of r - 2 into r3 gives the other factor r2 - 6r + 13, and

r2

-

6r + 13

=0

=}

r

= 6 ± J35 - 52 = 3 ± 2i. 2

-

8r 2 + 25r - 26

96

CHAPTER 4. HIGHER-ORDER LINEAR EQUATIONS

Hence, P has the real root r1 = 2 of multiplicity 1 and the complex-conjugate roots 3 ± 2i of multiplicity 1. Three real, independent solutions are therefore given by 3x Y1 = e 2x, Y2 = e 3x cos(2x) and y3 = e sin(2x), and y = c1e 2x

+ c2 e 3x cos(2x) + C3 e3x sin(2x)

is the general solution. Example 4.11. Consider the equation y ( 4)

-

7y"' + 27y" - 47y'

+ 26y = 0.

P(r) = r4 - 7r3 + 27r2 - 47r + 26 = 0. Since r = 1 and r = 2 are roots, (r - 1) and (r - 2) are factors. Long division of (r - l)(r - 2) = r2 - 3r + 2 into r4 - 7r3 + 27r2 - 47r + 26 gives the other factor r2 - 4r + 13, and





52 Jl6 = 2 ± 3i. 2 Hence, four real, independent solutions are given by y1 = ex , Y2 = e2x, y3 and y4 = e 2x sin(3x), and r2 - 4r + 13

y

=0

r

=

= e 2x cos(3x)

= c1ex + c2 e 2x + C3 e2x cos(3x) + C4e 2x sin(3x)

is the general solution. Example 4.12. Consider the equation 4 y ( ) - 8y"' + 26y" - 32y'

+ 13y = 0.

P(r) = r4 - 8r3 + 26r2 - 32r + 13 = 0.

Since r = 1 is a root of both P(r) and P'(r) = 4r3 - 24r2 + 52r - 32, (r - 1)2 is a factor of P(r). Long division of (r - 1)2 = r2 - 2r + 1 into r4 - 8r3 + 26r2 - 32r + 13 gives the other factor r2 - 6r + 13, and r2

-

= 6 ± J35 - 52 = 3 ± 2i. 2 multiplicity 2 and 3 ± 2i are two complex-conjugate

6r + 13

=0



r

roots Hence, r1 = 1 is a root of of multiplicity 1. Four real, independent solutions are therefore given by y1 = ex , x 3x Y2 = xe , y3 = e cos(2x) and Y4 = e3x sin(2x), and y

= c1ex + C2Xex + C3 e3x cos(2x) + C4e3x sin(2x)

is the general solution.

4.1. HOMOGENEOUS EQUATIONS

97

Example 4.13. Consider the equation 4 y( ) y

=e ⇒

+ y" + y = 0.

rx

P(r) = r4

+ r 2 + 1 = 0.

Since no real roots are immediately evident, write

r4 + r 2 + 1

(r 2 + ar + l)(r 2 + br + 1) r4 +(a+ b)r3 +(ab+ 2)r 2 +(a+ b)r + 1

and solve for a and b to obtain a= ±1 and b

r4 Choosing a

=

-1 and b

=

= =fl.

Choosing a= 1 and b

+ r 2 + 1 = (r2 + r + l)(r2 - r +

1).

1 gives the same result.

r2

+ r + 1 = 0 => r= -l±Jf=4 =-!±i \/'3 2 2 2 '

r2

-

r+1

= -1 gives

l±vT=°4 2

1 2

and

.v'3

= 0 => 1· = --- - = -±i-. 2

Hence, P(r) has two pairs of complex-conjugate roots of multiplicity 1. Four real, independent solutions are therefore given by y 1 = e-lx cos

and the general solution is y

( '7x)

= C1Y1 + C2Y2 + c3y3 + C4y4.

Example 4.14. Consider the equation y ( 4)

P(r)

-

4y111

+ 14y" - 20y 1 + 25y = 0.

= r4 - 4r3 + 14r2 - 20r + 25 = 0.

Since no real roots are immediately evident,

r4

-

4r3

+ 14r 2 - 20r + 25

(r 2

+ ar + 5)(r2 + br + 5)

r4 +(a+ b)r3 +(ab+ 10)r 2

+ 5(a + b)r + 25

98

CHAPTER 4. HIGHER-ORDER LINEAR EQUATIONS

⇒ a+b = -4 and ab+ 10 = 14 ⇒ b = -4- a ⇒ -a2 - 4a = 4 ⇒ a2 + 4a + 4 = 0 ⇒ (a+ 2)2 = 0 ⇒ a= -2 ⇒ b = -2 ⇒ r4

-

4r3 + 14r2

and

r 2- 2r + 5

-

20r + 25

= (r2 - 2r + 5)2,

2 ± ✓4- 20 = 0 ⇒ r = ----= l ± 2,;"· 2

Hence, 1 ± 2i are complex-conjugate roots of multiplicity 2. Four real, independent solutions are therefore given by Y1

= ex cos(2x),

Y2

= xex cos(2x),

y3

= ex sin(2x),

y4

= xex sin(2x),

and the general solution is y = c 1 y 1 +c2y2+c3y3+c4 y4. Note that, if P(r) is expressed as r4

-

4r3

+ 14r2 - 20r + 25

then the equations a+b lead to a contradiction. 4.1.2

(r2

+ ar + l)(r2 + br + 25)

r4 +(a+ b)r3 +(ab+ 26)r2

= -4, 25a + b = -20 and ab+26 =

+ (25a + b)r + 25,

14 are inconsistent and

Cauchy-Euler Equations

Definition 4.5. An equation of the form an xn y ( n )

+ an-1X n-l y ( n- l) + · · · + a2x2y " + a1xy ' + aoy = 0, n 2: 2,

(4.6)

where ai, 0 :::; i :::; n, are constants, is called a Cauchy-Euler or Euler equation. Definition 4.5 is a generalization of Definition 3. 7 given in Section 3.2.2 for n = 2 to equations of order n 2: 2. The solutions are analogous to the ones obtained therein for n = 2, and are given by the following:

Theorem 4.6. Let P(r) = 0 denote the indicial equation which results from the substitution of y = xr into Equation (4.6) for x > 0. If r = r0 is a real root of multiplicity m 2: 1, then m independent solutions of Equation 4.6 for x > 0 are given by xr0 , xr0 ln(x), xr0 [ln(x)]2, •··, xr0 [ln(x)J m -l_ If a± i/3 are complex-conjugate roots of multiplicity m 2: 1, then 2m independent solutions of Equation 4.6 for x > 0 are given by xa cos[/3 ln(x)], xa ln(x) cos[/3 ln(x)],

x0 [ln(x)] m -l cos[/3 ln(x)],

xa sin[/3 ln(x)], xa ln(x) sin[/3 ln(x)],

x0 [ln(x)]m-t sin[/3 ln(x)] .

The solutions for x =/- 0 are obtained from the ones for x > 0 by replacing x by lxl.

4.1. HOMOGENEOUS EQUATIONS

99

Proof. For x > 0, make the change of variables l = ln(x) and y(x) = z(l). Then, by the chain rule, dy dx d2 y dx2 d3 y dx3

,

dz dl l dz -- = -- ⇒ xy (x) = z'(t), dl dx x dt l dz 1 d2 z l d2 z dz [ = ⇒ x2 y"(x) = z"(l) z'(t), x2 dt2 x2 dl x2 dl2 dt l d3 z dz d2 z 2 d2 z dz 1 d3 z d2 z [ [ ] [ ] 2 ] = 3 + 2 3 3 3 2 2 3 3 x dt dt x dt dt dl dt x dl 111 3 111 = ⇒ x y (x) z (t) - 3z"(l) + 2z'(t),

]

-

(

etc., i.e., every term xk y( k) x) in the Euler equation transforms into an expression with constant coefficients. Thus, the Euler equation for y(x) transforms into an equation for z(t) with constant coefficients. Moreover, the indicial equation P(r) = 0 for y(x) is identical to the one for z(t) because z(t) = e rl if and only if y(x) = e r In(x) = xr . The forms of the solutions for y(x) then follow from the corresponding forms of the solutions for z(l) as given by Theorems 4.4 and 4.5 in Section 4.1.1. Replacing x by lxl then gives the solutions for both x > 0 and x < 0. Example 4.15. Consider the Euler equation

For x

> 0, y = xr

⇒ P(r)

r(r - l)(r - 2) + 3r(r - 1) - 2r + 2 r ( r - 1) ( r - 2) + 3r ( r - 1) - 2 (r - 1) (r - 1) [r ( r - 2) + 3r - 2]

1) ( r2 + r - 2) (r - 1) 2 (r + 2) = 0 (r -

⇒ r1 = 1 is a root of multiplicity 2 and r2 = -2 is a root of multiplicity 1. Hence, Yi= x, Y2 = x ln(x) and y3 = x- 2 for x > 0, and is the general solution for x =/ 0. Example 4.16. Consider the Euler equation 111 -

x3 y

4x2 y" + l3xy'

- l3y = 0.

100 For x

CHAPTER 4. HIGHER-ORDER LINEAR EQUATIONS

> 0, y = xr

⇒ P(r)

r(r - l)(r - 2) - 4r(r r(r - l)(r - 2) - 4r(r -

1) + 13r - 13 1) + 13(r - 1) 13]

(r - l)[r(r - 2) - 4r + ( r - 1)(r2 - 6r + 13) = 0

2 r - 6r + 13

Hence, Y1 = x, Yz

=0 ⇒

= x3 cos[2 ln(x)] y

r

and y3

=

6 ± J36 - 52 2

= x3 sin[2 ln(x)]

= 3 ± 2i. for x

> 0, and

= c1x + c2 x3 cos (2 ln lxl) + c3x3 sin(2 ln lxl)

is the general solution for x # 0. Example 4.17. Consider the initial-value problem x3 y'" - x2 y" - 2xy' For x

> 0, y = xr

+ 6y = 0, y(l) = 0, y'(l) = 1, y"(l) = 16.

⇒ r(r

P ( r)

- 1)(r - 2) - r ( r - 1) - 2r + 6 2 r(r - l)(r - 2) - (r + r - 6) r(r - l)(r - 2) - (r + 3)(r - 2) (r - 2)[r(r - 1) - (r + 3)]

(r - 2)(r2 - 2r - 3)

(r - 2)(r - 3)(r + 1) = 0

⇒ y = c1x2 + c2 x3 + c3x-1 is the general solution for x # 0. y' y" and y(l)

= 0,

y'(l)

=

=

1 and y"(l) y ( 1) y'(l) y"(l)

Then

2c1x + 3c2 x2 - c3 x-2 , 2c1 + 6c2 X + 2C3X-3 '

=

16



+ C2 + C3 = 0, 2c1 + 3c2 - c3 = 1, 2c 1 + 6c2 + 2c3 = 16.

Cl

Subtracting twice the first equation from third gives c2 = 4. Adding the first two gives 3c1 + 4c2 = 1 ⇒ c1 = -5, and the first equation then gives c3 = 1. Hence, y = -5x2 + 4x3 + x-1 is the solution of the initial-value problem.

4.1. HOMOGENEOUS EQUATIONS

101

Example 4.18. Consider the Euler equation

x4y ( 4) For x > 0, y = x r



-

x3 y"' + 4x2 y" - 8xy' + 8y

= 0.

= = =

r(r - l)(r - 2)(r - 3) - r(r - l)(r - 2) + 4r(r - 1) - 8r + 8 r(r - l)(r - 2)(r - 3) - r(r - l)(r - 2) + 4r(r - 1) - 8(r - 1) (r - l)[r(r - 2)(r - 3) - r(r - 2) + 4r - 8] = (r - l)[r(r - 2)(r - 3) - r(r - 2) + 4(r - 2)]

P(r)

= (r - l)(r - 2)[r(r - 3) - r + 4] = (r - l)(r - 2)(r2 - 4r + 4) (r - l)(r - 2)3 = 0

Example 4.19. Consider the Euler equation

x4y (4) For x > 0, y

r(r - l)(r - 2)(r - 3) + 7r(r - l)(r - 2) + 13r(r - 1) - 26r + 26 r(r - l)(r - 2)(r - 3) + 7r(r - l)(r - 2) + 13r(r - 1) - 26(r - 1) = (r - l)[r(r - 2)(r - 3) + 7r(r - 2) + 13r - 26]

P(r)

=

Hence, y

= xr ⇒

+ 7x3 y"' + 13x 2 y" - 26xy' + 26y = 0.

(r - l)[r(r - 2)(r - 3) + 7r(r - 2) + 13(r - 2)] (r - l)(r - 2)[r(r - 3) + 7r + 13] (r - l)(r - 2)(r2 + 4r + 13) = 0

-4 ± ✓16 - 52 r2 + 4r + 13 = 0 ⇒ r = ----- = -2 ± 3i. 2

= c1x + c2x2 + c3x-2 cos(3 ln !xi) + c4x- 2 sin(3 ln lxl)

for X-=/ 0.

is the general solution

Exercises 4.1 1. F ind the general solution of the equation y"' - 6y" + lly' - 6y = 0. 2. Solve the initial-value problem y"' - 2y" - y' y"(0) = 0.

+ 2y =

0, y(0)

=

6, y'(0)

=

0,

102

CHAPTER 4. HIGHER-ORDER LINEAR EQUATIONS

9 111 3. Find the general solution of the equation y 4. Find the general solution of the equation y111

+ 9y" - y' - y = 0.

-

5y" + 8y' - 4y

= 0.

5. Find the general solution of the equation y"' - 6y" + l2y' - 8y 6. Solve the initial-value problem y"' y"(0) = 2.

+ y"

- y' - y = 0, y(0) = 0, y'(0)

7. Find the general solution of the equation y"' - 5y" 8. Find the general solution of the equation y ( 4) 9. Solve the initial-value problem y( 4) y"(0) = 0, y'"(0) = 8.

-

= 0.

-

3,

+ l 7y' - l3y = 0.

5y" + 4y

2y'" - y" + 2y'

= 0.

= 0,

y(0)

= 0,

y'(0)

= 2,

+ 6y = 0.

10. Find the general solution of the equation y ( 4)

-

7y111 + l7y" - l7y'

11. Find the general solution of the equation y(4)

-

y"' - 3y" + 5y' - 2y

12. Find the general solution of the equation y( 4)

-

4y111 + 6y" - 4y'

13. Find the general solution of the equation y( 4)

-

6y111 + l3y" - 12y'

= 0.

+ y = 0. + 4y = 0.

14. Find the general solution of the equation y( 4) + 2y"' + 4y" - 2y' - 5y = 0. 15. Find the general solution of the equation y( 4) + lOy"' + 4ly" + 76y' + 52y = 0. 16. Find the general solution of the equation y ( 4) + 5y" + 4y = 0. 17. Find the general solution of the equation y ( 4)

+ 4y = 0.

18. Find the general solution of the equation y( 4)

-

8y"' + 26y" - 40y'

19. Find the general solution of the equation x3 y"' + x2 y" - 2xy' + 2y

+ 25y = 0.

= 0.

20. Find the general solution of the equation x3 y"' - 2x2 y" + 3xy' - 3y = 0. 21. Find the general solution of the equation x3y"' + xy' - y = 0. 22. Solve the initial-value problem x3 y"' + 8x2 y" + l3xy' - l3y = 0, x > 0, y(l) = 0, y'(l) = 6, y"(l) = -2.

23. Find the general solution of the equation x4 y( 4 ) + x3 y"' + x2 y" - 2xy' + 2y = 0. 24. Find the general solution of the equation x4 y(4 ) - x3y"' + 5x2 y" - lOxy' + lOy = 0. 25. Find the general solution of the equation x4y( 4) + 6x3 y111 + 6x2 y" - 2xy' + 2y = 0. 26. Find the general solution of the equation x4y( 4) + 2x3 y"' + 3x2 y" - 3xy' + 4y = 0.

103

4.2. NONHOMOGENEOUS EQUATIONS

4.2

Nonhomogeneous Equations

Theorem 4. 7. The general solution of the linear, nonhomogeneous equation of order n 2'. 2,

is y = YP + Yh, where YP is any particular solution, and Yh = C1Y1 is the general solution of the associated homogeneous equation

+ C2Y2 + · · · + C,,,iYn

where Yk, 1 ::; k ::; n, are n linearly independent solutions. Theorem 4.7 is a generalization of Theorem 3.3 given in Section 3.3 for n = 2, and the proof is similar. The general solution of Equation (4.8) is obtained by the methods discussed in Sections 4.1.1 and 4.1.2 for equations with constant coefficients and Cauchy-Euler equations. A particular solution of Equation (4.7) is obtained by variation of parameters or, where applicable, the method of undetermined coefficients, suitably generalized to equations of order n 2'. 2.

4.2.1

Variation of Parameters

Definition 4.6. The Wronskian of n solutions y 1, Y2, · · ·, Yn of Equation (4.8) is

W(x)

=

W[y1, Y2, · · · , Yn ](x)

=

Y1 y�

Y2 y;

(n-l nY1( l ) Y2 )

Yn y� n-l )

Yn(

Theorem 4.8. Consider the standard form of Equation (4.7), y(

+ Pn-1(x)y (n-l ) + · · · + P2(x)y" + P1(x)y' + Po(x)y = f(x),

n)

(4.9)

and the associated homogeneous equation n)

y(

+ Pn-1 (x)y (n-l) + · · · + P2(x)y" + P1(x)y' + Po(x)y = 0.

=

(4.10)

If Pk, 0 ::; k ::; n - 1, are continuous on an interval 1, and Yk, 1 ::; k ::; n, are solutions of Equation (4.10), then either W 0 on I or W(x) -=I= 0 for any x in J. The solutions Yi, Y2, · · ·, Yn are linearly independent on I if and only if W -=I= 0 on I.

104

CHAPTER 4. HIGHER-ORDER LINEAR EQUATIONS

The method of variation of parameters for equations of order n � 2 is given by the theorem which follows, and is an extension of the one for n = 2. n be n linearly independent solutions of Equation 2, · · · , Y Theorem 4.9. L et y1, y (4.10) and let W(x) be the Wronskian. Let �j (x), 1 :S j :Sn, denote the Wronskian with its j th column replaced by the column vector [0, 0, • • • , 0, J(t)J t , i.e.,

0 0 (n-1)

Y1

(n-1)

Y2

J(t)

(n-1)

n Y

9 Then a particular solution Y P of Equation (4 . ) is given by where

Example 4.20. Consider the nonhomogeneous equation II/ / 2y= 27e -X . y - 3y-

For the associated homogeneous equation yII/ - 3y/ - 2y= 0 ' y = erx ⇒ P(r) = r3 - 3r - 2 = (r + l)(r 2 - r - 2) = (r + l)2 (r - 2) = 0 ⇒ Y1 = e-x , 2x y2 = xe- x and y are three independent solutions. W ith J(x) = 27e- x , 3 = e

W(x) =

e- x xe-x e2x x x x e - xe -e 2e2x x x e - -2e- + xe-x 4e 2X x e- (sex - 6xex ) - xe-x (-6ex)

+ e2x ( e- 2x) = 9,

0 xe- x e 2x e-x - xe-x 0 2e2x = 27e- x (3xex - ex ) = 27(3x - 1), x x x 27e- -2e- + xe- 4e2x e- x e 2x 0 x -e 0 2e 2x = -27e-x (3ex ) = -81, x x e- 27e- 4e 2x e- x xe- x 0 x x e - xe- x -e = 27e-x ( e-2x) = 27e-3x . 0 x x e- -2e- + xe-x 27e- x

4.2. NONHOMOGENEOUS EQUATIONS

J J J

Then u1 u2 U3

and

(

YP

or YP = -;x2 e- x , since -3xe-x Hence, Y

=

-9dx

3e- 3x dx

-

--x2 e-x

2

= ;x2 - 3x,

3(3x - 1) dx

;x2 - 3x) e-x

9

105

-9x,

= -e-3x ,

-

9x(xe-x )

3xe- x

- (

e-3x e2x )

e- x

e- x is a solution of the homogeneous equation.

= --9 x2 e-x + C1e-x + C2 Xe-x + C3e2x

2 is the general solution of the nonhomogeneous equation. Example 4.21. Consider the initial-value problem y"'

- y' =

8xex , y(0)

= 0,

y' (0)

= 0,

y" (0)

= 0.

For the associated homogeneous equation

-



y"' -y'

= 0,

rx 1) = r(r - l)(r + 1) = 0 P(r) = r3 - r = r(r2 y=e x x y3 = e- are three independent solutions. With J(x) = 8xe ,

W(x)

=

1 ex

e-x -e-x e-x

= 0 ex x 0 e 0 ex e -x x 0 e -e-x x

x

8xe

1

0

e

0 0

0 8xex 1 ex

e

-x

e-x -e-x e-x

0 0 ex 0 0 ex 8xex

⇒ Y1 = 1, Y2 = e

2, = -16xex , = 8x,

= 8xe2x .

x

and

106

CHAPTER 4. HIGHER-ORDER LINEAR EQUATIONS

Then

J J J

U1

u2

U3

-8xexdx 4xdx

= -8xex + 8ex,

= 2x2 ,

4xe 2xdx = 2xe 2x - e 2X ,

and or YP

= 2x2 e x - 6xex,

since 7ex is a solution of the homogeneous equation. Hence,

is the general solution of the nonhomogeneous equation. Then 2x2 e x - 2xex - 6ex + C2 ex - C3e- x , 2x2 ex + 2xe x - 8e x + C ex + C3e-x,

y' y"

and y (0)

= 0,

y '(0)

2

= 0 and y"(0) = 0 ⇒

⇒ c1 = -8, c2 = 7,

c3 = 1, and the initial-value problem.

= 2x2 ex - 6xe x - 8 + 7ex + e-x

y

is the solution of

Example 4.22. Consider the nonhomogeneous equation y

111

-

3y " + 7y ' - 5y

= 32e x sin(2x).

For the associated homogeneous equation y

= e rx ⇒ P(r) = r3 2±F1] r = --- = l ± 2i y

2

-

3r2

⇒ y1

111

-

3 y" + 7 y ' - 5y

= 0,

+ 7r - 5 = (r - l)(r 2

=

ex, Y2

=

-

2r

+ 5) = 0

e x cos (2x) and y3

=



r1

=

1 and

e x sin(2x) are three

4.2. NONHOMOGENEOUS EQUATIONS independent solutions. With J(x)

=

107

32ex sin(2x),

e x sin(2x) ex e x cos(2x) x x x x e e cos(2x) - 2e sin(2x) e sin(2x) + 2ex cos(2x) x x x e -3e cos(2x) - 4e sin(2x) 4ex cos(2x) - 3ex sin(2x) e x (10 e 2x ) - e x (4e 2x ) + e x (2e 2x ) = 8e 3x,

W(x)

0 0 32ex sin(2x)

�coo(2x) x e cos(2x) - 2ex sin(2x) -3ex cos(2x) - 4ex sin(2x)

32ex sin(2x) ( 2e 2x ) ex ex ex

O O 32ex sin(2x)

=

64e3x sin(2x),

e x sin(2x) e sin(2x) + 2ex cos(2x) 4ex cos(2x) - 3e x sin(2x) x

=

-32ex sin(2.1:) [2e 2x cos(2x)]

-64e 3x sin(2x) cos(2x),

ex e x cos(2x) x x e e cos(2x) - 2ex sin(2x) e x -3ex cos(2x) - 4ex sin(2x) 32ex sin(2x) [-2e 2x sin(2x)] Then u1 u2 u3

J J J

e x sin(2x) e x sin(2x) + 2ex cos(2x) 4ex cos(2x) - 3ex sin(2x)

0 0 x 32e sin(2x)

= -64e 3x sin 2(2x).

8sin(2x)dx = -4cos(2x), -8sin(2x) cos(2x)dx = -4 -8sin 2(2x)dx = 4

J

J

sin(4x)dx = cos(4x),

cos(4x) - 1 dx = sin(4x) - 4x,

and YP

-4ex cos(2x) + ex cos(2x) cos(4x) + [sin(4x) - 4x]ex sin(2x) -4ex cos(2x) - 4xex sin(2x) + e x [cos(2x) cos(4x) + sin(2x) sin(4x)] -4ex cos(2x) - 4xe x sin(2x) + ex cos(2x) -3ex cos(2x) - 4xe x sin(2x),

or YP = -4xex sin(2x), since -3ex cos(2x) is a solution of the homogeneous equation. Hence, is the general solution of the nonhomogeneous equation.

108

CHAPTER 4. HIGHER-ORDER LINEAR EQUATIONS

Example 4.23. Consider the nonhomogeneous equation y(4) - 6y111 + 1ly" - 6y' = 36e3x .

For the associated homogeneous equation y(4) - 6y"' + 1ly" - 6y' y = e rx =} P(r) = r4 - 6r3 + 1lr 2 - 6r = 0 =} r(r3

Y1 = 1, Y2 Wronskian =}

=

e

x,

-

y3

6r 2

=e

+ 1lr - 6)

2x

and y4

W (x)

=

= 0,

l)(r2 - 5r + 6) r(r - l)(r - 2)(r - 3) r(r -

0

e

3x

are four independent solutions, with the

e3x l ex e2x x 2 2X 3e3X e 0 e 0 ex 4e2x 9e3x 0 ex 8e2X 27e3X

=

Computing the determinant along the fourth row, we obtain 1 e2x e3x 1 ex e3x x 2x 2x 3x 3e - 8e W(x) = e O 2 e O ex 3e3x + 27e3 0 4 e2x 9e3x O ex 9e3x ex (6e5 ) - 8e2 (6e4 ) + 27e3 (2e3x ) = 12e6 _

l ex e2x 0 e x 2 e 2x 0 ex 4 e2x

x

x

With f(x)

x

x

x

= 36e3x ,

ex e2x e3x ex 2e 2x 3e3X = ex 4e 2x 9e 3x f(x) ex 8e2x 27e3x - f(x) [ex (6e5x ) - ex (5e5x )

0 0 0

1 0

0 0 fl 3 (x)

x

0

J(x)

1 ex 0 ex 0 ex 0 ex

1 0 0 0

e2x

0 0

2 e2x 4 e2x 8e 2x 0 0 0

J(x)

e3x 3e 3x 9e3x 27e 3X

- f(x) + ex ( e5

= f(x)

x

)]

=

-2 e6x f(x)

1 e2x e3x 0 2e2 3e3x 0 4 e2 9e3x x

x

e3x 3e3x 9e3x 27e3x

x e e2 0 x 2 2x e e 0 e x 4 e 2x 0 ex 8e2x f(x) "'

e x e 2x e3x ex 2 e2x 3e3x 9e3x

=

-72 e9x ,

= 6e5x f(x) = 216e = -6e4x f(x) =

=

1 ex e2x f(x) 0 ex 2 e2x 0 e x 4 e 2x

8x

,

-216e7x ,

= 2e3x f(x) = 72e6

x

_

109

4.2. NONHOMOGENEOUS EQUATIONS Then

J

U1

-6

u2

18

U3

-18

U4

6

e3x dx

je j

J

2x

dx

=

-2e3x ,

= 9e 2x ,

ex dx = -18e x,

1 dx

= 6x.

Hence, YP = -2e3X + 9e2X ex - 18ex e2X + 6xe3x = -1le3x + 6xe3x, or YP -lle3x is a solution of the homogeneous equation. Thus,

= 6xe3x , since

is the general solution of the nonhomogeneous equation. Example 4.24. Consider the nonhomogencous equation 111 -

x3 y

3x2 y"

+ 6xy' - 6y = 8x

3

ln(x), x > 0.

The associated homogeneous equation is the Euler equation 111 -

x3 y

for which y

=

xr

=}

P(r)

= r(r -l)(r -

(r - l)[r(r - 2) - 3r + 6]

⇒ Y1 = x, Y2 = x2 and y = x 3

W(x)

�1(x)

=

3x 2 y"

+ 6xy' - 6y = 0, 2) - 3r(r - 1)

= (r - l)(r 2

3

- 5r + 6)

= (r -

x2 x3 1 2x 3x2 = x (6x2 ) - 4x3 0 2 6x x2 x 3 0

f(x)

0 0

f(x)

x3

3x2

l)(r - 2)(r - 3)

6x

= - f(x)

[x4 ]

=

2x3,

= 8x4 ln(x),

[2x3]

=

-16x3 ln(x),

2

0 x 1 2x 0 0 2 f(x)

X

= f(x)

2x 3x2 2 6x

0

1

=}

are three independent solutions. With f(x)

x

X

+ 6r - 6 = 0

=

f(x) [x2 ]

= 8x2 ln(x).

=

0

= 8ln(x),

110

CHAPTER 4. HIGHER-ORDER LINEAR EQUATIONS

Then

u1

41 xln(x)dx=2x ln(x)-x

u2

-8

u3

4

2

j j

2

,

ln(x) dx = -8x ln(x)+ 8x,

ln x) � dx = 2[ln(x)] 2 .

Hence, [2x2 ln(x) - x2 ] x+ [-8xln(x)+8x] x2 +2[ln(x)] 2 x3 2 3 -6x3 ln(x) + 7x3 +2x [ln(x)] ,

YP

or YP = 2x3 [ln(x)] 2 - 6x3 ln(x), since 7x 3 is a solution of the homogeneous equation. Thus,

is the general solution of the nonhomogeneous equation for x > 0.

Example 4.25. Consider the nonhomogeneous equation

x3 y'" + 2xy' - 2y = x, x > 0. The associated homogeneous equation is the Euler equation

x3 y"' + 2xy' - 2y = 0, for which y

=

xr

=}

P(r)

=

r(r - l)(r - 2) +2r - 2

(r - l)[r(r - 2)

+ 2]

⇒r1 =1and r2 -2r+2 = 0⇒r =

=

0

=}

= (r - l)(r2 - 2r + 2) 2± \1-4 2

=

l±i⇒y 1

=

=

0 x,y2

=

xcos [ln(x)]

111

4.2. NONHOMOGENEOUS EQUATIONS and y3

= xsin[ln(x)] are three independent solutions.

W(x)

X

1

= 2, X

xsin[ln(x)] xcos[ln(x)] sin[ln(x)] + cos[ln(x)] cos[ln(x)] - sin[ln(x)] sin[ln(x)] - cos[ln(x)] ¾ cos[ln(x)] - ¾ sin[ln(x)]

x 1 0

=

With f(x)



[�] -

¾

1[1]

= 1,

xsin[ln(x)] xcos[ln(x)] sin[ln(x)] + cos[ln(x)] cos[ln(x)] - sin[ln(x)] � cos[ln(x)] - � sin[ln(x)] -� sin[ln(x)] � cos[ln(x)] (x) f 1 0 0

J(x)[x] = -, X

xsin[ln(x)] x O 1 0 sin[ln(x)] + cos[ln(x)] 0 J(x) ¾ cos[ln(x)] - ¾ sin[ln(x)] cos[ln(x)] - J(x){x cos[ln(x)]} = ----,

6._3(X)

X

0 xcos[ln(x)] cos[ln(x)] - sin[ln(x)] 0 cos[ln(x)] J(x) sin[ln(x)] - ¾ sin[ln(x)]. f(x){-xsin[ln(x)]} = X x 1 0



Then u1 u2

u3

.

/ 1 dx

-/ .

- ./

= ln(x),

cos[l (x)] dx : sin[l (x)] dx :

= - sin[ln(x)], = cos[ln(x)].

Hence, yP = xln(x) - xcos[ln(x)]sin[ln(x)] +xsin[ln(x)]cos[ln(x)] = xln(x), and y

= x ln(x) + c1x + c2xcos[ln(x)] + C3X sin[ln(x)]

is the general solution of the nonhomogeneous equation for x > 0.

4. 2. 2

Undetermined Coefficients-Annihilators

In Section 3.3.1, the method of undetermined coefficients was employed to determine particular solutions YP of second-order nonhomogeneous equations . The same method can be applied to the higher-order equations considered in the present chapter, with

112

CHAPTER 4. HIGHER-ORDER LINEAR EQUATIONS

a slight difference. Instead of listing the form of YP for every type of nonhomogeneous term g(x), the form of YP will be determined by g(x). The unknown constants in YP are then determined by substitution into the nonhomogeneous equation, as before. Definition 4. 7. Let D denote the linear operator defined by DJ= n[J] where f is a differentiable function. Let D2 f

= n 2 [J] =

=

d�

= J,

f

'

n2 denote the linear operator defined by

D[n[f ]]

=

n[J']

= f",

and, more generally, for any integer n 2': 1,

the derivative of f of order n. If n = 0, n ° f is defined by n° f = f. The abuse of notation n[J(x)] is customary, and denotes the value of the function nf = f' at x, i.e.' n[J(x)] = (nf)(x) = J'(x), and, more generally,

Let P(r) be a polynomial of degree n 2': 0, i.e.,

where ak, 0 � k � n, are constants. The linear operator P(n) is defined by P(n)J

° 2 n (an nn + an -ln -l + · · · + a2 n + a1n + aon ) f an nn f + an -l nn-l f + · · · + a2 n2 f + a1nf + aon° f l anf( n ) + an _if ( n- ) + · · · + a2J" + aif ' + aof.

If P(n)J(x) = 0 for all x, then P(n) is called an annihilator of the function the function f is said to be annihilated by P(n). Example 4.26. Let P(r) = r - 1. Then P(n) any differentiable function f. If f ( x) = ex , then Thus,

= n - 1,

n - 1 is an annihilator of the function e . x

and (n - l)f

J, and

= f' - f for

4.2. NONHOMOGENEOUS EQUATIONS

113

Example 4.27. Let P(r) = r2 + 4. Then P(D) = D2 + 4, and (D2 + 4)J for any differentiable function f. If f(x) = sin(2x), then P(D)[sin(2x)]

= f" + 4J

= (D 2 +4) sin(2x) = D2 sin(2x)+4 sin(2x) = -4 sin(2x)+4 sin(2x) = 0.

Thus, D2 + 1 is an annihilator of the function sin(2x). Example 4.28. Let

f(x) = cos(3x).

Df(x) = -3 sin(3x)

Hence, (D2 + 9) cos(3x) f(x) = cos(3x).

=

Then

D2 f(x) = -9cos(3x) = -9f(x). P(D) = D2 + 9 is an annihilator of the

and

0, i.e.,

function

Example 4.29. Let f(x) = e3x sin(2x). How can an annihilator off be determined? Recall that e 3x cos(2x) and e3x sin(2x) are solutions of a second-order homogeneous equation for which the indicial equation has the roots r = 3 ± 2i. Then

r - 3 = ±2i

=>

(r - 3)2 = -4

P(r) = r2 - 6r + 13 = 0

=>

is the indicial equation. Hence, e3x cos(2x) and e 3x sin(2x) are the solutions of the equation y" - 6y' + 13y = 0, i.e., they are annihilated by P(D) = D2 - 6D + 13. Example 4.30. Let J(x) = x2 e2x . Recall that e 2x , xe 2x and x2 e 2x are solutions of a third-order homogeneous equation for which the indicial equation has the root r = 2 of multiplicity 3, i.e., the indicial equation is P(r) = (r-2)3. Hence, P(D) = (D-2)3 is an annihilator of x2 e 2x , as well as e 2x and xe 2x . Consider a nonhomogeneous equation of order n � 2 with constant coefficients, n n an y ( ) + an -lY ( -J) + · · · + a2y" + a1y' + aoy = g(x),

where the nonhomogeneous term is denoted by in standard form. With

g(x)

(4.11)

since the equation need not be

Equation (4.11) can be expressed as

P(D)y = g(x).

If Q ( D) is an annihilator of g ( x), then

P(D)y = g(x)

Q(D )P(D)y = Q(D)g(x) = 0, i.e., if y is a solution of the nonhomogeneous equation P(D )y = g(x), then it is a solution of the homogeneous equation Q(D)P(D)y = 0. From the solutions of the =>

latter equation, the form of yP can be deduced by means of the following:

114

CHAPTER 4. HIGHER-ORDER LINEAR EQUATIONS

Theorem 4.10. Consider the nonhomogeneous equation (4.11), and let Q be a polynomial of the least degree m � 1 such that Q(D) is an annihilator of g. Let

be n+m independent solutions of the equation Q(D)P(D)y = 0 of order n+m, where y 1, y2, · · · , Yn are independent solutions of P(D)y = 0. Then there exist constants c1, c2 , · · ·, Cm such that (4.12) Equation (4.12) gives the form of Yp, and the constants Ci, 1 � i � m, are determined by substitution of yP into Equation (4.11). Proof. Since Q(D)P(D)zi = 0, 1 � i � m, P(D)zi are solutions of Q(D)y = 0. They arc linearly independent because if d1P(D)z1 + d2P(D)z2 + · · · + dmP(D)zm = 0, then P(D)(d1z1 + d2 z2 + · · · + dm zm ) = 0, i.e., z = d1z1 + d2 z2 + · · · + dmZm is a solution of P(D)y = 0. Hence, there exist constants b1, b2 , · · ·, bn such that

Then



· · · + dmzm = b1Y1 + b2Y2 + · · · + bn Yn d1z1 + d2z2 + · · · + dm zm - b1 Y1 - b2 y2 - · · · - bnYn = 0 d1z1 + d2z2 +

⇒ di = 0, bj = 0, 1 � i � m, 1 � j � n, smce are linearly independent. Thus, P(D)zi, 1 � i � m, are m linearly independent solutions of Q(D)y = 0. Since g(x) is a solution of Q(D)y = 0, there exist constants c1, c2, · · · , Cm such that

g(x) which shows that YP

c1P(D)z1 + c2P(D)z2 +

· · · + em P(D)zm P(D )(c1z1 + C2Z2 + · · · + CmZm ),

= c1z1 + c2 z2 + · · · + CmZm satisfies P(D)yp = g(x).

Example 4.31. Consider the nonhomogeneous equation y

111 -

3y' - 2y

= 27e-x,

which has been solved by variation of parameters in Section 4.2.1, Example 4.20. The indicial polynomial is P(r) = (r + 1)2 (r - 2). Hence, the nonhomogeneous equation lS

P(D )y

= (D + 1)2 (D - 2)y = 27e-x.

4.2. NONHOMOGENEOUS EQUATIONS

115

The annihilator of g(x) = 27e-x of minimal degree is

Q(D) = D + l.

Thus,

Q(D)P(D)y = (D + 1)3 (D - 2)y = (D + 1)[27e-x ] = 0 Since e-x , xe-x and e2x are solutions of the homogeneous equation YP = C3x 2 e-x, or YP = cx2 e- x , where c is to be determined. Then y�

c[2xe-x - x2 e-x ],

Y; y;'

c[-6e-x + 6 xe-x - x2 e-x ],

c[2e-x - 4xe-x

P(D)y

0,

+ x2e-x],

⇒ -6c = 27 ⇒ c = - 29 ⇒ YP = - 29 x2e -x ⇒ 9 y = --x2 e-x 2

+ c 1 e-x + C2 Xe-x + C4e

2x

is the general solution of the nonhomogeneous equation. Example 4.32. Consider the nonhomogeneous equation y"' - y' = 8xex , which has been solved by variation of parameters in Section 4.2.1, Example 4.21. The indicial polynomial is P(r) = r(r - l)(r + 1). Hence, the nonhomogeneous equation IS

P(D)y = D(D - l)(D + l)y = 8xex . The annihilator of g(x) = 8xex of minimal degree is Q(D) = (D - 1)2. Q(D)P(D)y

Thus,

(D - 1)2 D(D - l)(D + l)y D(D - 1)3 (D + l)y = (D - 1)2 [8xex] = 0

Since 1, ex and e-x are solutions of the homogeneous equation

P(D)y = 0,

116

CHAPTER 4. HIGHER-ORDER LINEAR EQUATIONS

where a and bare to be determined. Then

y� y� y�'

[ax2

(2ax + b)ex + (ax2 + bx)ex [ax2 + (2a + b)x + b]ex , [2ax+ 2a + b]ex + [ax2 + (2a + b)x + b]ex [ax2 + (4a + b)x + (2a + 2b)]ex, [2ax+ 4a + b]ex + [ax2 + (4a + b)x + (2a + 2b)]ex [ax2 + (6a + b)x + (6a + 3b)]ex ,

+ (6a + b)x + (6a + 3b)]e x - [ax2 + (2a + b)x + b]ex = 8xex ⇒

⇒ a

= 2 and b= -6

⇒ YP

4ax + ( 6a + 2b) = 8x

= (2x2 - 6x)ex



is the general solution of the nonhomogeneous equation. Example 4.33. Consider the nonhomogeneous equation

y

111

-

3y" + 7y' - 5y

= 32ex sin(2x),

which has been solved by variation of parameters in Section 4.2.1, Example 4 2. 2. The indicial polynomial is P(r) = (r - l)(r 2 - 2r + 5) . Hence, the nonhomogeneous equation is P(D)y = (D - l)(D2 - 2 D + 5)y = 32ex sin(2x).

Since ex cos(2x) and ex sin(2x) are solutions of a homogeneous equation for which the indicial equation has the roots r = 1 ± 2i, r - 1 = ±2i ⇒ (r - 1)2 = -4 ⇒ r2 - 2r + 5= 0 is the indicial equation. Hence, the annihilator of g(x) = 32ex sin(2x) of minimal degree is Q(D) = D2 - 2D + 5. Then

Q(D)P(D)y ⇒

2D + 5)(D - l)(D2 - 2D + 5)y (D - l)(D2 - 2D + 5)2 y = (D2 - 2D + 5)[32ex sin(2x)]

( D2

-

=0

y = c 1 ex + c2 cx cos(2x) + c3ex sin(2x) + C4Xex cos(2x) + C5Xex sin(2x).

Since c1 ex +c2 ex cos(2x) +c3ex sin(2x) satisfies the homogeneous equation P(D)y = 0,

117

4.2. NONHOMOGENEOUS EQUATIONS

where a and b are to be determined. Then y�

=

ex[acos(2x) + bsin(2x)] + xex [acos(2x) + bsin(2x)] +xex[-2asin(2x) + 2bcos(2x)] e"[acos(2x) + bsin(2x)] + xex[(a+ 2b) cos(2x) + (b- 2a) sin(2x)], x x y; - e [acos(2.T) + bsin(2x) ] + e [-2asin(2x) + 2bcos(2x)] +ex [(a + 2b) cos(2x) + (b- 2a) sin(2x)] +xex [(a+ 2b) cos(2x) + (b- 2a) sin(2x)] +xex [(-2a- 4b) sin(2x) + (2b- 4a) cos(2x)] ex[(2a + 4b) cos(2x) + (2b- 4a) sin(2x)] +xex[(4b- 3a) cos(2x) + (-4a- 3b) sin(2x)], Ill ex[(2a + 4b) cos(2x) + (2b- 4a) sin(2x)] Yp +e x[(-4a- 8b) sin(2x) + (4b- 8a) cos(2x)] + ex[(4b- 3a) cos(2x) + (-4a- 3b) sin(2x)] +xex[(4b- 3a) cos(2x) + (-4a- 3b) sin(2x)] +xex[(-8b + 6a) sin(2x) + (-8a- 6b) cos(2x)] ex[(12b - 9a) cos(2x) + (-12a- 9b) sin(2x)] +xex[(-lla- 2b) cos(2x) + (2a- llb) sin(2x)], 11 1 Ill Yp - 3 Yp + 7 Yp - 5 YP =

ex cos(2x)[(12b- 9a) - 3(2a + 4b) + 7a] + ex sin(2x)[(-12a- 9b) - 3(2b- 4a) + 7b] +xex cos(2x)[(-lla - 2b) - 3(4b- 3a) + 7(a + 2b) - 5a] +xex sin(2x)[(2a - llb) - 3(-4a- 3b) + 7(b- 2a) - 5b] 32 ex sin(2x)

-8aex cos(2x) - 8bex sin(2x) ⇒ a= 0 and b = -4 ⇒ YP = -4xex sin(2x) ⇒ =}

= 32 ex sin(2x)

y = -4xex sin(2x) + c1ex + c2ex cos(2x) + c 3 ex sin(2x)

is the general solution of the nonhomogeneous equation. Example 4.34. Consider the nonhomogeneous equation

y ( 4)

-

6y111 + lly" - 6y'

= 36 e3x

)

which has been solved by variation of parameters in Section 4.2.1, Example 4.23. The indicial polynomial is P(r) = r(r - l )(r - 2 )(r - 3 ). Hence, the nonhomogeneous equation is P(D)y = D(D - l)(D - 2)(D - 3)y = 36e3x.

118

CHAPTER 4. HIGHER-ORDER LINEAR EQUATIONS

The annihilator of g(x)

= 36e3x

of minimal degree is Q(D)

= D - 3.

Thus,

(D - 3)D(D - l)(D - 2)(D - 3)y D(D - l)(D - 2)(D - 3)2y = (D - 3)[36e3x ]

Q(D)P(D)y

=?

Y

=0

= C1 + C2ex + C3e 2x + C4e 3x + C5Xe3x .

Since c 1 + c2ex + C3 e 2x + c4e3x is a solution of the homogeneous equation P(D)y 3x 3x YP = c5xe , or YP = axe , where a is to be determined. Then a[e3x + 3xe3x ],

Y�

a[6e3x + 9xe3x ],

Y;

a[27e3x + 27xe3x ],

y;' 4 Yt )

4 Yt ) - 6y�' + lly � - 6y�

= 0,

a[108e 3x + 81xe 3x ],

=

a[108e3x + 81xe3x ] - 6a[27e 3x + 27xe3x ] +lla[6e 3x + 9xe3x ] - 6a[e3x + 3xe 3x ] 36e 3x

⇒ 6a = 36 ⇒ a = 6 ⇒ Yv = 6xe 3x ⇒

is the general solution of the nonhomogeneous equation.

Exercises 4.2 In Exercises 1-16, employ variation of parameters to find the general solution or to solve the initial-value problem. 1.

y "'

- 3y " - y ' + 3y = 8e 2x

2.

y "'

- 3y " + 2y' = 8xe 2x

3.

y "'

- 2y" + y '

= -2e x cos(x)

4.

y "'

- y " + 2y

=

lOex

5. y111 - y" + y' - y = 4sin(x) + 4cos(x) 6.

y '"

- y " = 20x3 ,

7. y"' - 2y"

y (0)

= 5sin(x)

= 20,

y '(0)

= 0,

y "(0)

= -20

CHAPTER 4 EXERCISES

=2

8. y( 4)

-

9. y( 4)

+ y" = 6x

y"

119

10. x 3y'" - x 2y" + 2xy' - 2y

= x,

11. x3 y"' + 5x2 y" + 2xy' - 2y

+ 2x2 y" = 1,

12. x3 y"'

13. x3 y"' + x2 y" 14.

:r;3 y'"

15. x3 y111

x

>0

x

1

= -,

x

>0

> 0, y(l) = 5, y'(l) = 1, y"(l) = 0

- xy' = 4x2 ,

x

>0

+ 2x2 y" - 2xy' = x3 cos(x),

x

>0

+ 3x2 y" + 2xy' = 1, x > 0

16. x 3 y111 + 3x 2 y"

= x 2 sin(x), x

>0

In Exercises 17-25, employ annihilators to determine a particular solution.

+ 3y = 8e2x

1 7. y"' - 3y" - y' 18. y"' - y" + 2y 19. y'" - y"

= lOex ( see Exercise 4)

= 20x3

20. y"' - 2y"

-

(see Exercise 6)

= 5 sin(x) (see Exercise 7)

21. y 111 - 3y" + 2y' 22. y111

( see Exercise 1)

= 8xe 2x

(see Exercise 2)

y" + y' - y = 4sin(x)

23. y"' - 2y" + y'

=2

=

24. y(4)

-

25. y(4)

+ y" = 6x

y"

+ 4 cos(x)

(see Exercise 5)

-2ex cos(x) (see Exercise 3)

(see Exercise 8) (see Exercise 9)

Chapter 4 Exercises 1. Consider the equation y"' - y" - 2y'

=

12.

(a) Employ variation of parameters to solve the equation. (b) Employ annihilators to solve the equation. (c) Integrate the equation once and solve the resulting second-order equation. 2. Solve the equation x3 y"' + x2y"

= x,

x

> 0.

120

CHAPTER 4. HIGHER-ORDER LINEAR EQUATIONS

3 3. Solve the equation x y

/II

4 . 2) . + x2 y/I = - , x > 0 ( see Exerc1se X

4. Integrate the equation y"' - y" = 20x3 twice and solve the resulting first-order equation (see Exercise 6 in Section 4.2 on page 118). 5. Solve the equation x4y( 4)

6. Solve the equation x3 y111

-

4x3 y"' + 12x2 y" - 24xy'

+ 24y = x4, x > 0.

x2 y" + lOxy' = 156x2 sin[3 ln(x)], x > 0.

7. Let P(r) be a polynomial with real coefficients. Show that if z is a complex root, then its complex conjugate z is also a root.

8. Let p(x) and q(x) be continuous on the interval I = (-1, 1). Is it possible for the equation y" + p(x)y' + q(x)y = 0 to have the solutions y 1 = 1, y2 = x and 3 y3 = x on I? 9. Solve the equation y"' + 3y" + 3y' + y = 2xe-x cos(x) by variation of parameters.

10. Employ annihilators to solve the equation in Exercise 9.

Chapter 5 Linear Systems Definition 5.1. A set of n first-order differential equations which involve n unknown functions of an independent variable l, where n 2'. 2, is called a system of first-order equations. A system is linear if it has the form x: x;

+ a12(l)x2 + · · · + a1 n (t)xn (l) + !1 (t) a21(t)x1 + a22(t)x2 + · · · + a2n (l)xn(t) + h(l)

au (l)x1

(5.1)

=

where x 1, x 2, · · · , X n are unknown functions oft, and aij (l), 1 :S i,j :Sn, and Jj (t), 1 :S j :S n, are given functions of t. A linear system is homogeneous if fj (t) 0, 1 :S j :Sn. Otherwise, it is nonhomogeneous. It has constant coefficients if aij (t) = aij is constant, 1 :S i, j :S n. For example, if n = 2, writing x 1 = x and x 2 = y for convenience, the system x' y'

3x + 2y 2x -y

is linear and homogeneous with constant coefficients, whereas X

1

y'

X -

3y + t 2

6x + 4y + e l

is linear and nonhomogeneous with constant coefficients. Definition 5.2. A solution of a system is a set of n differentiable functions x 1 (t), x2 ( t), · · · , X n ( l) which satisfy every equation in the system for all t in an interval I.

121

122

CHAPTER 5. LINEAR SYSTEMS For example, x(t) = 2e t , y(t) = et is a solution of the system

x'

y'

because

x'(l)

y' ( t)

for all t in the interval I

3x - 4y x-y

(5.2)

2i = 3(2et ) - 4(et ) = 3x(t) - 4y(t) et = 2et - et = x (l) - y ( t)

= JR.

In the following sections, methods will be developed for the determination of all solutions of linear systems with constant coefficients. This is achieved most efficiently, especially for large n, by expressing a linear system in matrix form. For example, with

the system (5.2) can be expressed as

and the solution x(t) function

= 2e t , y(t) = et can be expressed as the vector (or vector-valued)

Similarly, the system of three nonhomogenous equations x'

y' z'

2x + 3y - z + t 2 x - 4z - et -y + 6z + 5

can be expressed as x' =Ax+ f(t) by letting

More generally, the system (5.1) can be expressed as x'

= A(t)x + f(t)

by letting

123

5.1. HOMOGENEOUS SYSTEMS au(l) a 1 2(l)

a2 1(t) a2 2(t)

=

For a homogeneous system, f(t) 0, and, for a system with constant coefficients, A (l) = A has constant entries aij, 1 :S i, j '.S n.

5.1 5.1.1

Homogeneous Systems General Theory

Theorem 5.1. Let A(t) be an n x n matrix with entries aij(t) and consider the homogeneous system x' = A(t)x. If x1(l), x2(t), · · ·, xk(t), k � 1, are solutions, then any linear combination

is also a solution.

x'(l)

(c1XJ (t) + C2 X2 (t) + · · · + Ck Xk(t))' C1X� (t) + C2x;(t) + · · · + Ckx�(t) c 1 A(t)x1 (t) + c2A(t)x2 (t) + · · · + ckA(t)xk(t) A(t)(c1x1(l) + c2x2(t) + · · · + ckxk(l)) A(t)x(t).

Definition 5.3. The vector-valued functions x 1 (l), x2 (t), · · ·, xk(t), k � 1, are linearly dependent on an interval I if there exist constants c 1, c2 , · · ·, Ck, not all zero, such that

for all t in I. Otherwise, they are linearly independent on I. Definition 5.4. Let x1 (l), x2 (t), · · ·, xn (l) be n-vector-valued functions, i.e.,

1 :S i :S n.

124

CHAPTER 5. LINEAR SYSTEMS

The Wronskian W(t)

= W[x1(t),x2(t), · · · ,x (l)] is defined by n

W(t) = det(x 1 (t) x2(t) · · · Xn(t)) =

xu(t) X1 2(t) Xn(t) X22(t) Xnn (l)

Theorem 5.2. Suppose that A(t) = (aij(t)) is continuous on an interval I, i.e., aij(t) is continuous for 1 � i,j � n, and let x1 (t),x2(l), · · · ,xn(t) be solutions of x' = A(t)x. Then either W(t) 0 on I, or W(t) i= 0 for any t in I. The solutions x1(t),x2(t), · · · ,xn (t) are linearly independent on I if and only if W(t) i= 0 on I.

=

Definition 5.5. Let x1 (t),x2(l), · · · ,xn (t) be n linearly independent solutions of x' = A(t)x. The set {x1 (t),x2(t), · · · ,xn (t)} is called a fundamental set of solutions. The matrix

xu(t) X12(t)

X2 1 (t) X22(t)

Xnn(t)

l



is called a fundamental matrix for the system. The general solution of the system is

x(t)

= C1 X1 (t) + C2X2(t) + · · · + Cn Xn (t),

where c 1 , c2, · · ·, Cn are arbitrary constants, and may be expressed more concisely as

x(t) = (x 1 (l) x2 (l) · · · xn(t)) ( : A solution in which c 1 , c2,

· · ·,

solution.

= X(t)c,

where c

= (:

Cn are assigned particular values is called a particular

Theorem 5.3. A fundamental matrix X(t) for a system x' = A(t)x, where A(t) is continuous on an interval I, is nonsingular on I, i.e., det(X(t)) i= 0 on I, and satisfies the matrix equation X'(t) = A(t)X(t), where X'(t) = (x�/t)).

Proof. Since x1 (t), x2 (t), · · ·, xn (t) are linearly independent, det(X(t)) on I, and, since x� = A(t)xi, 1 � i � n, X' ( t)

( x't ( t) x; ( t) · · · x� ( t)) (Ax1(l) Ax2(t) · · · Axn(t)) A(x1(l) X2(t) · · · Xn (l))

AX(t).

=

W(t) i= 0

125

5.1. HOMOGENEOUS SYSTEMS

Theorem 5.4. Let A(l) be continuous on an interval J. If X(t) is a fundamental matrix for the system x' = A(l)x, then X(t)C is also a fundamental matrix for any nonsingular n x n matrix C. If X ( t) and Y(t) arc any two fundamental matrices, then there exists a nonsingular constant n x n matrix C such that Y(l) Equivalently, X(t)

= X(l)C.

= Y(t)c- 1.

= X(t)C satisfies the matrix equation Z'(t) = AZ(t) because Z'(l) = [X(l)C]' = X'(t)C = [AX(l)]C = A[X(l)C] = AZ(t). Since det(X(l)) =I= 0 and det(C) =I= 0, det(Z(t)) = det(X(t)) det(C) =I= 0. Hence, Proof. Z(t)

the columns of Z(l) arc linearly independent, and each column Zj is a solution of the system because (z1(t) z2 (l) · · · Z n (l))' A(z1(t) z2(t) · · · Z n (l)) (Az1(l) Az2(l) · · · Az n (t))



z�(t) = Azj(l), 1 :::; j :::; n. Thus, Z ( l) is a fundamental matrix. If X ( l) and Y ( t) are any two fundamental matrices, let C(t) = X(t)- 1 Y(t). Then X(t)C(t) = Y(t)



X' (t)C(l) + X(t)C'(t) = Y' (l) ⇒ A(l)X(l)C(t) + X(l)C'(t) = A(l)Y(t) ==> A(l)Y(l) + X(t)C'(t) = A(t)Y(t) ⇒ X(t)C'(t) = 0

= X(l)- 1 X(t)C'(t) = 0. Hence, C(t) = C is a constant matrix, and Y(t) = X(t)C. ==> C '(l)

Definition 5.6. An initial-value problem for a homogeneous system consists of a system x' = A(t)x, together with an initial condition x(lo) = x0, which requires that the solution x(l) be equal to x0 at l = lo, and determines the arbitrary constant vector c. Theorem 5.5. Let A(t) be continuous on an interval J containing the point t0 in its interior. The solution of the initial-value problem x' is given by x(t) Proof. x(t)

= X(t)X(lo)-

1

= A(t)x,

x(lo)

= xo,

x0, where X(t) is a fundamental matrix.

= X(t)c and x(to) = xo ==> x0 = x(lo) = X(l0)c ==> c = X(t0 )-1 x0 x(t) = X(t)c = X(l)X(tot 1 xo.

==>

126

5.1.2

CHAPTER 5. LINEAR SYSTEMS

Systems with Constant Coefficients

Let A = ( a ij) be an n xn matrix, where a ij, 1 � i, j � n, are constants, and consider the system x'= Ax. (5.3) Analogous to single equations with constant coefficients, seek solutions in the form (5.4)

=

where A is a scalar and v =/- 0 is a vector, to be determined. The case v = 0 is omitted because it yields only the trivial solution x(l) 0. Then

x'(t) = Ae

Al

v,

and x'(t) = Ax(t) if and only if Ae At v = Ae Alv, or Av = Av, i.e., A is an eigenvalue of A and v is a corresponding eigenvector. The equation Av= Av is equivalent to (AI - A)v = 0,

(5.5)

where I is the n x n identity matrix. Equation (5.5) is a homogeneous algebraic system, and admits nontrivial (i.e., nonzero) solutions v if and only if the coefficient matrix Af - A is singular, i.e., det(AI - A) = 0.

(5.6)

The eigenvalues of A are determined by solving Equation (5.6) for A, and for each eigenvalue Ai , the eigenvectors which correspond to Ai are obtained by setting A = Ai in Equation (5.5) and solving for v. If then x n matrix A admitsn real, linearly independent eigenvectors v1, v2, · · ·, vn , corresponding to the eigenvalues A1 , A2 , · · · , An (not necessarily distinct), then n linearly independent solutions of the system x' = Ax are given by X1(l) = e

A1

lV1, X2 (t) = e

A2 l

V2, · · · , Xn (t) = e An lVn .

This is automatically true in case A possesses n real, distinct eigenvalues, because eigenvectors which correspond to distinct eigenvalues are linearly independent. The cases where A has complex eigenvalues or fewer than n independent eigenvectors will be di cussed later. Example 5 .1. Consider the system

{ :: : �: -

:

��

}

.

(5.7)

1 27

5.1. HOMOGENEOUS SYSTEMS This system may be expressed in matrix form as x' x= ( : )

Then

I

det(>J - A)

=} ,\. 1

= 3 and

,\.2

For A1 = 3, (,\. 1 f

A ;

2

and A = ( _ � � ) .

(A- 2)(,\.-7)+4=,\. 2 -9,\.+18 A-_\ l=

(,\. - 3)(,\. - 6) = 0

= 6 are the eigenvalues of A. -

A)v= 0 with v= ( � ) ⇒

⇒ a - 2b = 0, and b = l ⇒ v= v1 = (

the eigenvalue ,\.1

= 3.

⇒ 2a -b = 0, and a= l ⇒ v= v2 =

the eigenvalue ,\. 2

= Ax, where

= 6.

� ) is an eigenvector which corresponds to

( � ) is an eigenvector which corresponds to

Thus, two independent solutions of the system (5.7) are given in vect9r form by x 1 ( t) = e 3t

(�)

2 t = ( �: ) e

3t 2e and X(t) = (x1(t) x2(t)) = ( 3 et solution in vector form is

and x2 ( t) = e 6t

(�)

= (

;:;t

),

e 6t 6 ) is a fundamental matrix. The general 2e t

from which the general solution x(t) = 2c 1 e 3t + c2e 6t and y(t) =c1 e 3t + 2c2e 6t of the system (5.7) is obtained in scalar form. Note that, if vis-an eigenvector of A, then

128

CHAPTER 5. LINEAR SYSTEMS

so is cv for any constant c =/- 0. Thus, the choice of an eigenvector is not unique, 4 2 4e3t 2e6t ) .1s also and we may take v 1 = ( ) and v2 = ( ) . Then Y(t) = ( 3 6 t t 2 4 4e 2e

a fundamental matrix. By Theorem 5.4, X(t)- 1 Y(t) = ( � � ) is a constant matrix. Example 5.2. Consider the system

{ x ; :_ y

-x -4x

y 3 } 6y

+ +

(5.8)

This system may be expressed in matrix form as x' = Ax, where

Then det(>-.J - A)

I ,\: 1 ,\

-=-\

I = (>-. + 1)(,\ - 6)

(,\ - 2)(,\ - 3) = 0

+ 12 = ,\ 2 _ 5,\ + 6

⇒ ,\ 1 = 2 and ,\2 = 3 are the eigenvalues of A. For ,\ 1 = 2, (,\if - A)v = 0 with v = ( � ) ⇒

3b = 0 , and a= l ⇒ v = v 1 = ( � ) is an eigenvector which corresponds � to the eigenvalue ,\ 1 = 2. ⇒ 3a

For ,\2 = 3, (,\2 J - A)v = 0 with v = ( � )



( : =� ) ( � )

= ( � )

⇒ 4a - 3b = 0, and a = 3 ⇒ v = v2 = ( � ) is an eigenvector which corresponds

to the eigenvalue >-.2 = 3.

129

5.1. HOMOGENEOUS SYSTEMS Thus, two independent solutions of the system (5.8) are given in vector form by 1 ) 1

=

(

e2t e2t

(x 1 (l) x2 (l))

=

(

e2t 3e3t ) is a fundamental matrix. The general e 2 t 4e 3 t

x1(t) and X (t)

=

= e2t

(

solution in vector form is

)

and x2 (l)

from which the general solution x(t) = c 1 e2t system (5.8) is obtained in scalar form.

= e3t

+ 3c2 e3t

(

3 ) 4

and y(t)

=

(

3e3t 4e 3t

)

,

= c 1 e2t + 4c2 e3t

of the

Example 5.3. Consider the system (5.9) This system may be expressed in matrix form as x'

Then

det(>.J - A)

=I

= Ax,

where

>.; 1 � I = (>. + 1) >. 1

⇒ >. 1 = -1 is the only eigenvalues of A, with multiplicity 2. For >.1

= -1,

(>. 1 1 - A)v

= 0 with v = ( � )

2

=0



= 1 and b = 0 gives the eigenvector = 0 and b = 1 gives the eigenvector v2 = ( � ) . Since v1 and

which states that both a and b are free. Hence, a

v1

= (

t

),

and a

v2 are linearly independent, two independent solutions of the system (5.9) are given in vector form by

130

CHAPTER 5. LINEAR SYSTEMS e

( ;

t

solution in vector form is

from which the general solution x(t) obtained in scalar form.

� ) is a fundamental matrix. The general e t

= c 1 e-t

and y(t)

= c2 e-t of the system (5.9) is

Note that, even though the 2 x 2 matrix A has only one eigenvalue A 1, the general solution is obtained as in the preceding examples because there are two independent eigenvectors which correspond to ,\ 1 . A system such as (5.9) is said to be decoupled because each equation involves only one of the unknown functions, and can be solved independently of the other(s). Thus, x' = -x ⇒ x(t) = c1 e-t, and y' = -y ⇒ y(t) = c2 e-t, as above.

=

Example 5.4. Consider the initial-value problem

{

x; y

-

3x +

- 4X

-

2y } ' x(0) y

= 1, y(0) = 5.

This problem may be expressed in matrix form as x' X

= (: ) ' A=

Then

3 I ,\ �

det(,X.J - A)

(>. ⇒ A1

=

-1 and >.2

⇒ 4a + b

= 0,

=2

>-1

,\�\

and

Xo

=(

!).

where

I = (>. - 3)(>. + 2) + 4 = ,\2 - A - 2

+ 1)(>. - 2) = 0

are the eigenvalues of A.

and a= l ⇒ v

to the eigenvalue

( -� -� )

= Ax, x(0) = xo,

(5.10)

= -1.

= v1 = (

_! )

is an eigenvector which corresponds

131

5.1. HOMOGENEOUS SYSTEMS

⇒ -a - b = 0, and a = l ⇒ v = v2 = (

to the eigenvalue >. 2

= 2.

-i )

is an eigenvector which corresponds

Thus, two independent solutions of the system (5.10) are given in vector form by

e -t e2t ) is a fundamental matrix. The solution -t ( - 4e -e 2t of the initial-value problem (5.10) is then given by x(t) = X(l)X(o)- 1 xo. and X(t)

= (x 1 (l) x2(l)) =

Note that the inverse of a 2 x 2 matrix ( � � ) is obtained simply as

(

1 1 d -b ) 1 d -b ) a b )( ( a c d - det(A) -c ad - be -c a ·

Thus, X(o)- 1 = (

x(i)

_! -� )- 1 ( -! 1

=

=

i.e., x(t) = -2e-t

(

+ 3e2t

and y(t)

{ -=

�),and

-1 -1 1 ) ( 5 ) 4 1

=

(

e2t e -t -4e-t -e2t

)(-

2 3 )

= 8e-t - 3e2t .

Example 5.5. Consider the system x' y' z'

=

- 2y -3x + y -x + y

+ + +

2z 3z } 3z

This system may be expressed in matrix form as x' = Ax, where and A= ( -� -1

� � ) . 1 3

(5.11)

132

CHAPTER 5. LINEAR SYSTEMS

Then

det(,H - A)

>2 -2 3 >- - 1 -3 1 -1 >- - 3

>-[(>- - 1)(>- - 3) - 3] - 2[3(>- - 3) + 3] - 2[-3 - (>- - 1)] >-[>-2 - 4>-] - 4), + 16 >-2 (>- - 4) - 4(>- - 4) (>- - 4)(>-2 - 4) (>- - 4)(>- - 2)(>- + 2) = 0

⇒ >-1 = 2, >-2 = -2 and A3 = 4 are the eigenvalues of A. For>.1

=

2, (>. 1 1 - A)v = 0 withv = ( � ) =>

Row reduction of the coefficient matrix (R2 � denotes row i) then gives

=> b

=

0 and a

+ b- c

-

R2

=

3R3 and R3

-

R1

-

2R3, where

= 0 => a = c, and a = 1 => v1 = ( i ) is an eigenvector

which corresponds to the eigenvalue >- 1 = 2. For>.,

-

-2, (>.2 1-A)v=Owithv= ( �) =>

Row reduction of the coefficient matrix gives

5.1. HOMOGENEOUS SYSTEMS

133

⇒ c = 0 and -a + b - c = 0 ⇒ a= b, and a= 1 ⇒ v2 = (

i)

is an eigenvector

which corresponds to the eigenvalue >-. 2 = -2. For ,\3

= 4, (,\3/

- A)v

=0

= (�) ⇒

with v

=� ) ----+ ( � � =� ) ----+ ( � -� -� ) 0 -i :DU) 0)

Row reduction of the coefficient matrix gives

(1



1 -1

1

1 -1

1

0

3 -3



⇒ -b+c = 0 and 2a+b-c = 0 ⇒ a= 0, and b = 1 ⇒ v3 = (:)

is an eigenvector

which corresponds to the eigenvalue >-.3 = 4. Thus, three independent solutions of the system (5.11) are given in vector form by

and X(l)

=

(x1(t) x2(l) x3 (t))

,c�

=

e2t

0

(

e

2t

general solution of the system (5.11) is

x(t)

C1X1(t)

X(t)c e t 2

o)

e- 2t 2

e4t

O

e4t

e-

t

is a fundamental matrix. The

+ C2X (t) + C3X3(i) 2

e- 2 l e- 2 t

0

e�' )( �:)

e

4l

C3

c,e"He-") C2 e- 2l C1 e

2l

+ C3e + C3e

4l

4t

,

134

CHAPTER 5. LINEAR SYSTEMS

{ -

Example 5.6. Consider the system x'

y' = z'

(n

=

n

(5.12)

This system may be expressed in matrix form as x' = Ax, where

x= Then det(,\J - A) = =? ,\ 1

and A=

(J

,\-1 0 0 ,\-1 0 0 ,\-2 0

0 1 0

= (,\ - 1) 2 (,\ - 2) = 0

= 1, of multiplICity 2, and ,\ 2 = 2 are the eigenvalues of A.

For A 1 = 1, (A 1 I - A)v = 0 with v = ( � ) =>

=> a - c = 0 => a = c and b is free. Thus, a = 1 and b = 0 => v 1 = ( � ) , and a = O and b = 1 => v 2 = ( [ ) axe independent eigenvectors which correspond to the eigenvalue ,\ 1 = 1. For A 2 = 2, (A 2 J - A)v = 0 with v = ( � ) =>

5.1. HOMOGENEOUS SYSTEMS => a - 0 and b - 0 , and c - I => v 3

- (�)

135 is an eigenvector which corresponds

to the eigenvalue A2 = 2. Thus, three independent solutions of the s ystem (5.12) are given in vector form by

et 0 0 ) and X(l) = (x1(l) x 2(l) x3(t)) = ( 0 et O is a fundamental matrix. The et 0 e2t general solution of the system (5.12) is

x(t)

Complex Eigenvalues Let A be an n x n matrix with real entries, and suppose that A = a+ i(3, (3 =/- 0, is a complex eigenvalue. Then "X = a - i(3 is also an eigenvalue since det( AI - A) is a polynomial with real coefficients . If v_is an eigenvector corresponding to A, then vis an eigenvector which corresponds to A because Av= AV



Av= Av



Av= "Xv ⇒

Av= Av.

Then z(t) = e>.tv and z(t) = e>-t v are two complex solutions of the system x' = Ax. Let v = a + ib, where a and b are real vectors. Then v = a - ib and, employing Euler's identity, we obtain z(l)

z(l)

°'t i e(a+i,B)t[a + ib] = e e ,Bt [a + ib] e°'t [cos(,6t) + i sin(,6t) ][a + ib]

e°'t{[cos(,6t)a - sin(,6t)b] + i[sin(,6t)a + cos(,6t)b]}, ° e 't{[cos(,6t)a - sin(,6t)b] - i[sin(,6t)a + cos(,6L)b]}.

CHAPTER 5. LINEAR SYSTEMS

136

Since the system is linear and homogeneous, any linear combination of solutions is a solution, by Theorem 5.1. Thus,

Xz(t)

z(t) + z(t) = �(z(t)) = e'�t [cos(,Bt)a - sin(,Bt)b], 2 z(t) z(t) = Ss(z(t)) = e0t [sin(,8t)a + cos(,Bt)b] ;

are two real solutions which are linearly independent. Example 5. 7. Consider the system

{ :_ x

y;

2x -3x

+ +

3y 2y

}

(5.13)

This system may be expressed in matrix form as x' = Ax, where

Then det(>J - A) = \

>- 2 ;

-\ I = (>- - 2) 2 >-

+ 9 = >-2

-

4>- + 13 = 0

. = 2 ± 3·i are the comp1ex e1genva1ues of A. ⇒ >-, >- = 4 ± J=36 2 -

For >- = 2 +. 3i, (>-I - A)v = 0 with v = ( � )

⇒ 3ia -



3b = 0, and a= l ::::::}

is an eigenvector which correspo,nds to the eigenvalue >- = 2 + 3i. Then z(t)

e2t [cos(3t) + i sin(3t)] [ ( � )

+i (

� )]

e 2t { [cos(3t) ( � ) - sin(3t) ( � ) ]

+i

[sin(3t) ( � )

+ cos(3t) (

� )] }

3

1 7

5.1. HOMOGENEOUS SYSTEMS is a complex solution, and x1(l)

R(z(t))

=

3

3

e 2t [cos( t) ( �) -sin( l) ( �)]

3 3

e 2t cos(3t) cos( t) = ( e 2t ( ) -e2t sin( t) ) ' - sin( l) x 2(l)

3

3

�(z(t))= e 2 t [sin( t) ( �)

+ cos(3 t)

3

( �) ]

2t sin(3 l) t = ( e2 sin( t ) e2 ( ) e tcos(3l) ) cos( t)

3

are two real, independent solutions of the system (5.13). X()t

=

3 3

e 2tcos(3l) = cos(3l) sin( t ) (x i()l x 2(t))= e 2t ( ( -e 2t sin(3t) -sin( l) cos( t ) )

3

is a fundamental matrix, and the general solution is

3

3 3

e 2t sin( l) e 2tcos( t) )

3

. . c1 cos( t) sin( t) c1x1(t) + c 2x 2(t) = X(l)c = e 2t ( -sin(3t) cos(3l) ) ( c 2 )

x(t)

3

3 3

c1 cos( t) + c 2 sin(3 l) c1e 2t cos(3t) + c 2 e 2t sin( t) = ( ) -c1 e 2t sin( t) + c 2 e 2t cos( t) ) ' -c1 sin(3 t) + c 2 cos(3t)

3 i.e., x(t) = e l[c1 cos(3 t) + c2 sin(3 t)] and y(t) = e t[-c1 sin(3 t ) + c e 2t (

2

2

3 t)].

2 cos(

Example 5.8. Consider the system

+ { ;: : �:

13y

}

.

(5.14)

This system may be expressed in matrix form as x'= Ax, where

Then det(>J - A)=



-

>-, >- =

For >- = 2



I

>-



4

-1 3

1

3

= >-(>- - 4) + 1 = >- 2 - 4 >-

\/-36 = 2 ± 3i are the complex eigenvalues of A. 2

+ 3i, (>-[ - A)v= 0 with v = ( � ) ⇒

( -2 + 3i 1

3

-1

2+

0)

a )= ( ) ( 3i b 0

+ 13 = 0

138

CHAPTER 5. LINEAR SYSTEMS

⇒ a+ (2 + 3i)b = 0, and b = -1 ⇒ is an eigenvector which corresponds to the eigenvalue ,\ = 2 + 3i. Then e 2l[cos(3t) + i sin(3t)] [ (

z(t)

{[

!1 ) - sin(3t) ( � ) ] +i [sin(3t) ( ! ) + cos(3t) ( � ) ] 1 } e 2l

cos(3t) (

is a complex solution, and

x 1 (t) = �(z(t)) = e 2l [cos(3t) ( e2

t(

t(

!1 ) -sin(3t) ( �)]

2cos(3t) - 3sin(3l) e 2l [2cos(3l) - 3sin(3t)] = ) ( )' -cos(3t) -e 2l cos(3t)

x 2(t) = �(z(t)) = e 2l [sin(3t) ( e2

!1 ) + i ( � ) ]

!1 ) + cos(3t) ( � ) ]

2sin(3t) + 3cos(3t) e 2t[2sin(3t) + 3cos(3t)] = ( ) ) -sin(3t) -e 2t sin(3l)

are two real, independent solutions of the system (5. 1 4). X(t)

(x1(t) x 2(t)) 2cos(3t) - 3sin(3l) 2sin(3l) + 3cos(3t) e 2t ( ) -cos(3t) -sin(3t) e 2t[2cos(3t) - 3sin(3t)] e 2t[2sin(3t) + 3cos(3t)] = ( ) -e2t sin(3t) -e 2t cos(3t)

is a fundamental matrix, and the general solution is x(l)

( (

c1 x1(t) + c 2 x 2(t) = X(t)c c1 2cos(3t) - 3sin(3t) 2sin(3t) + 3cos(3t) e 2l ) ( c2 ) -cos(3l) -sin(3t) ci[2cos(3t) - 3sin(3t)] + c 2 [2sin(3t) + 3cos(3l)] e2 t )' -c1 cos(3l) - c 2 sin(3t)

i.e.'

x(t) e 2l [(2c 1 + 3c 2) cos(3t) + (2c 2 - 3c1) sin(3t)], y( t) = e 2l[-c1 cos(3t) - c2 sin(3t)].

139

5.1. HOMOGENEOUS SYSTEMS Example 5.9. Consider the initial-value problem

{

X

1

y'

=

+ _ x + y} = 5 5 y

X

, x(0) = 1,

y (0) = 3.

(5.15)

This problem may be expressed i n matrix form as x' = Ax, where

The n

x = ( ; ) , A= ( � � ) -

det(>J - A) =

I

>. 1 ;

and x(O) = ( � ) .

1 -_ = (>. - 1)(>. - 5) + 5 = >.2 5

>.

1

-

6>. + 10 = 0

⇒ >., ->. = 6±R = 3 ± i are the complex eigenvalues of A. 2

For >. = 3 + i, (>.I - A) v = 0 with v = ( � )



-1.)(a o 2+i )=( ) ( 5 -2+i b 0

⇒ (2 + i)a - b = 0, and a = 1 ⇒

i s an eigenvector which corresponds to the eigenvalue >. = 3 + i. Then z(t)

= e3t [co s(l) +i sin(t)] [ ( � ) + i (� )] e3t {[cos(t) ( � )- sin(t) ( � )] +i [ sin(l) ( �) +co s(t) ( �) } ]

is a complex solution, and x 1 (t)

= R(z(l)) = e 3t [co s(t) ( � ) - si n(t) ( �)] e 3t

x 2 (t)

(

2cos(t) � si n(l)

(

) 2sin(�\ f �os(t)

s

)

)

'

3t �(z(t)) = e [sin(t) ( � ) + co s(t) ( �)]

e

3t

n

140

CHAPTER 5. LINEAR SYSTEMS

are two real, independent solutions of the system (5.15) . Thus,

X(t) _ (xi (t ) x2 (t) ) _ - e 3t is a fundamental matrix,

(

cos(t ) sin(t ) 2cos(t) - sin(t ) 2sin(t ) + cos(t ) )

X(0) = ( � � ) , X(0)- 1 = (

-

� � ),

and x(O) = ( ; ) ⇒ x(t)

X(t)X(o)- 1 x(0) cos(t) sin(t ) 1 1 0 = e 3t 2cos(t ) - sin(t ) 2sin(t ) + cos(t ) ) ( -2 1 ) ( 3 )

- e3t = e3 i.e., x(t)

t

( ( (

2cos (t) � sin(t ) 2sin(�� f �os(t ) s

)

cos(t) + sin(t ) 3cos(t) + sin(t) ) '

n

) ( ) �

= e3t [cos(t ) + sin(t )] and y(t) = e3t [3cos(t ) + sin(t ) ].

Example 5.10. Consider the system

{ �;z' =: -4x �: +: 4y4� +: 3:z } .

(5.16 )

This system may be expressed in matrix form as x' = Ax, where

Then det(AI - A) =

-3 A+ l -4 -2 A - l -1 4 -4 A - l

= (A+ l)[(A - 1)2 - 4] + 4[-2(.A. - 1) + 4] - 3[8 - 4(.A. - l)] (A+ 1)[.A.2 - 2,,\ - 3] + 4A - 12 (A+ l)(A - 3)(A + 1) + 4(.A. - 3) (A - 3)[(.A. + 1)2 + 4] = (A - 3)[.A. 2 + 2.A. + 5] = 0

5.1. HOMOGENEOUS SYSTEMS

-

=}

A1 = 3 and A2, A2 =

-2± J=R 2

141 = -1 ± 2i are the eigenvalues of A.



For .\ 1 = 3, (.\,I - A)v = 0 with v = ( � )

Row reduction of the coefficient matrix gives

⇒ c = 0 and a-b = 0, and a= 1 ⇒ v 1

= (

i)

is an eigenvector which corresponds

to the eigenvalue A 1 = 3. Hence, e

3l

( e 3l )

0 is one solution of the system (5.16). For.\2

=

-1+2i, (.\2 /-A)v

=

Owithv

(

=

!)



( 2i -4 -2 2i- 2 4 -4 Row reduction of the coefficient matrix gives ( 2i -4 -3 ) -2 2i - 2 -1 4 -4 2i - 2

---+

(

-4 2i 0 -2i- 6 0 -8i-4

142

CHAPTER 5. LINEAR SYSTEMS

==> 2b + c = 0 ==> c = -2b and 2ia - 4b - 3c = 0 ==> 2ia + 2b = 0, and b = 1 ==> c = -2 and a= i ==>

is an eigenvector which corresponds to the eigenvalue A.2 = -1 + 2i. Then

is a complex solution, and

- sin(2t) -e-l sin(2t) x 2(t) = R(z(t)) = e-t ( cos(2t) ) = ( e-tcos(2t) ) , -2 cos(2t) -2e-l cos(2t) cos(2t) e-tcos(2l) e-t s�n(2t) ) s�n(2t) ) = ( x3(t) = S'(z(t)) = e-l ( -2c t sm(2t) -2 sm(2t)

are two real, independent solutions of the system (5.16). Thus, three independent solutions are x1(t), x2(t) and x3(t), e3t -e-l sin(2t) e-tcos(2t) X(t) = (x1(t) x 2 (t) x3 (l)) = ( e3l 0 -2e-tcos(2l)

is a fundamental matrix, and x(l)

143

5.1. HOMOGENEOUS SYSTEMS i.e.'

x(l)

c 1 e3l

y(l)

c 1 e + c2 e-l cos(2l) +c3 e-t sin(2t), -2c2 e-l cos(2t) - 2c3e-t sin(2l)

-

3l

z(l)

c2 e-t sin(2t) + c3e-l cos(2l),

is the general solution of the system (5.16).

Example 5.11. Consider the system

(5.17)

U ( i)

This system may be expressed in matrix form as x' = Ax, where

X=

Then

and A=

0 0 A 0 -1 0 -8 ,\

det(,\J - A)

0 -4 0 0 0 0 0 8

4 0 A 0

0 2 0 ,\

-�)

0 2 0 ,\ 0 -8 0 ,\ ,\

=.A.

,\[,\(,\ 2 + 16)] + 4[,\2 + 16] (,\2 +16)(.A.2 +4) = 0 => .A. 1, >: 1 = ±2i and A 2 , >:2 = ±4i arc the eigenvalues of A.

For>. 1 = 2i, (A 1 J - A)v = 0 with v = ( ; ) =>

(t

0 2i 0

-1 0 -8

4 0 2i

0

0 0

+4

,\ 2 0 0 0 -1 0 -8 ,\

144

CHAPTER 5. LINEAR SYSTEMS

Row reduction of the coefficient matrix gives

(

0 4 0 2i O 2 -1 0 2i O ) 0 -8 0 2i �

=> d = 0 => b

=0

--t

2i O 4 0 2i 0 4 0 2i O 2 0 2i 0 => 0 0 0 O O O O ( ) ( O O O 6 0 0 0

and 2ia + 4c

= 0,

=

and c

1 => a

= 2i =>

l 0 n ! ! )l( ! l

is an eigenvector of A which corresponds to the eigenvalue >. 1

Z1

(t) = [cos(2t) + sin(2t)] i

= [cos(2t) (

[

) +

i

=

2i. Then

(

) + cos(2t) ( � ) ]

sin(2t) ( � ) ] + i [sin(2t) (

is a complex solution, and

2

0

!!l(z 1 ( t)) = cos(2t) [

- sin(2t)

(

( � )

)

+ cos(2t)

( � )

are two real, independent solutions of the system (5.17).

ForA2

= 4i, (,\ 2 1 - A)v = 0

with v

=

2

0

'J(z 1 (t)) = sin(2t) [

( [ )

= (;

) =>

=

(

-2 sin(2l)

cos!2t) 2 cos(2t)

sin:2t) )

) ,

145

5.1. HOMOGENEOUS SYSTEMS Row reduction of the coefficient matrix gives 4i o 4 0 4i 0 ( -1 0 4i 0 -8 0

i

*

u �

u n1

is an eigenvector of A which corresponds to the eigenvalue >.2 = 4i. Then z,(l) - [coo(4l) + isin(4l)] [

-

) +i(

[cos(4l) (�)-sin(4l) (;)1 +i [sin(4t) (�)+co (4t)

is a complex solution, and � x 3(t) = �(z2(l)) = [cos(4t) ( )-sin(4t) ( � 2

x4(t) = 8'(z2(l)) = [sin(4t) (

� 2

0

)1

) + cos(4t) ( � ) 0

1

(;)

1

si�(4t) = (), 2cos(4t)

= (

coo�4t) 2sm(4t)

)

are two real, independent solutions of the system (5.17). Thus, x 1 (t), x2(t), x3(l) and X4 (l) are four real, independent solutions, 0 0 ) -2sin(2t) 2cos(2l) cos( 4t) 0 sin(4l) O X(t) - ( cos(2t) 0 sin(2l) 0 2cos(4t) 2sin(4t) 0 0

146

CHAPTER 5. LINEAR SYSTEMS

is a fundamental matrix, and

x(t)

i.e.,

C1X1(l)

X(l)c

(

+ C2X2(t) + C3X3(t) + C4X4(t)

-2 sin(2t) 2 cos(2t)

0

0

cos(2t)

sin(2t)

0

0

0

0

- sin(4t) cos(4t) 0 0 2 cos(4l) 2 sin(4t)

-2c1 sin(2t) + 2c2 cos(2t), -c3 sin(4t) + c4 cos(4t),

x(l) y(t)

c1 cos(2l) + c2 sin(2t), 2c3 cos(4l) + 2c4 sin(4t)

z(t) w(t)

is the general solution of the system (5.17). Generalized Eigenvectors Consider the system x' = Ax, where A is a 2 x 2 matrix. If A possesses two distinct eigenvalues >-1 and >-2 , with corresponding eigenvectors v 1 and v2, then two linearly independent solutions of the system are x1 (t) =

e

.\ 1 t

v1

and x2(l) =

e

.\ 2 t

v2.

If A possesses only one eigenvalue >.1 of multiplicity two, i.e., det(>.I -A)= (>. - >.1) 2 , and if A has two independent eigenvectors v1 and v2 corresponding to >.1, then two independent solutions of the system are x1(t) = e .\ 1 tv1 and x2(t) = e .\ 1 t v2. More generally, if an n x n matrix A possesses n independent eigenvectors (real or complex), then n independent solutions of the system x' = Ax can be determined by the methods already discussed.

If an n x n matrix A admits only k linearly independent eigenvectors, where k < n, which occurs whenever an eigenvalue of A of multiplicity m > 1 has fewer than m independent eigenvectors, then the methods employed thus far yield only k linearly independent solutions of the system x' = Ax, and additional independent solutions are required in order to obtain all n independent solutions. Given an n x n maLrix A, det(>J - I\) is a polynomial of degree n, called the characteristic polynomial of A. If >.1, >.2, ·· ·, Ar are the distinct eigenvalues of A, with multiplicities m 1, m2, ···, mr , respectively, then

5.1. HOMOGENEOUS SYSTEMS with

m1

147

+ m2 + · · · + mr = n.

Thus, n independent solutions of the system x' = Ax can be determined provided that, for every eigenvalue Ai of multiplicity mi, a total of mi independent solutions of the system can be found. Several cases are possible, and will be considered for n = 2, n = 3 and n = 4, from which the suitable procedures for any n 2:: 2 can be deduced. Let A be a 2 x 2 matrix with only one eigenvalue A 1 of multiplicity 2, and with only one independent eigenvector v1. Then one solution of the system x' = Ax is x1(t) = e >-1t v1. A second, independent solution x2(l) is sought in the form (5.18) where u0 and u 1 arc vectors to be determined. Then

and, hence, x;(t) = Ax2 (t) if and only if A 1 e >-1t (tuo Division by e >-1t

+ u 1 ) + e >-1t uo = Ae>- 1t (tuo + u 1 ).

=f. 0 yields the condition

which holds for all real l if and only if (5.19) The first condition in (5.19) is equivalent to

i.e., uo is an eigenvector of A corresponding to the eigenvalue A 1 . We may, therefore, take uo = v 1 , or uo = cv 1 for any constant c =f. 0. The second condition in (5.19) is equivalent to (5.20) A solution u1 of Equation (5.20) is called a generalized eigenvector of the matrix A corresponding to the eigenvalue A 1 . Note that Equation (5.20) is a nonhomogeneous algebraic system with a singular coefficient matrix A - A 1 I, and such a system may have no solution, in general. In the present case, however, it can be shown that, since u0 is an eigenvector which corresponds to A 1 of multiplicity m 1 > 1, a solution u 1 exists.

148

CHAPTER 5. LINEAR SYSTEMS Thus, two independent solutions of the system x' = Ax are given by x1(t) = eA 1 t v1 and x2(t) = e A 1 t (tu0

+ u 1),

where uo is an eigenvector and u 1 is a generalized eigenvector corresponding to the eigenvalue ,\ 1 of multiplicity 2. Example 5.12. Consider the system

{

x: y =-

5x 4x

-+ } y y

(5.21)

This system may be expressed in matrix form as x' = Ax, where

Then

= (>. - 5)(>. - 1) + 4 = ,\2 _ 6,\ + 9 I ,\�/ ,\ � 1 I (,\. - 3) 2 = 0

det(>.I - A)

=} ,\ 1

=}

= 3 is the only eigenvalue of A, of multiplicity 2.

-2a + b = 0, and a

=

l

=}

v = v 1 = ( � ) is an eigenvector which corresponds

to the eigenvalue ,\ 1 = 3. Thus, one solution of the system (5.21) is

x 1(t) = e3l

(�)

= (

;::t

).

Since A has only one independent eigenvector which corresponds to the eigenvalue ,\ 1 of multiplicity 2, a second, independent solution of the system must be sought in the form x2(t) = e 3t (tuo + u1), where u0 = v 1 is an eigenvector corresponding to ,\ 1 and u 1 is a solution of the algebraic system (A - >. 1 I)u 1 = u0, i.e., a generalized eigenvector. Employing the fact that (A - >-1!) = -(>.1I - A), (A - >-1I)u1 = uo with u1 = ( � )

=}

149

5.1. HOMOGENEOUS SYSTEMS => 2a - b= l, and a= 0 => b= -l => u 1 = ( � ) . Hence, l x�(t)= e 3t

[t ( � ) + (

�l ) ] = e

3l

. ( 2l � 1 )

Thus, two independent solutions of the system (5.21) are given by

and is a fundamental matrix. The general solution is x(l)

c1x1(t) + c2x2(l) = X(l)c

e

3

t

(�

2t � 1 ) ( �� )

Example 5.13. Consider the system

{

x' y'

= =

-5x -x

l

=(

-

+

det(>J - A) => ,\.1

=

A

-

l e )

3 t

)'

4y } . y

= Ax,

(5.22) where

\ l=(,\.+5)(,\.+1)+4=,\.2 +6,\.+9 A� (,\. + 3) 2 = 0 I

;

S

-3 is the only eigenvalue of A, of multiplicity 2.

For ,\. 1 = -3, (,\.1/

3l

2c1e�i� c:( �; �

This system may be expressed in matrix form as x'

Then

t

A)v= 0 with v = ( � ) =>

150

CHAPTER 5. LINEAR SYSTEMS

⇒ a - 2b = 0, and b = l ⇒ v = v 1 = ( � )

is an eigenvector which corresponds to

the eigenvalue A1 = -3. Thus, one solution of the system (5.22) is

Since A has only one independent eigenvector which corresponds to the eigenvalue ). 1 of multiplicity 2, a second, independent solution of the system must be sought in the form

x2(t) = e- 3t(tuo + u1),

where u0

=

v1

is an eigenvector corresponding to ). 1 and u 1 is a solution of the

algebraic system (A- ,,\ 1 I)u 1 = u0, i.e., a generalized eigenvector. With u 1

⇒ -a+ 2b = 1, and b = 0 ⇒ a= -1 ⇒ u 1 = (



l

).

= ( � ),

Hence,

Thus, two independent solutions of the system (5.22) are given by

and

is a fundamental matrix. The general solution is

If A is a 3 x 3 matrix with fewer than three independent eigenvectors, then there are three possibilities, depending upon the number of distinct eigenvalues and the number of independent eigenvectors which correspond to each eigenvalue.

5.1. HOMOGENEOUS SYSTEMS

151

Let A be a 3 x 3 matrix with only two distinct eigenvalues, >.. 1 of multiplicity m 1 = 1 and >.. 2 of multiplicity m 2 = 2, and with only two independent eigenvectors v1 and v2 , corresponding to >.. 1 and >.. 2, respectively. Then three independent solutions of the system x' = Ax are given by x1(l) = e >qt v1, x2(t) = e >- 2t v2 and x3(t) = e >- 2 t(tuo

+ u1),

where u0 is an eigenvector which corresponds to >..2 and u 1 is a generalized eigenvector which corresponds to >.. 2, i.e., u 1 is a solution of the algebraic system (A->..2 J)u 1 = u0.

{

Example 5.14. Consider the system x' = 2x y' = z' =

+ 3y 3y y

--

9z 4z } . z

(5.23)

=� ) .

This system may be expressed in matrix form as x' = Ax, where and A = ( � � 0 1 -1 Then det(>..I

- A)

>..

-2

-3

9 4

= (>.. - 2)[(>.. - 3)(>.. + 1) + 4] >.. - 3 -1 >.. + 1 2 (>.. - 2)(>.. - 2>.. + 1) = (>.. - 2)(>.. - 1)2 = 0

⇒ >.. 1 = 2 of multiplicity 1

0 0

and >..2

=

1 of multiplicity 2.

For,\, = 2, (,\1J - A)v = 0 with v = ( ; ) cc>

cc> -b + 4c = 0 and -b + 3c = 0 cc> b = c = 0, and a= l cc> v = v1 = ( �) 1s an eigenvector which corresponds to the eigenvalue >.. 1 = 2. Thus, one solution of the system (5.23) is

152

CHAPTER 5. LINEAR SYSTEMS

For >. 2

= I,



+ 2c =

v

-b

(>.2 1 - A)v

= 0 with v = ( �)

0 and -a - 3b

= v, = ( � )

+ 9c =

0

=>

⇒ b = 2c and

-a+ 3c

=

0, and c

is an eigenvector which corresponds to the eigenvalue >.2

=

=

1



I. Thus,

a second, independent solution of the system (5.23) is

Since A has only one independent eigenvector which corresponds to the eigenvalue .A. 2 of multiplicity 2, a third, independent solution of the system must be sought in the form where u0 = v2 is an eigenvector corresponding to >-2 and u 1 is a solution of the algebraic system (A- A2 J)u1 = Uo, i.e., a generalized eigenvector. With u, = ( � ) ,

=> a+ 3b - 9c

=3

and b - 2c

=

I, and c

=0

=> b

=

I and a= 0 => u,

Hence,

Thus, three independent solutions of the system (5.23) are given by

= (

!}

5.1. HOMOGENEOUS SYSTEMS

153

and

is a fundamental matrix. The general solution is x(t) =

Let A be a 3 x 3 matrix with only one eigenvalue )q of multiplicity m 1 = 3, and with only two independent eigenvectors v 1 and v2 corresponding to ,\. 1 . Then two independent solutions of the system x' = Ax are given by

A third, independent solution is sought in the form

where, as before, uo is an eigenvector which corresponds to ,\. 1 and u 1 is a generalized eigenvector which corresponds to A 1, i.e., u1 is a solution of the algebraic system (A - ,\. 1 I)u 1 = u0. In the present case, since there are two independent eigenvectors v 1 and v 2 corresponding to A 1 , we must set u0 = av 1 + /3v2, and choose a and /3 in order that the nonhomogeneous algebraic system

be consistent and have a solution u 1 .

r

Example 5.15. Consider the system = X = 3x y' z' = -x

+ y + 5y + y

-

-

2z

6z }

=� ) .

This system may be expressed in matrix form as x' = Ax, where and A = ( -� � -1 1

0

(5.24)

154

CHAPTER 5. LINEAR SYSTEMS

Then

det(>J - A)

,\-1 -1 2 ,\-5 6 3 1 -1 ,\

(,\-1)[,\(,\ - 5) + 6] + [3,\ - 6] (>. - 1)(,\ 2 - 5,\ + 6) + (,\ - 2)

+ 2[-3 - (,\ -5)]

(,\ - 1)(,\ - 2)(,\ - 3) + (,\ - 2)

(>. -2)[(>. - 1)(,\ - 3) + 1] (,\ -2)[>.2 - 4,\ + 4] = (,\ - 2)3 = 0

⇒ A1 = 2 is the only eigenvalue of A, of multiplicity 3. For ,\ 1 = 2, (A 1 I - A)v = 0 with v = ( � ) =>

=> a - b + 2c = 0. c = 0 => a = b => v = v 1 = (

i),

and a = 0 => b = 2c

=> v = v 2 = ( � ) are two independent eigenvectors which correspond to the eigenvalue ,\1 = 2. Thus, two independent solutions of the system (5.24) are

Since A has only two independent eigenvectors which correspond to the eigenvalue ,\ 1 of multiplicity 3, a third, independent solution of the system must be sought in the form

!)

where u 0 = av 1 + /Jv 2 is an eigenvector corresponding to ,\1 and u 1 is a solution of the algebraic system (A-,\ 1 l)u 1

= uo, i.e., a generalized eigenvector.

With u 1

= (

155

5.1. HOMOGENEOUS SYSTEMS

n

which is consistent if and only if a = (3. Then -a+ b - 2c = a, and a a= c

=0

⇒ b = 1 ⇒ u, =

(

and uo

=

�)

(

Hence,

=

1 and

Thus, three independent solutions of the system (5.24) are given by

and

x 1 ( t)

= e"

(

i),

X (t)

x, ( t)

= e"

(

!)

and x, ( t)

= (x 1 ( t) x, (t) X3 ( t)) = e21

(

is a fundamental matrix. The general solution is x(t)

C1X1(t)

= e

2t

i.e., x(l)

+ C2X2 (t) + C3X3(t)

(� � 0 1

= X(t)c

2t 3t � 1 ) ( �: ) = e l C3

= (c1 + c3t)e2 t. y(t) =

[c1

+ 2c2 + c

3

Example 5.16. Consider the system

{ �: : -� +

i;

= e"

3t 7 I )

( c + 2�

1

1

C2

: � t3t 3

+ C3 l

(3t + l)]e2t and z(t)

y

+

This system may be expressed in matrix form as x'

J

( 3t � 1 ) ,

= Ax, where

=

+ l)

(c2

) ,

+ c3t)e2t . (5.25)

156

CHAPTER 5. LINEAR SYSTEMS

Then



det(>.I - A) =

,\-1 0 0 ,\ - 1 0 0 1 -1 ,\-1

= (,\ - 1) 3 = 0

>-1 = 1 is the only eigenvalue of A, of multiplicity 3.

For .\1 = I, (.\ 1 1 - A)v = 0 with v = ( � ) =>

0. a = I and c = 0 => v = v 1 = (

=> a - b => v

=

v,

= (�)

i ),

and a

O and c = I

are two independent eigenvectors which correspond to the

n

eigenvalue ,\1 = 1. Thus, two independent solutions of the system (5.25) are x1(t)

= e'

( D

and x,(t)

= e'

(

Since A has only two independent eigenvectors which correspond to the eigenvalue ,\ 1 of multiplicity 3, a third, independent solution of the system must be sought in the form

where uo = av1 + f3v2 is an eigenvector corresponding to ,\1 and u 1 is a solution o� the algebraic system (A- ,\if)u 1 = u0, i.e., a generalized eigenvector. With u 1 = ( � )

157

5.1. HOMOGENEOUS SYSTEMS

which is consistent if and only if a= 0. Then -a+ b = (3, and (3 = 1 and a= c = 0

o) n.

=> b = 1 => u, = 0 ) and Uo = ( � ) Hence, x(, t) = e'

[l

( D

+

J = e' (

Thus, three independent solutions of the system (5.25) are given by

and

0 0 X(t) = (x1(t) x2(t) x,(t)) = e' ( � 0 1 )

is a fundamental matrix. The general solution is

1 t

Let A be a 3 x 3 matrix with only one eigenvalue >. 1 of multiplicity m1 = 3, and with only one independent eigenvector v 1 corresponding to >. 1 . Then one solution of the system x' =Axis given by A second, independent solution is sought in the form where, as before, u0 is an eigenvector which corresponds to ,\ 1 and u 1 is a generalized eigenvector which corresponds to >. 1, i.e., u 1 is a solution of the algebraic system (A - >. 1 I)u1 = u0. A third, independent solution is sought in the form

CHAPTER 5. LINEAR SYSTEMS

158

where uo, u1 and u2 are vectors to be determined. Then xI3(t) and, hence, x;(t) >-1e >.1t (

(2 1

= A 1eA1l

= Ax3(t)

t 2 uo

+ tu1 + u2 ) + e .), t( tu0 + u1 ) , 1

if and only if

}t uo + tu1 + u ) + e 2

2

>. 1 t

(tuo

(

+ u1) = Ae >.it }t 2u0 + tu1 + u2 ).

Division by e >.it =/- 0 yields the condition >-1

(�t 2uo + tu1 + u ) + (tuo + u1) = A (�t 2uo + tu1 + u ), 2

2

which holds for all real t if and only if >-1 uo

= Auo,

-

>-1u1 + uo

= Au1

and >-1u2

-

+ u1 = Au2.

(5.26)

The first two conditions in (5.26) are equivalent to (>.1I

A)uo = 0 and (A

>.iI)u1 = uo,

respectively, as before. The third condition in (5.26) is equivalent to

(A - >-1I)u2

= u1,

a solution u2 of which is also called a generalized eigenvector corresponding to the eigenvalue >.1 of multiplicity 3.

{

Example 5.17. Consider the system

= X + 2y = -x + 4 y y + =

z z z

x-(n A-(-i

i

X

1

y' z'

}

(5.27)

=D

This system may be expressed in matrix form as x' = Ax, where and

Then det(>.I - A)

-

1 >. 1 -2 1 >.-4 1 0 -1 >.-1 (>. 1)[(>.-4)(>.- 1) + 1] - [-2(>. - 1) (>. 1)(>.2 - 5>. + 5) + (2>. - 3) >.3 - 6>.2 + 12>.- 8 (>.

2) 3

=0

+1]

5.1. HOMOGENEOUS SYSTEMS

159

⇒ >. 1 = 2 is the only eigenvalue of A, of multiplicity 3. For ,\ 1

= 2, (,1 1/ - A)v = 0 with v = ( � ) ⇒

⇒ -b + c = 0 and a-2b + c = 0 ⇒ b = c and a = c, and c = 1 ⇒ v = v 1 = ( : ) the only independent eigenvector which corresponds to the eigenvalue >. 1 one solution of the system (5.27) is

= 2.

1s

Thus,

Since A has only one independent eigenvector which corresponds to the eigenvalue >. 1 of multiplicity 3, a second, independent solution of the system must be sought in the form x2(l) = e2\luo + u1), where u0

=

algebraic system ( A-,\ 1 I) u 1

(A - >-d)u1

!),

v1 is an eigenvector corresponding to >.1 and u 1 is a solution of the

= no, i.e., a generalized eigenvector.

With u 1

= (

= uo ⇒ a -1 2 -1 (-1 2 -1 ) ( b ) 0 1 -1 C

1 ( 1 ) 1

⇒ -a + 2b - c = 1 and b - c = 1, and c = 0 ⇒ b = 1 and a = 1 ⇒ U1 =

A third, independent solution must be sought in the form x3 (t) = e2

t ( tl2 uo + tu1 + u2 ) ,

(

i)



160 where Uo

= v1 = ( : )

and u 1

algebraic system (A - A 1 I)u2

= (

= u 1.

i)

CHAPTER 5. LINEAR SYSTEMS

are as above, and u 2 is a solution of the

With u2

= ( � ) , (A - A 1 I)u2 = u 1 cc>

n

=> b - c = 0 and -a+ 2b - c = 1 => b = c, and c = 1 => b = l and a = 0 =>

u,

= ( cc>

Thus, three independent solutions of the system (5.27) are given by

t+l

x 2 (t)=e2t ( t+l )

t

and x3 (t)=e 2t

and

is a fundamental matrix. The general solution is

t2 t ½2 + ( ½t +t+l ) , 2 +1 lt 2

161

5.1. HOMOGENEOUS SYSTEMS i.e.,

x(l)

[c1+c2 (t +1)+c3

(�t2 +t)] e

y(t)

[c1+c2 (l+l)+c3

(�t2 +t+1)] e

z(l)

[c1+c2 t+c3

(�t +1)] e 2

2t

, 2 t,

2t .

Example 5.18. Consider the system

{ :: : =�: : �� = �; } . z' =

(n

-x

+

y - 4z

5

(5.28)

This system may be expressed in matrix form as x' = Ax, where

x= Then

det(>.I - A)



and A=

c

2

-3)

-7 3 -8 -1 1 -4

3 >- + 5 -2 7 8 >- -3 1 -1 >-+4 (>-+5)[(>- -3)(>-+4)+ 8] +2[7(>-+4) - 8] +3[-7 - (>- -3)] (>-+5)(>-2 +>- - 4)+1U +28 (>.3 +6>. 2 +>- - 20)+ lU +28 >.3 +6>-2 +12>. +8 3 (>-+2) =0

>. 1 = -2 is the only eigenvalue of A, of multiplicity 3.

o =n H n o ) o -! =D u ) o ) ⇒

c; b-3c = 0 and 3a-2b+3c = 0 c> b = 3c and a= c, and c = 1 c> v = v1 = ( � ) is the only independent eigenvector which corresponds to the eigenvalue >-1 = -2. Thus,

162

CHAPTER 5. LINEAR SYSTEMS

one solution of the system (5.28) is

Since A has only one independent eigenvector which corresponds to the eigenvalue >. 1 of multiplicity 3, a second, independent solution of the system must be sought in the form where u0 = v 1 is an eigenvector corresponding to >.1 and u 1 is a solution of the algebraic system ( A - A 1 I) u 1 = Uo, i.e., a generalized eigenvector. With u 1 = ( � }

(A - >.1I)u1 = uo



A third, independent solution must be sought in the form

where u0 = v 1 = ( � ) and u 1 = ( � ) are as above, and u2 is a solution of the algebraic system (A- A,J)u2 = u 1. With u2 = ( � ) , (A - A,I)u, = u, =>

163

5.1. HOMOGENEOUS SYSTEMS ⇒ -b + 3c = 1 and -3a + 2b - 3c = 1, and a = 0 ⇒ b = 2 and c = 1 ⇒ u2 =

( 0�)

Thus, three independent solutions of the system (5.28) are given by

and

l+l 3t + 2 l

is a fundamental matrix. The general solution is

i.e.'

2

(tt2 t)]

2

e- i,

x(t)

[c1 + c (t + 1) + c3

y(l)

[3c1 + c (3l + 2) + c3 (�l + 2l+ 2)] e-

z(t)

[c1 + c l+ c3 (�l + 1)] e-

2

2

+

2

2

2

2

t,

t.

Let A be a 4 x 4 matrix which has only one pair of complex-conjugate eigenvalues and "X1 of multiplicity 2, and with only one pair of independent eigenvectors v 1 and v1, corresponding to >.1 and °X1 , respectively. Then two complex-conjugate solutions of the system x' = Ax are given by >.1

164

CHAPTER 5. LINEAR SYSTEMS

from which two real, independent solutions can be obtained as

as demonstrated earlier. In the present case, it is necessary to obtain two additional real, independent solutions. Hence, a second pair of complex-conjugate solutions is required, and they are obtained by means of complex generalized eigenvectors. Thus, seek a complex solution in the form

where, as before, uo is an eigenvector and u 1 is a solution of the algebraic system (A - >. 1 J)u 1 = u0, i.e., a generalized eigenvector. Then z 2 and z2 are two complex solutions, and provide the two additional real, independent solutions

for a total of four real, independent solutions. Example 5.19. Conside the system

r,

y' z'

w'

= -y + z = X + w = -w = z

}

This system may be expressed in matrix form as x' = Ax, where

Then, expanding the 4 x 4 determinant along the fourth row, we obtain det(>.I - A)

1 -1 0 >. 0 -1 >. 1 0 0 -1 >. >. 1 -1 >. 1 0 0 -1 >. -1 + >. -1 >. >. 0 0 1 0 0 2 2 (>. + 1) + >.[>.(>. + 1)] 2 (>. + 1) 2 = 0 >. -1 0 0

(5.29)

5.1. HOMOGENEOUS SYSTEMS

165

⇒ .\ 1 = i and .\ 1 = -i are the only eigenvalues of A, of multiplicity 2. For

)'1

= i,

(!. 1 1 - A)v

=0

with v

= (;)

=>

Row reduction of the coefficient matrix gives i 1 -1 0 i 1 -1 0 i 1 -1 -1 i 0 -1 0 0 -1 -i --) 0 0 -1 ( ) --) ( ) ( 00 i 1 00 i 1 0 0 0 00 -1 i 0 00 0 0 0 0

⇒ d =0 ⇒ c =0 and ia + b =0, and a= i ⇒ b = 1 ⇒

is an eigenvector which corresponds to the eigenvalue .\ 1. Then

166

CHAPTER 5. LINEAR SYSTEMS

are two complex-conjugate solutions and, hence,

- sin(t) cos t x 1 (t) = !R(z 1 (l)) = ( �( ) )

cos(t) sin t and x2 (t) = '.:r(z 1 (t)) = ( �( ) )

are two real, independent solutions of the system (5.29). A generalized eigenvector corresponding to )11 is a solution u 1 of the algebraic system

(A� .\ 1 I)u 1 = tJo = v 1, i e , with u 1 = ( ; ) ,

Row reduction of the augmented matrix gives

c� c�

-1 -1,

0 0 -1

1 0 -1,

1

0 1 -1 -1,

1 0 i 1 0 0 0 -2 0 0 0

I ;) c� I ) I 1 I o I o

I I

I

-1 0 0

0

1 1 -1,

0

0 -1

0

2:) 0

0

-

2: -2

o

⇒ d = l, c +id= 2i and -ia - b + c = i ⇒ c = i and ia + b = 0, and b = 0 ⇒ a = 0

!)

167

5.1. HOMOGENEOUS SYSTEMS

=> u1 = (

is a generalized eigenvector, and

and z2 (t) are two additional complex-conjugate solutions. Hence,

are two additional real, independent solutions of the system 5 ( .29),

X(t) = (x 1 (t) x2 (t) x 3 (t) x.,(t)) =

- sin(l) cos(t) -tsin(t) tcos(t) sin(t) t cos(t) tsin(t) cost ) ( - sin(t) cos(t) 0 �( ) sin( t) cos( t) 0

168

CHAPTER 5. LINEAR SYSTEMS

is a fundamental matrix, and x(t)

C1X1(t) + C2X2 (t) + C3X3(t) + C4X4(t) X(t)c c1 -sin(t) cos(t) -tsin(t) icos(t) cos(t) sin(t) tcos(t) tsin(t) ) ( :: ) ( c 0 0 -sin(t) cos(t) cos(t) sin(t) 0 0 4 -C1 sin(t) + C2 cos(t) -C3 t sin(t) + c4t cos(t) c1 cos(t) + c2 sin(t) + c3t cos(t) + c4 t sin(t) ) ( -c3 sin(t) + c4 cos(t) C3 cos(t) + C4 sin(t)

i.e.'

x(t) · y( t) z(t) w(t)

-c1 sin(t) + c2 cos(t) - c3t sin(t) + c4 t cos(t), c 1 cos(t) + c2 sin(t) + c3 t cos(t) + c4 t sin(t), -c3 sin(t) + c4 cos(t), c3 cos(t) + C4 sin(l)

is the general solution.

Exercises 5 .1

= -5x + 6Y } . 1. Find the general solution of the system { x; y=- 3x+ 4 y

2. Find the general solution of the system { :: : ��; y } . 3. Find the general solution of the system x'

= Ax, where A=

(

t � ).

4. Find a fundamental matrix for the system x' = Ax, where A = ( � 5. Find a fundamental matrix for the system x'

= Ax, where A=

= x - Y } , (O) 6. Solve the initial-value problem { x; x y = 2x+ 4 y 7. Find a fundamental matrix for the system x'

= 2,

. � ) ,

( �

y(O)

=

� )

.

1.

= Ax, where A = ( _ �

-� ) .

5.1. HOMOGENEOUS SYSTEMS 8. Find the general solution of the system

{

' = 2y }. x y' = 2x

t:

10. Find a fundamental matrix for the system x'

= Ax, where A = (

1

= Ax, where

1 . Find a fundamental matrix for the system x'

1 3.

}

�2x + 3y z' = -x+y+ 2z

9. Find the general solution of the system {

1 2.

169

A

=

= 3 Find the general solution of the system { x; - x + Y } . y=- 4x- 3y Find the general solution of the system x'

= Ax, where A=

14. Find a fundamental matrix for the system x'

-!

-1 -3 5

(

( � 1

= Ax, where A=

(

=! ! � ) .

Solve the initial-value problem x' = Ax, x(O) = x0, where xo = (

1 6.

x' =x Find the general solution of the system { y' = 2x + 2y - 2z } . 1 Z = X +y

1 7.

1 8.

furr:r

Find the general solution of the system x'

Find a

A=

(

0 1 -

1

.o

=

r trix for the system x'

1 0 O 0

)

{

X

1

y'

!1 ) and

0 0 0 1 Ax, where A= ( -1 2

= Ax, where

.

19. Find the general solution of the system

=� ) .

i -� ) .

1 5.

A= ( � -� ).

=� ! ) .

=

X

=y

+ 3y

}

170

20. Find the general solution of the system x'

= Ax, where A=

21. Find a fundamental matrix for the system x'

(

= Ax, where A=

x' = 3x -y 22. Find the general solution of the system { y' = .1: + 2y - z z' = 3y - z

}

23. Find the general solution of the system x' = Ax, where A= (

24. 25.

i daf

:: i

::

(

(

a

�rr�r

r-r�1 -1 1 -4

2 -1

matrix for the system x'

= Ax, where

matrix for the system x'

= Ax, where

)

-4

)

26. Find the general solution of the system ( ;: ) z' 27. Find the general solution of the system x'

(

i d th n of the system x' � :� · r r-rr) (

0 0 -1 1

(



� �

)

.

i �).

=� � � ) ( ; ) . 0 0 -1

= Ax, where A=

= Ax, where

z

. (-� -1-0� 0 � ) �-� � 1 1 1 -

28. Find the general solution of the system x' = Ax, where A = ( 29.

=1 ; ) .

CHAPTER 5. LINEAR SYSTEMS

).

5.2. NONHOMOGENEOUS SYSTEMS

5.2

171

Nonhomogeneous Systems

Theorem 5.6. The general solution of the nonhomogeneous system x' = A(t)x +

f(t)

· (5.30)

is x(t) = xv(i) + x1i(t), where xv(t) is any particular solution of (5.30), and x1i(t) is the general solution of the associated homogeneous system

x' = A(l)x,

(5.31)

X1i(l) = C1X1(l) + C2X2(l) + · · · + CnXn (l), where x1(l), x2(t), · · ·, xn (l) are n independent solutions of (5.31).

Proof. x(l) = xv(l) + x1i(t) is a solution of the system (5.30) because x'(l)

[xv(t) + x1i(t)]' x�(t) + x;i (t) A(t)xv(t) + f(i) + A(l)x1i(t) A(l)[xv(l) + x1i (l)] + f(l) A(l)x(t) + f(t).

Since x(t) contains n arbitrary constants, it is the general solution of (5.30). Let X(t) be a fundamental matrix for the homogeneous system (5.31), and seek a particular solution of the nonhomogeneous system (5.30) in the form

xv(t) = X(t)u(t),

(5.32)

where u(t) is a vector function to be determined. Then

x�(l) = X'(t)u(l) + X(t)u'(l), and, hence, (5.32) is a solution of (5.30) if and only if

X'(t)u(l) + X(t)u'(l) = A(t)X(t)u(t) + f(t).

(5.33)

Since X'(t) = A(t)X(t) by Theorem 5.3 in Section 5.1.1, (5.33) is equivalent to

A(l)X(t)u(l) + X(t)u' (t) = A(t)X(l)u(l) + f(t), i.e.'

X(t)u'(t) = f(t),

172

CHAPTER 5. LINEAR SYSTEMS

left multiplication of which by X(t)- 1 gives u'(t) = X(tt 1 f(t) and, hence, u(t) =

J

X(t)- 1 f(t) dt.

(5.34)

Thus, Equation (5.32), with u(t) defined by (5.34), gives (5.35) and determines a particular solution xp (t) for the nonhomogeneous system (5.30). Analogous to single nonhomogeneous equations, this method is called variation of parameters for systems. Note that the integrand in (5.35) is an n-vector. The integral of a vector-valued function v(t) is defined as the vector-valued function the components of which are the integrals of the components of v(t), i.e., v(t) =

=>

Vn (t)

j

v(t) dt =

f V1 (t) dt f v2(t) dt

The general solution of the nonhomogeneous system x' = A(t)x + f(t) is then Xh(t) + X p(t)

x(l)

X(t)c + X(t) J X(t)- 1 f(l) dt

X(l) [c +

J

X(lt 1 f(t) dl]

As in variation of parameters for single equations, the addition of arbitrary constants of integration is not necessary, since they may be incorporated into the arbitrary constant vector c. Example 5.20. Consider the nonhomogeneous system

{

x' y'

= -5x + 4y + 2t = -4x + 5y + t }

(5.36)

This system may be expressed in matrix form as x' =Ax+ f(t), where x=

( ) X y

, A=

( -5 _

4)

4 5

and f(t) =

( 2t ) . t

5.2. NONHOMOGENEOUS SYSTEMS

173

For the associated homogeneous system x'

=

det(-XI - A)

=

I A: 5

A

-=._\

(-X - 3)(-X + 3)

= Ax,

I = (-X + 5)(-X - 5) + 16 = ,X

2

-

g

=0

⇒ -X1 = 3 and A2 = -3 are the eigenvalues of A.

⇒ 2a - b = 0, and a= l ⇒ v = v1 = ( � ) the eigenvalue -X1

= 3.

⇒ a - 2b = 0, and b = l ⇒ v = v2 = ( � ) the eigenvalue -X2

is an eigenvector which corresponds to

=

is an eigenvector which corresponds to

-3.

Hence, two independent solutions of the homogeneous system are given by 2e-3t 1 e3t 2 ) , x1 (t) = e3t ( ) = ( 3 ) and x2(t) = e-3t ( ) = ( e-3t 1 2e t 2 and is a fundamental matrix. Then

J

O 2e-3t 2t ) ( l ) = ( 3t ) ' te -e3t

X(t)- 1 f(t)

0

X(t)- 1 f(l) dt ::::}

Xp(l)

2e-3l ), -e3t

X(t) � 3t e g

J

1

0

3l ( dt = ( e ) ) = g 3t - 1 ½te3t - ! e 3t

1 e3t 2e-3t O X(t)-1f(t) dt = ( 3t ) e3t ( ) 3 2e e- t 3t _ 1 9 2(3t - l)e-3t � ( 6t - 2 ) ( 3 ) = (3t - 1 )e- t 9 3t - 1 '

174

CHAPTER 5. LINEAR SYSTEMS

and the general solution of the nonhomogeneous system (5.36) is xh(t) + Xp (t) = X(t)c + Xp (t) 2e t ) ( �: ) + t ( ( ;::t e _�:

x(t)

�!

=� ) '

i.e.'

x(t) y(t) Example 5.21. Consider the nonhomogeneous system x' = x - 2y { y' = 2x + y

+ e t cos(2t) + e t sin(2l) }

(5.37)

This system may be expressed in matrix form as x' =Ax+ f(l), where x=

( ) y

X

, A=

(

1 2

-2 ) 1

and f(t) = e

For the associated homogeneous system x' = Ax, det(.U - A)

I ,,\-- 1 2

,,\

-

2

-

2 ,,\-1

(

COS (2t)

sin(2t) ) .

I = (,,\ - 1)2 + 4

2,,\ + 5 = 0

⇒ ,,\, ,,\ =

2±FI6 = 1 ±2i are the eigenvalues of A.

For,,\= 1

+2i,

2

t

(,,\1 J - A)v = 0 with v = ( � ) =>

=> ia + b = O,· and a= 1 ⇒ v = ( � ) is an eigenvector which corresponds to the i �igenvalue ,,\ = 1 +2i .

. Hence, a complex solution of the homogeneous system is given by z(l) = e t [cos(2t) +i sin(2t)] [ ( �)

+i ( �l ) ] ,

175

5.2. NONHOMOGENEOUS SYSTEMS two real, independent solutions are given by x 1 (l)

i [cos(2l) ( � ) - sin(2t) ( �l ) ]

x2 (t)

e

l

[ sin(2i) ( � )

and X(t)

=e

+ cos(2l) ( �l ) ] = e

.

= (xi(l) x2 (i)) = e

l

(

e

J

X(l)- 1 f(l) X(l)- 1 f(l) di ⇒

Xp(t)

e

cos(2t) sin(2l) l ) sin(2t) -cos(2l) e

-t

X(t) e

t

(

(

cos(2t) - sin(2l) -l = - sin(2t) cos(2t) ) e

-t

J

dl = ( �)

X(tt 1 f(l) dt =

t

cos(2t) sin(2t) sin(2t) -cos(2l) )

is a fundamental matrix. Then

_ (( J ( �)

( ��;g:j ) , ( _ ;�:g�� ) ,

t

e

l

(

(

cos(2t) sin(2t) sin(2t) - cos(2l) ) '

1 cos(2l) = ( ) ) 0 ' sin(2t)

t sin(2t) c�s(2t) ( ) sm(2t) -cos(2l) ) 0

l cos(2t) t sin(2t) ) '

and the general solution of the nonhomogeneous system (5.37) is x(i)

(

x1i (i) + Xp (l) = X(t)c + Xp(t) c1 t cos(2t) sin(2t) ) ( c ) e sin(2t) -cos(2t) 2

t

+ e

t

(

cos(2t) sin(2t) ) '

1.e.' l [c1 cos(2t) + c2 sin(2t) + t cos(2t)], et e [c1 sin(2l) - c2 cos(2t) + t sin(2l)].

x(l)

y( l)

Example 5.22. Consider the nonhomogencous system {

x' = -5x + 9y + y' = -4x + 7y +

=� � )

e

t

e

t

}

(5.38)

This sys tem may be expressed in matrix form as x' =Ax+ f(t), where x = ( � ) , A= (

and f(t) =

e

t

( �) .

CHAPTER 5. LINEAR SYSTEMS

176 For the associated homogeneous system x'

I

det(,U -A)

>, ;

5

;,

-_\

= Ax,

I = ( A + 5) (A - 7) + 36 = ,),2 - 2A + 1

(>--1)2 =0 ==> >-1

= 1 is the only eigenvalue of A,

==> 2a - 3b

= 0,

and a= 3 ==> v

to the eigenvalue A 1

of multiplicity m 1

= v1 = ( � )

= 2.

is an eigenvector which corresponds

= 1.

Hence, one solution of the homogeneous system is given by

A second, independent solution takes the form

where u0 = v 1 is an eigenvector which corresponds to >- 1 , and u1 is a solution of the algebraic system (A - >- 1 I)u1

==> 3b-2a

= 1,

and a

=

= uo.

l ==> u 1

With u 1 = ( � ) , (A - >-1I)u1

=( �)

==>

and

X(l)

= (x1(t) x2(t)) = et

(

3 3i + l ) 2 2t + 1

= uo

==>

5.2. NONHOMOGENEOUS SYSTEMS

177

is a fundamental matrix. Then

J

X(l)- 1 f(t) X(t)- 1 f(l) dt

and the general solution of the nonhomogcneous system (5.38) is x1i(t)

x(t)

=

+ xp(t) = X(t)c + xp (t)

t i ( � �� : � ) ( �: ) + e

(

i.e.'

�s:/ ) '

x(l)

el [3c1

+ c2 (3t + 1) + �t 2 + t]

y(t)

e l [2c 1

+ c2(2t + 1) + t2 + t] .

,

Definition 5. 7. An initial-value problem for a nonhomogeneous system consists of a system x' = A(t)x+f(l), together with an initial condition x(t0 ) = x0, which requires that the solution x(t) be equal to x0 at t = t0, and determines the arbitrary constant vector c. Theorem 5. 7. Let A ( l) and f ( l) be continuous on an interval J containing the point t0 in its interior. The solution of the initial-value problem x'

= A(l)x + f(t),

x(to)

is given by

x(t)

= xo,

X(t)X(t0)-1 x0 + X(l)

1l

X( l) [ X ( t 0 )- x0 +

X ( s )- 1 f ( s) ds]

1

1:

lo

X(s)- 1 f(s) ds

where X(t) is a fundamental matrix for the associated homogeneous system.

178

CHAPTER 5. LINEAR SYSTEMS

Proof. The general solution of the nonhomogeneous system may be expressed in terms of a definite integral as 1

x(t) = X(t)c + X(t) [ X(s)- 1 f(s) ds. lto

Then x(to) = xo ==>- xo = X(lo)c ==>- c = X(t0 )-1x0. Note that the definite integral of a vector-valued function is defined in a manner analogous to the definition of the indefinite integral:

1: v1(s) ds 1: v2(s) ds

v(t) = Vn (t)

Example 5.23. Consider the initial-value problem {

x ' = -x y' =

+ 2Y + et } x(O) = 2 y(O) = 1. ' y + et '

(5.39)

This problem may be expressed in matrix form as x' =Ax+ f(t), x(O) = xo, where

For the associated homogeneous system x' = Ax, det(,U - A) =

⇒ >- 1

I

>,; 1 -_2 >, 1

I = (>- + 1)(>- - 1) = 0

= 1 and >- 2 = -1 are the eigenvalues of A.

⇒ a= b, and a= 1 ⇒ v = v 1 = ( � ) is an eigenvector which corresponds to the eigenvalue >. 1 = 1.

5.2. NONHOMOGENEOUS SYSTEMS

⇒ b = 0, and a = 1 ⇒ v = v 2 = (

eigenvalue >-2 = -1.

t)

179

is an eigenvector which corresponds to the

Hence, two independent solutions of the homogeneous system are given by

and is a fundamental matrix. Then

Q -l -e -t = ) ( el � et ) ' el

Q -( - el

( �t t

��l ) ( :: ) t

t)

(

ds = ( ; )

t)'

i: ( � -i ) ( � ) i ) '

1

(

=

=

( � ),

(

(i)+(i)=( : ) t

⇒ x(t)

X(l) ( :

:

[x(o)- 1 x(O) + 1 ; ) (

e

t

t

;

1

(t + l)et + e -t ( ) (t + l)et

l

t

X(s)- 1 f(s)ds]

)

i.e., x(t) = (l + l)et + e-t, y(t) = (l + l)et is the solution of the initial-value problem (5.39). Example 5.24. Consider the initial-value problem

x' = -2x + 4y + 4e-2 t { y' = -x - 2y + 2e- 2t }

'

x(O) =2 ' y(O) = -1.

(5.40)

180

CHAPTER 5. LINEAR SYSTEMS

This problem may be expressed in matrix form as x' =A x+ f(t), x( O) = x 0, where x=

( ) y X

, A=

(

-2 2 e-2t _ 1 _ 42), f ( t) = (42) and xo =(_ 1) .

For the associated homogeneous system x' = Ax,

-

4±FT6 ⇒ ,\, ->,. = ---= -2± 2i are the eigenvalues of A. 2 ⇒ For,\= -2 + 2i, (>-.J -A)v = 0 with v = � ( )

⇒ a+ 2ib = O, and b = i ⇒ v =

eigenvalue,\= -2 + 2i.

: is an eigenvector which corresponds to the ( )

Hence, a complex solution of the homogeneous system is given by

z(l) = e-2t[cos(2t) + i sin(2t)] [(� ) + i ( � ) ] , _2

l

2cos(2 ) O , =e _2t -sm(2t) t[cos(2·t)(02). (_ sin(2l)) (1 )]

x 1 (t)

e

x 2 (t)

e- 2 t [sin(2t) � + cos(2t)(� )] = e () -

2

t

( ��:g�� ) 2

arc two real, independent solutions, and X(l) = (x1(t) x2(t)) = e _2

t(

2cos(2t) -sin(2l)

2sin(2t) cos(2t))

181

5.2. NONHOMOGENEOUS SYSTEMS is a fundamental matrix. Then

�e2t 2

( cos(2l)

-2sin(2t) sin(2l) 2cos(2t)) cos(2t) -2sin(2t) = �e2 t sin(2t) 2 cos(2t)) 2 2 cos(2t) - 2sin(2t) ( 2sin(2t) + 2cos(2t)) 2cos(2s)-2sin(2s) 2 sin(2s) + 2cos(2s)) 0

(

lt(

'

e-2t

(

4 2)

d8

sin(2s) + cos(2s) t ( -cos(2s) + sin(2s)) I0 sin(2t) + cos(2t) - 1 ( -cos(2t) + sin(2t) + 1 )

1 ( � � ) ( !1 ) = ( !1 ) ' X(o)-1x(O) +

1t

X(st1f(s) ds =

⇒ x(t)

sin(2t) + cos(2t) - 1 1 ) + ( -cos(2t) + sin(2t) + 1 ) ( -1 sin(2t) + cos(2t) ( - cos(2t) + sin(2t))

X(t) [ X(o)- 1 x(O) + e _2t e-2t

(

1t

X(s)-1f(s) ds]

sin(2t) + cos(2t) 2cos(2t) 2sin(2t) ) ( - sin(2t) cos(2l) -cos(2t) + sin(2t))

( !1

) ,

i.e., x(t) = 2e-2t, y(t) = -e- 2 t is the solution of the initial-value problem (5.40). Example 5.25. Consider the initial-value problem

{ :: : x _ �� : � } , x(O) = 3, y(O) = 2.

=� ) '

(5.4 ) 1

This problem may be expressed in matrix form as x' =Ax+ f(t), x(O) = x0, where X

= ( :) ' A = ( �

f(t) = ( � )

and

Xo

= ( �) .

182

CHAPTER 5. LINEAR SYSTEMS

For the associated homogeneous system x' = Ax, det(,\J - A)

I :1 ,\

!

2 / 2 (,\ + 1) = 0

= ,\(,\ + 2)

+ 1 = ,\2 + 2,\ +

1

=> A.1 = -1 is the only eigenvalue of A, of multiplicity m1 = 2. For A.1 = -1, (,\if - A)v = 0 with v = ( � ) =>

=> a= b, and a= l => v = v 1 = ( � ) is an eigenvector which corresponds to the eigenvalue ,\ 1 = -1. Hence, one solution of the homogeneous system is given by

A second, independent solution takes the form where u0 = v 1 is an eigenvector which corresponds to ,\ 1 , and u 1 is a solution of the algebraic system (A - ,\ 1 J)u 1 = uo. With u 1 = ( � ) , (A - A1I)u1 = uo =>

=> a - b = l, and b = 0 => u 1 = ( � ) =>

and

5.2. NONHOMOGENEOUS SYSTEMS is a fundamental matrix. Then X(l)- 1

= -el

X(l)- 1 f(l) 1

l

(

i

(



183

1

l l = �/ ) ( � ) � l

= (t -t)e l -(2s-l)e8 2

= (l2

1

1

l

(2 - s)e8 ds

), i ( ; �: ) , �

(s2 -s)e8 [- (2s-l)e8 ds 1

(s2 -s)e8 ds

l

(

- -l = et \ )

-

= (t2 (l2

X(s)- 1 f(s) ds

l

e8

(l 2

el

1

ds

3l 3)el 3 ) (3 - l)et 3 . -

( � -� ) ( � )

=

( � ) , (

+

(t2

(3 - l)e

= e -l

- -

i.e, x(t) = 6- l 3e- l problem (5.41).

(

[x(o)- 1 x(O) +1 1 t+ 1 t

1

) (

- -

(l2

t

-

3t + 3)e l 3 ) (3-t)e l 3 -

--

2 el ( ( t - 3l + 3) t

x(l) = X(t)

i:

ds

2 ( 1 )



t

t)et (2t-l)el -1 +2et -2 - 3l 3)el 3,

-

(2-t)el 2 (3-t)e l -3,

1

�/

+2 o e8 ds f

(2t-l)el -1 + 2e8

(2 - s)e8

(

X (o)- 1 x(O) =

+ i: + fo\s - + ( � �: ) + --

l)e l

i:

t

l

2

1

)

X(s)- 1 f(s) ds]

3t + 3)et -1 ) (3 - t)e l 2 -

)

-

( 6 - l 3e- t 2te- t ' 3 - e l - 2lc-t 2le- t , y(t) = 3- e- t 2te- l is the solution of the initial-value

184

{

CHAPTER 5. LINEAR SYSTEMS

Example 5.26. Consider the initial-value problem

= y' = z' =

X

1

X

- z + tcos(t) , x(O) = 1, y(O) = 2, z(O) = 1. (5.42) 3x -x + y + tsin(t) }

This problem may be expressed in matrix form as x' =Ax+ f(l), x(O) = x0, where x=

� � -� , f(t) = : , A= tco�(l) ( ) ( ) ( ) z -l 1 0 tsin(l)

For the associated homogeneous system x' = Ax, det(>.J - A)

>. - 1 0 0 >. 1 -3 1 -1 >.

⇒ >.1 = 1 and >.2 , "X2 = ±i are the eigenvalues of A. For ,\ 1 = 1, (,\ 1/

-

A)v = 0 with v =

( � ) cc>

cc; b = 2c and a = c, and c = 1 cc> v = v 1 =

(i-1 -3 1

(� )

cc>

0) Jl 0) 0)

n

cc>

and Xo =

(

�l) ·

5.2. NONHOMOGENEOUS SYSTEMS

185

⇒ a = 0 and b = ic, and c = 1 ⇒ v = v2 =

(: )



z(l) = [cos(t) + isin(t)] [ ( � )

+

is a complex solution of the homogeneous system, x2 (t)

i(

! )]

cos(t) ( � ) - sin(t) ( � ) = (- s�n(t) ) , 1 cos(t) 0

x3 (l) = sin(t) ( � ) 1



+ cos(l) ( � ) = ( c� (l) ) 0

sm(t)

are two real, independent solutions, and

is a fundamental matrix. Then (see Appendix A.3 for inversion of matrices) X(lt

1

=

X(l)- 1 f(l) =

0

� ( 2sin(t): cos(t) - sin(t) co (l) ) , - sin(l) - 2cos( t) cos(t) sin(t) 0 2sin(t): cos t sin(t) co�(l) ) ( t co�( t) ) = ( � ) , ( ) ( - sin(t) - 2cos(t) cos(t) sin(t) l sin(t) t

186

CHAPTER 5. LINEARSYSTEMS ⇒ x(t)

1 1 1.e., x(t) = et , y(t) = 2et + t2 cos(t), z(t) =et + t2 sin(t) is the solution of the

initial-value problem (5.42).

2

2

Exercises 5. 2

' = 2x + 3e2t 1. Find the general solution of the system { x, y =x- y+et

}.

x'=4y+4cos(2 t) } 2. Find the general solution of the system { , . . y = -X - 2 Slll(2t )

x' = -x + 4y + t2 et 3. Find the general solution of the system { , y = -x + 3y + tet }. 4. Find the general solution of the system { x; = 2x - 2Y y =x+4y+ 2

x =x+e 5. Solve the initial-value problem { ; t y = 2y+ 2e2 6. Solve the initial-value problem {

n'



t } , x(0) = 3, y(0) = 2.

x; = -y 1 + } , x(0) = 2, y(0) = 1. Y = x+ 1

n'

x' = -2x - y + e- 4 t 7. Solve the initial-value problem { , _ t y = 4x- 6y+ 2e 4

UJ

}

,

x(0) = 0, y(0) = 1.

8. Solve the initial-value problem x' =Ax+ f(l), x(0) =xo, where X

= (

A=

(

-

� -[

Chapter 5 Exercises

f(l) = e'

and Xo

= (

D

1. A differential equation of order n 2:: 2 for the unknown function y(i) can be transformed into a system of n first-order equations for the unknown functions X1(t), X2(t), · · ·, Xn (l) by letting X1 = y, X2 = y', X3 = y", ·· ·, Xn = y( n -l)_ Consider the equation y" - 2 y'+y =0.

CHAPTER 5 EXERCISES

187

(a) Find two independent solutions y 1 (l) and y2 (t) of the equation, and the general solution y(l). (b) Transform the equation into a system of two first-order equations. (c) Find two independent solutions x 1 (l) and x2 (l) of the system in part (b), and the general solution x 1 (l), x 2 (l). Confirm that x 1 (t) is equivalent to y( l).

(d) Compute the Wronskians W[y1 (l) y2 (t)] and W[x 1 (t) x2 (l)] and compare the results. 2. Let X (t) be a fundamental matrix for the system x' = Ax. (a) Find a fundamental matrix Y(l) such that Y(O) = I, the identity matrix, and express the solution of the initial-value problem x' = Ax, x(O) = x0, in terms of Y (t).

(b) Find a fundamental matrix Z (l) such that Z ( t0 ) = I, where t0 is a real number, and express the solution of the initial-value problem x' = Ax, x(t 0) = xo, in terms of Z(l). 3. Consider the system

{

x' , y

= 2x = x- 2 y

}

.

(a) Find the general solution of the system. (b) Solve the first equation in the system for x(t), substitute into the second equation for x, and solve the second equation for y(l). Confirm that your answer is equivalent to the one in part (a). (c) Differentiate the second equation in the system and eliminate x and x' in order to obtain a second-order equation for y(t). (d) Solve the equation in part (c) for y(t), determine your solution is equivalent to the one in part (a).

x(l),

and confirm that

=:

4. An n x n matrix A is diagonalizable, i.e., there exists a nonsingular matrix P such that p- 1 AP = D is a diagonal matrix, if and only if A admits n independent eigenvectors. Then D has the eigenvalues of A along its diagonal, and the columns of Pare the eigenvectors of A. Consider the system x' = Ax, 1 � ). where A = ( (a) Find the general solution. (b) Determine a new unknown y in terms of x such that the system x' for x transforms into an equivalent system y' = Dy for y. (c) Solve the system in part (b) for y(l).

=

Ax

188

CHAPTER 5. LINEAR SYSTEMS (d) Determine x(t) from y(t) in part (c) and confirm that your answer 1s equivalent to the one in part (a).

0 1 5. Find a fundamental matrix for the system x' � Ax, where A � ( � 0 0 by generalizing the method employed in Examples 5.15 and 5.16.

6. Find a fundamental matrix for the system x'

= Ax, where A = (

H

0 0 1 0

n ),

0 0 0 1 by generalizing the method employed in Examples 5.17 and 5.18 and taking

In general, the subscripts of the vectors

from left to right, are n = 0, 1, 2, tn 3, · · •, and their coefficients, from right to left, are , n = 0, 1, 2, 3, · · ·. The 1 n. latter sequence of functions has a special significance, and will be encountered in Part II of this book. Un,

7. Let >. = a + i/3 be a complex eigenvalue of an n x n matrix A, with the corresponding complex eigenvector v = a + ib. Show that the corresponding two real solutions are linearly independent on R

Part II Infinite Series

189

Chapter 6 Sequences and Series Infinite sequences and series arise in such numerous applications that neither science nor engineering would be possible in its present form without them. Sequences and series are defined in the present chapter. More specialized forms of series and some of their applications will be discussed in subsequent chapters. All of the linear differential equations solved in Part I of this book are equations with constant coefficients, with the exception of Cauchy-Euler equations, which are of such a special form that they can be transformed into equations with constant coefficients. In general, an equation with variable coefficients such as

y" + xy' + y = 0

(6.1)

can be solved only by means of infinite series. In the final chapter of this book, we shall have gathered sufficient knowledge to solve Equation (6.1).

6.1

Sequences

Definition 6.1. Let f be a real-valued function and k an integer. The ordered set of values off at the integers k, k + l, k + 2, · · ·, i.e., {J(k), f(k +

1),

J(k + 2),

· · ·},

is called a sequence of real numbers. The integer k is the initial value of n. It is customary to let J(n) = an (or bn or en ), called the general term of the sequence, and to express the sequence' as or, more concisely, as {an } �=k . 191

192

CHAPTER 6. SEQUENCES AND SERIES

Example 6.1. The ordered set of real numbers

! ! ! ... }

{1 ' 2' 3' 4' is a sequence with a n =

� n

and k

= 1.

00

It may be expressed as { � } n n=l

Example 6.2. The ordered set of real numbers

is a sequence with an =

1 1 . and k = 0. It may be expressed as { n } 2 n=O 2 n

Example 6.3. The ordered set of real numbers { -4, -2, 0, 2, 4, 6, · · ·} is a sequence with an = 2n and k

= -2.

It may be expressed as { 2n}�=- 2.

Example 6.4. The ordered set of real numbers {2, 3, 4, 5, · · ·} is a sequence with an = n and k = 2, or an = n + 1 and k = 1, or an = n + 2 and k = 0. Thus, the representation of a sequence as {an}�=k is not unique. Hence,

but { n}�=2 is the simplest. Example 6.5. The ordered set of real numbers {1 ' -1 ' 1 ' -1 ' 1 ' -1 ' · · ·} is a sequence with an = (-lt and k and k = 1.

= 0,

or a n= (-lt and k

Example 6.6. The sequence {an}�=o, where an =

n v'n+I' is n+l

= 2, or an

=

(-1r- 1

193

6.1. SEQUENCES Example 6. 7. The Fibonacci sequence is defined by

The first few terms are

{1, 1, 2, 3, 5, 8, 13, 21, 34, · · · }. Such a sequence, where the terms arc not given explicitly in terms of n, but in terms of preceding terms, is said to be defined inductively. Given a sequence { an }�=k' it is frequently the case that the initial value k of n is irrelevant. In that case, the sequence is expressed simply as {an }. One such instance is when it is necessary to determine whether or not the terms an approach a unique, finite number as n is increased indefinitely. Definition 6.2. Given a sequence { an }, lim an= L

n-+oo

means that, the absolute value of the difference (i.e., the distance) between an and L can be made arbitrarily close to O by taking n sufficiently large. More rigorously, for any E > 0, however small, there exists an integer N, which depends upon E, such that n 2:'. N => Ian - LI < E.

If lim an= L, the sequence { an } is said to converge to L, and Lis called the limit of n-+oo the sequence. The notation an ---+ L as n ---+ oo is also employed. If no such L exists, then the sequence is said to diverge, and lim an does not exist. n-+oo

Example 6.8. It is clear that the number I_ can be made arbitrarily close to O by n taking n sufficiently large. For example, in order to make I_ < 10- 1 0 , take n > 10 10 . n 100 100 In order to make I_ < 10- , take n > 10 , etc. Hence, n

.

1

hm - = 0. n-+oo n Formally, for any

E

> 0, I I_ - 0 I = � < n

n

let N be any integer such that N > 1

n >­

E

!.

E

if and only if n

Then 1 n

- �. Hence, it suffices to E

194 which proves that the sequence {

CHAPTER 6. SEQUENCES AND SERIES

¾}

converges to 0.

Notice the manner in which this proof would fail if we made the false assumption that lim a n= L where L-/=- 0. For example, if we take L = 17, then n�oo

for all n -/=- 0, and, hence,

I¾-

/¾-

171

11/ � 16

< c is impossible for any c � 16.

Example 6.9. Let c be a constant. In the present context, a constant is any number independent of the "variable" n. The sequence { C}

converges to c, i.e., lim c n->oo

= { C,

C, C, · · · }

= c since, for any

c > 0, le - cl

= 0 < c for all n � 1.

Example 6.10. As in Example 6.8, the sequence { �2 } converges to O since �2 can

be made arbitrarily close to O by taking n sufficiently large. More generally, hm -=0 n->oo nP

.

for any real number p

1

. > 0 smce, for any c > 0, let N > l/p. Then c

n� N If p

1

0, then lim � = lim 1 = 1. If p < 0, then lim � does not exist because n-+oo nP n-+oo nP n-+oo P n- , with -p > 0, increases indefinitely as n increases indefinitely. Thus, the

=

-P = n

1

sequence {

:P} converges for p � 0 and diverges for p < 0.

Example 6.11. Consider the sequence { ( -1 )"} of Example 6.5. lim (-lt n-+oo

doeti not exist because a n = ( -1 )" takes both of the values 1 and -1 regardless of how large n is and, hence, the terms an do not approach a unique number. Thus, the sequence { (-1 )"} diverges.

6.1. SEQUENCES

195

Example 6.12.

lim sin(n) does not exist because sin(n) oscillates and does not n......,oo remain close to any one number regardless of how large n is. Hence, the sequence {sin(n)} diverges. Example 6.13.

The number

sufficiently large. Hence,

1 can be made arbitrarily close to O by taking n 2n

1

.

hm n---+oo 2 n i.e., the sequence {

= 0,

1 2

n} converges to 0.

Example 6.14. Let r be a real number, and consider the sequence {rn }�=O· We shall determine those values of r for which the sequence converges, and find the limit. If r > 1, then rn increases indefinitely as n increases indefinitely; hence, the sequence {rn } diverges. If r = 1, then rn = 1 for all n

2:: 0; hence, the sequence {rn} converges to 1.

If lrl < 1, i.e., -1 < r < 1, then rn can be made arbitrarily close to 0 by taking n sufficiently large; hence, the sequence {rn } converges to 0. If r � -1, then rn alternates between positive and negative values, with rn is even and rn � -1 if n is odd; hence, the sequence { rn} diverges. Theorem 6.1. Let {a n} and {bn} be convergent sequences, with lim an n......,oo lim bn = M. Then: n-,oo 1. The sequence {an + b,i} converges, and lim (an + bn ) n......,

of Theorem 6.1 with c

L and

= cL.

= LM.

an . an 4. The sequence {- } converges if M =I- 0, and lim n......,oo bn bn Example 6.15. By Example 6.10, lim � ,i-,oo n

=

= L + M.

2 . The sequence {can} converges for any constant c, and lim can n......,oo 3. The sequence { an bn } converges, and lim an bn n......,oo

2:: 1 if n

L M

= -.

= 0 and lim --; = 0. Hence, by item 2 ,i-,oo n

= -1, lim (--;) = 0 and, by item 1, lim (� - -;) = 0. n......,oo

n

n......,oo

n

n

196

CHAPTER 6. SEQUENCES AND SERIES

Example 6.16. Since the sequence {

¾}

¾}

converges to 0 by Example 6.8 and the

sequence {1} converges to 1 by Example 6.9 with c = 1, it follows by Theorem 6.1 that the sequence { 1 -

converges to 1.

Example 6.17. If either {an } or {bn } diverges, then Theorem 6.1 is not applicable,

and sums, differences, products and quotients of such sequences may converge or may diverge, depending upon the particular sequences.

Both of the sequences {n2 } and { n3 } diverge because their terms increase indefinitely as n increases indefinitely, and { n 3 - n2} diverges. The limit is not 0. and { fa} diverge, but Both of the equences {

vn+T}

� vln+l+fa � vn-t-1-fa= ( v n+l-fa) n+l+fa n vn+1

=

1

n+l+ fa n vnTI

converges to 0 since the denominator can be made arbitrarily large and, hence, the fraction arbitrarily close to 0, by taking n sufficiently large. The sequence {

¾}

converges to 0 and the sequence { n2} diverges. The product

{ ¾n2} = {n} diverges. The sequence { :3 } converges to 0 and the sequence { n2 } diverges. The product

{

2 : n } 3

={

¾}

converges to 0.

n2 + 1 }· In order to determine the 2n3 + n -l limit of such an expression, divide the numerator and the denominator by nr , where r is the greatest exponent of n in the denominator. In the present case, r = 3, and Example 6.18. Consider the sequence {

n3

-

n2 + 1 2n3 + n -l

n3

1

-

1

1

Since lim - = 0, lim 2 = 0, lim 3 = 0 and lim c = c for any constant c, n->oo n n->oo n n->oo n->oo n lim (1 - � + �) = 1 -= n �

and

n3 - n2 + 1 n->oo 2 n3 + n - l

by Theorem 6.1. Hence, lim

lim

�= 1 2

(2 + �� - �� ) = 2

6.1. SEQUENCES

197

2n + 1 Example 6.19. Consider the sequence { J }· Divide the numerator 3n2 + 3n + 5 and the denominator by n = Jni to obtain 2n + 1

J3n2 which converges to

2+1n

+ 3n + 5

2 J3 as n --+ oo.

Definition 6.3. Given a sequence {an }, lim an

n->oo

=

oo

means that an can be made arbitrarily large by taking n sufficiently large. More rigorously, for any K > 0, however large, there exists an integer N, which depends upon K, such that Similarly,

lim an

n->oo

=

-oo

means that lan l can be made arbitrarily large with a n < 0 by taking n sufficiently large. More rigorously, for any K > 0, however large, there exists an integer N, which depends upon K, such that n�N

==}

a n < - K.

In either case, the sequence { a n} diverges. Example 6.20. The terms of the sequence { n3 } can be made arbitrarily large by taking n sufficiently large. Formally, for any K > 0, let N > K 1 13. Then n� N ==}



n > K 113



n3 > K

lim n3 = oo and the sequence { n3 } diverges. Similarly, lim (-n3 ) = -oo and the

n---+oo

n---+oo

sequence { -n3 } diverges.

Definition 6.3 is useful in distinguishing sequences { an } with lim a n = ±oo from other divergent sequences such as {sin(n)} and

{(-lt}.

n->oo

In general, it is difficult to determine the convergence or divergence of a sequence directly from first principles, especially for more complicated sequences, and several theorems are available in order to facilitate this task.

198

CHAPTER 6. SEQUENCES AND SERIES

Theorem 6.2. For any sequence {an },

=0

lim a n

n-+oo

= 0.

lim lan l

if and only if

n-+oo

Proof. an is close to 0 if and only if lan l is close to 0. Example 6.21. Since lim

n->oo

I (-ir1 = n->oo lim � = 0, n

n

(-lt = 0. it follows, by Theorem 6.2, that lim -n->oo n Theorem 6.3. Let J be a real-valued function and let an = J(n). Then



limf(x) = L

x-+oo

liman= L.

n--+oo

In other words, if lim J(x) = L, then it is also true that lim an= L. x-+oo

n-+oo

The converse of Theorem 6.3 is false, i.e., if lim a n = L, then it may or may not n->oo

be true that lim J(x) = L. For example, let J(x) = sin(1rx). Then an= sin(1rn) = 0 X->00

for all n and, hence, lim an = 0, but lim J(x) does not exist. x-+oo

n-+oo

Example 6.22. Consider the sequence { the sequence, let J(x) =

ln(x) , and employ L'Hopital's rule to determine lim f(x). X

ln(x) lim --

X->00

ln n) · � } In order to determine the limit of

X

X->00

l.

= X->00 lim =0 ⇒ 1 .:£

lim

n->oo

ln(n) = 0' n

by Theorem 6.3. Example 6.23. Consider the sequence { ( 1 limit of the sequence, let J(x) = ( 1

f(x) =

+ �) x'

eln[J( x)]

=

ln

e

+

¾) n}.

and employ the fact that

[(1+¾rJ

=

e xln (l+¾)_

Since the exponential function is continuous, l im [x ln(l+ 1.1m ex In(l+l) = e x -oo X

X->00

In order to determine the

.l X

)] '

199

6.1. SEQUENCES and

lim x ln

X->00

(

1+

-X1)

= lim

X->00

ln (1

+ l) x

-1

,

which is of the form i and, hence, L'H6pital's rule can be applied. Thus, lim

X->00

by Theorem 6.3.



ln(l

+ l)

lX

x

= lim

X->00

1�1

( - ;2 )

x- 1 �

lim J(x) = e = e 1

X->00



1

= lim --1 = 1 X->00

lim

n->OO

1+

(1 + ]:_) n

X

n

= e,

Theorem 6.4. Let {an }, {bn } and {en } be sequences, and suppose that an :s; bn :s; Cn for all n � k, where k is an integer. Then lim an

n->oo

=

lim Cn = L

n->oo

=}

lim bn = L.

n->oo

This theorem is called the squeeze theorem because the terms of the sequence {bn } are "squeezed" between the corresponding terms of the sequences {an } and {bn }. Example 6.24. Consider the sequence

obtain the inequalities

1 n

00 sm n ) } { . (

-n-

sin(n) n

n=l •

Since -1 :s; sin(n) :s; 1, we

1 n

--oo n n->oo n

sin(n) = O. n

Alternatively, since O :s; I sin(n)I :s; 1, we obtain the inequalities

By Theorem 6.4, lim O = 0 and lim (�) = 0 ⇒ lim n->oo n->oo n->oo n sin(n) lim -- = 0 by Theorem 6.2. n-oo n

I sin(n) n

I

= 0 and, hence,

200

CHAPTER 6. SEQUENCES AND SERIES

2n Example 6.25. Consider the sequence {an }�=1, where an = . Then 1 n. 2-2·2···2· 2 2 2 2 2 2 an = -------- = - . - . -···--. -· ( -l)·n 1 2 3 n-1 n 1-2·3···n

4

4

If n = l, then a1 = 2:::; -. If n = 2,then a2 = 2:::; -. If n :2:: 3,then n n 2 oo n-->oo n n Example 6.26. Consider the sequence {an }�=l,where an an

=

· 1 ·2·3 ···(n-1) n -------n·n·n···n·n

=

=

=

0.

=

0.

nl �- Then n

l 2 3 n- l n - . - . -•••--. -, n n n n n

a1 = 1 and, for n :2:: 2, -

2 -···--· 3 n 1n -· -oo n-->oo n n-->oo

Definition 6.4. A sequence { a n } is n i creasing for n :2:: N, where N is an integer, if an+l :2:: an for all n :2:: N. It is decreasing for n :2:: N if an+l:::; an for all n :2:: N. A sequence which is either increasing or decreasing is called monotone. Example 6.27. The sequence {an }

=

{

1}

is decreasing for n 2: 1 because

1 1 an+ l = -- < - = an for all n 2: 1. n+ 1 n

201

6.1. SEQUENCES

Example 6.28. The sequence { an } = { 1 - �} is increasing for n 2: 0 because 2

1 1 ->-=> n+l 2n

2

i.e., an < an +l for all n 2: 0. n2

Example 6.29. Consider the sequence { an}�= 1, where an = ne-s. In order to x determine whether or not this sequence is monotone, let f(x) = xe- 8. Then 2

J'(x)

2

= e_xs +

2

xe_xs

(-�) =

2

e_x: ( 1 - :

)

< 0 for x > 2.

It follows that the function f is decreasing for x > 2 and, hence, the sequence {an } is decreasing for n 2: 3 > 2. Example 6.30. The sequences {(-l)n} and {sin(n)} are not monotone because none of the inequalities in Definition 6.4 can hold for all n 2: N for any integer N. Definition 6.5. A sequence { an }�=k is bounded above if there exists a real number K such that an :::; K for all n 2: k. It is bounded below if there exists a real number M such that an 2: M for all n 2: k. It is bounded if it is bounded above and bounded below, i.e., there exist real numbers K and M such that M :::; an :::; K for all n 2: k. A sequence which is not bounded is called unbounded. K is called an upper bound, and M a lower bound of the sequence. 00

n is bounded because Example 6.31. The sequence {- -} n + l n=O n > 0. 0< -- < 1 for all n n+ 1 -

Here, M = 0 and K = l. These numbers are not unique since any M :::; 0 and K 2: 1 are also lower and upper bounds of the sequence, respectively. Example 6.32. The sequence {2n }�=l is bounded below because 2n 2: 2 for all n 2: 1. It is not bounded above because lim 2n = oo, i.e., 2n can be made arbitrarily large n--+oo by taking n sufficiently large and, hence, it is impossible to have 2n ::; K for all n 2: 1, for any real number K. Thus, {2n }�=l is unbounded.

202

CHAPTER 6. SEQUENCES AND SERIES

Example 6.33. The sequence {1- n}�=l is bounded above because n 2: 1



-n S -1



an = 1 - n S 0.

It is not bounded below because lim (1 - n) = -oo. Thus, it is unbounded. n --+oo Theorem 6.5. A sequence { an }�=k is bounded if and only if the sequence {/an /}�=k is bounded above. Proof. If { an }�=k is bounded, then there exist real numbers M and K such that MSa n SK for all n 2: k. Let L = max{/M/, /K/}. Then M Sa n SK

=?

-L

S an S L

=?

/an / S L,

i.e., {/an /}�=k is bounded above. Conversely, if {/an /}�=k is bounded above, then there exists a real number K such that /an /S K for all n 2: k, and /an /SK

=?

-KS an SK,

i.e., {a n}�=k is bounded. Example 6.34. The sequence {sin(n)}�=O is bounded because -1 S sin(n) S 1, which is equivalent to I sin(n)/S 1, i.e., the sequence {I sin(n)/}�=O is bounded above. Theorem 6.6. If a sequence {an }�=k is increasing for n 2: N, where N is an integer, and bounded above, then it converges. If it is decreasing for n 2: N and bounded below, then it converges. If it is monotone for n 2: N and bounded, then it converges. 00

1 l 1 is decreasing for n 2: 1 since -- < - for Example 6.35. The sequence { - } n n= n+1 n l 1 all n 2: 1, and it is bounded below since - > 0 for all n 2: 1. Hence, by Theorem 6.6, n it converges, a result which has already been established in Example 6.8. Example 6.36. The sequence { 1 - e-n } :'=o is increasing for n 2: 0 because e n+l

> en

e-n

=?

>

e-(n+l)

and it is bounded above since e-n > 0



-e-n < 0

=?



a n = 1 - e- n < 1- e-(n+l) = an +l, a n = 1 - e- n < 1 for all n 2: 0.

Hence, by Theorem 6.6, it converges. Note that the sequence { e-n } :'=o converges to 0 by Example 6.14 with r = e- 1. Hence, { 1 - e-n } :'=o converges to 1 by Theorem 6.1.

6.1. SEQUENCES

203

Theorem 6. 7. Let P(n) denote a statement or a formula which involves an integer n. If P(l) is true and the implication P(n) =} P(n + 1) is true for all n 2 1, then P(n) is true for all n 2 1. More generally, if P(k) is true, where k is an integer, and P(n) =} P(n + 1) is true for all n 2 k, then P(n) is true for all n 2 k. This is called the principle of induction. In order to prove that the implication P(n) =} P(n + 1) is true, assume P(n) is true and employ P(n) to show that P(n + 1) is also true. The assumption that P(n) is true is called the inductive hypothesis. n

Definition 6.6. Let a 1, a2,

· · · ,

an be real numbers. The symbol Lai, read "The

sum of ai as i ranges from 1 to n," is defined by n

Lai = a1 i=l

i=l

+ a2 + a3 + · · · + an-I + an,

and is called sigma notation. Example 6.37. Consider the sum n

Li= 1 + 2 + 3 + · · • + (n -1) + n. i=I

We wish to prove the formula �i= n(n+l) � 2 i=l

(6.2)

for all n 2 1. Let P(n) denote the formula (6.2), and apply the principle of induction. If n = 1, then P(l) states that 1 = 1, which is true. In order to prove that the implication P(n) =} P(n + 1) is true, assume P(n) is true and employ P(n) to show that P(n + 1) is also true. Thus, n+l

Li i=l

1+2

+ 3 + · · • + n + (n + 1)

n i=l

n(n + 1) 2

n(n+l)

+ (n + 1) 2(n+l)

+ 2 (n + l)(n + 2) 2

2

(by the inductive hypothesis)

204

CHAPTER 6. SEQUENCES AND SERIES

which is P(n + 1), i.e., (6.2) with n replaced by n been proven by induction.

+ 1.

Thus, the formula (6.2) has

Example 6.38. Consider the sequence { an }�=l, defined inductively by

We shall employ induction to prove that the sequence is increasing for n 2: 1 and is bounded above, and, hence, that it converges, by Theorem 6.6. The first four terms of the sequence are

Since

-/7 < 3, we find that

a1

= 1 < 3,

a2

= v'7 < 3,

a3

=

J + v'7 < 6

3, a4

=

✓ + J + v'7 < 6

6

3,

and it seems that an < 3 for every n 2: 1. In order to confirm this, we shall employ induction, where P(n) is the statement "an < 3." Since a 1 < 3, P(l) is true. Assuming that P(n) is true (the inductive hypothesis), i.e., an < 3, we obtain

which is P(n + 1). Thus, it is proved that an < 3 for all n 2: 1, and the sequence is bounded above. Since a 1 < a2 < a3 < a4, it seems that the sequence is increasing. Let P(n) denote the statement "an < an+1•" Since a 1 < a2,

P(l)

is true. Assuming that P(n) is true, i.e., an < an+ i, we obtain

which is P(n + 1). Thus, it is proved that an < an+ l for all n 2: 1, and the sequence is increasing. By Theorem 6.6, the sequence converges. Let L an+l

= J6 + an

=>

lim an+l

n---+oo

= n->oo lim an.

Then

= n---+(X) lim J6 + an =

✓+ 6

lim an ,

n---+oo

205

6.1. SEQUENCES where the last equality follows by the continuity of the function g(x) n ---. oo ⇒ n + l ---. oo, lim a n+ l

n--+oo

= J6 + x.

Since

= n+lim an+l = lim an = L. n--+oo 1--+oo

Hence, L = J6 + L ⇒ L 2 = 6 + L ⇒ L 2 - L- 6 = 0 ⇒ (L- 3)(L + 2) or L = -2. Since an > 0 for all n 2'.: 1, L 2'.: 0 and, hence, L = 3.

=0⇒L=3

Exercises 6.1 In Exercises 1 8, express the given sequence as {an }�=k. 1. {5, 7, 9, 11, · · ·}

3 {1, .

5. {1,

2. {-6, -3, 0, 3, ··· }

1, -}, . ··} 3' 9' ·} 1 2'

1

1

4.

1 .. 27'

{1,

1 1 1 } 4' 9' 1 6' ...

1 6. { 0, 1-- 1-� 1-� ···} 4' 3' 2'

.

9. Wnte down the first five terms of the sequence

{

Jn2 + 1 --- } 2n + 1

00

. n=O

10. Write down the first five terms of the inductively-defined sequence a1 a2 = 3, an = a n-1 - an -2 for n ;::: 3. 11. Let {an }�=O and {bn}�=l be sequences such that bn = an+ J Determine bn+ l for n 2'.: 0 and bn-l for n 2'.: 2.

1

1 1 1 } , 3 5, 7, 9, · · ·

12. Express the sequence {

+;

2,

for all n 2'.: 1.

as {an}�=k in three different ways.

In Exercises 13-48, determine whether or not the given sequence { an} converges. If it converges, find the limit. If it diverges, determine whether or not lim an = ±oo. n--+oo

206

CHAPTER 6. SEQUENCES AND SERIES

1 13. an= -1r n

20. a n 101,000,ooo,ooo,ooo no.0000000001

21. a n

=

23. an

=

n+

25. a n

=

27. a n

=

=

3 r,;:; yn

2 22. a n = 3 - -2 n

1 3n

24. an

=

3n2 2n2

3n3 - 2n - 3 4n2 + n+ 1

26. an

=

3n2 2n3

3n2 - 2n - 3 ----;:::==== v2n4 + 3n + 1

28. an =

29. an =

3n - 2n - 3 ----;::==== 3

30. a n = n2

31. an

Jn2

=

2

v'2n

33. an = Jn3

35. a n

=

-

n+1

+n-

Jn2

-

n

-

+ -

+

3n2

5

+ -n 4

2n - 3 3n + 1 2n - 3 2n -1

-

2n - 3

----;:::==== v2n5 - 2n + 1 -

2y'n

+n

32. an

=

n - Jn2

36. an

=

1 - (-1r -­ -n

+1-n

1r (--­ -

-1

n

ln(n )] 2 37. an = [ n n P 40. a n = [ln( )] , p > 0, q > 0 nq

42. an 1 43. a n = [ln(n)] /n

=

[ln(n)] 1/ln(n)

l ln( n) 44. an = n /

6.1. SEQUENCES

207 sin(n2 )

cos(n) 45. an =-n

46. an=

en 47. an = n!

(2n)! -48. an 2nn2 n

fo

In Exercises 49 54, determine whether or not the given sequence { a11 } is increasing or decreasing for n 2::: N. oo

49.

50.

{2-�}

51.

52.

{ ln�n)} �= 2

53.

{

n=l

00

00

n ln(n) en } n =]

54.

sin n) { � } n n =l

In Exercises 55-58, determine whether or not the given sequence { an }�=l is bounded above (an :S K), bounded below (an 2::: M), or bounded (M :S an :S K). In each case, determine Kand/or M, if it exists (the bounds K and Marc not unique).

t

55. an = 57. an

=

1 56. an = 2n

n

vnTI -1 1 1-n+]

� . . 59. Prove, by mduct10n, that � i 2 t=l

- ln(n)

58. an=n2 e-n

=

n(n + 1)(2n + 1) . 6

60. Prove, by induction, that the sequence { an }�=l, defined inductively by a1 = 1 and an+l = 1 + .ja;,, for n 2::: 1, is increasing for n 2::: 1 and bounded above, hence convergent, and find the limit. 61. Consider the sequence { an }�=l, defined inductively by a1

=

1 and an+l

for n 2::: 1. Does the sequence converge? Justify your answer.

=

1-2_ an

62. Consider the sequence {an }�=l , defined inductively by a1 =2 and an+l =2 - 2_ an for n 2::: 1. Prove, by induction, that 1 :S an :S 2 and that {an }�=l is decreasing for n 2'. 1, and find the limit.

CHAPTER 6. SEQUENCES AND SERIES

208

6.2

Series

Definition 6. 7. Let { an }�=l be a sequence. Define the sequence { sn }�=l by S1

a1 a1 + a2

s2 s2

a1

a1

Sn

+ a2 + a3

+ a2 + a3 + · · · + an =·

Lai. n

i=l

The sequence {S1 , S2 ,

83 ,

···

, Sn,

{ a1, a1 + a,, a1

···}

+a,+ a3,

is denoted by the symbol

La 00

n= l

n

= a1 + a2 + a3 + · • •

(6.3) 00

and called an infinite series, or simply a series. The series Lan is said to converge n=l

if the sequence { sn } converges. Otherwise, the series diverges. If lim Sn = s, then s n-->oo

is called the sum of the series, and oo

n

� an = lim Sn = lim � ai = S. L n----+oo n----+ooL n=l i=l

La . 00

The number Sn is called the n th partial sum of the series

n=l

n

Since it is impossible to perform an infinite number of additions, Definition 6. 7 gives precise meaning to the concept of an "infinite sum" as the limit of a sequence of finite sums. Note that the symbol n in (6.3) is a "dummy index," analogous to a "dummy variable" in a definite integral, and may be replaced by any other symbol whatsoever. Thus,

L a = L ai = L= ak = La, = a1 + a2 + a3 + · · · 00

n

n=l

00

00

00

i=l

k l

(=l

209

6.2. SERIES

all represent one and the same series, determined entirely by the sequence {an } n=l. In Definition 6.7, the sequence {an } was employed with n = l, 2, 3, · · · for the sake of definiteness. More generally, if the initial value of n is k, then the corresponding senes 1s CX)

Lan n=k

For example,

La CX)

n=O

n

= ak + ak+l + ak+2 + · · ·

= ao + a1 + a2 + · · · ,

CX)

and

Lan

=

n=17

a11 + al8

+ a19 + · · ·

Since a series is a special type of sequence, the properties of sequences given by Theorem 6.1 in Section 6.1 are also shared by series. The first two are relevant for senes: CX)

Theorem 6.8. If Lan and n=l

CX)

1. The series L(an n =l

L ca

L CX)

CX)

+ bn ) converges, and L(an n =l

n=l

n

converges for any constant c, and

CX)

n=l

L (a CX)

n =l

n

L ca CX)

Note that, if Lan converges and because, if

CX)

+ bn ) = s + l.

CX)

2. The series

L

CX)

bn converge, with Lan = s and bn = t, then: n =l n=l n =l

L CX)

n =l

n =l

n

CX)

bn diverges, then L(an n=l

CX)

=

c Lan n=l

=

cs.

+ bn ) must diverge

+ b11 ) converges, then

CX)

Lbn n=l

=

CX)

L[(an n =l

CX)

+ bn ) - an ]= L(an n =l

+ bn ) - Lan n =l

converges, which is a contradiction.

'

Example 6.39. Let r be a real number, and consider the sequence corresponding series is CX)

Lrn , n =O

{rn }�=o·

The

210

CHAPTER 6. SEQUENCES AND SERIES

called a geometric series. In order to determine the values of r for which the series converges and to find the sum, consider the nth partial sum Sn =

Then

Lr = 1 + r + r i=O

rs n

and, hence,

i

2

+ r3 + • • • + rn-l + rn.

r (1 + r + r2 + r3 + • • • + rn-l + rn) r + r 2 + r3 + ... + rn-1 + rn + rn+l 1 + r + r 2 + r 3 + ... + r n-l + r n -r_r 2 _ r 3_ ..._r n-1 _ r n _ r n+l

1 - rn+l 1 - rn+l _

⇒ (1-r)s n

If r-=/ 1, then

S n= --1-r By Example 6.14 in Section 6.1, the sequence {rn } converges to O if -1 diverges for r ::; -1 or r > 1. Hence,

< r < 1 and

1-r n+l lim Sn = lim ---

1 if -1 < r < 1, n-+oo 1 - r n-+oo 1-r and { s n} diverges for r ::; -1 or r > 1. If r = 1, then rn = 1 for all n, and Sn = 1 + 1 + 1 +

⇒ { s n} diverges. In summary,

··· + 1 = n+1

n = �r � 1-r' n=O 00

1

and the series diverges for lrl � 1.

Example 6.40. Consider the series

1

lrl < 1,

l

= 2 I: 2n I: =O ( ) =O n 00

00

n

1 1 1 This series is geometric with r = - and, since - = - < 1, the series converges and 2 2 2 11 n

-2 �2n- 1_.!2 n =O 00



1

1

211

6.2. SERIES

Example 6.41. Consider the series

00

00 (

2n � (-3)n = �

This series is geometric with r and

=

-f

2 n )

-3

1-11 = f

and, since

< 1, the series converges

Example 6.42. Consider the series oo 8 n 8 n n n 8 ) ( ( -2 9 =-2 -2.3-2 = � 9) ' � �

00

00

where the last equality follows by item 2 of Theorem 6 8 This series is geometric . . with r = and, since = < 1, the series converges and

i

Ii I i 00

'""""' -2



· 8n · 3 -2n

n=O

Example 6.43. Consider the series

�5 (-3) 2n+> r 3n =

Th.1s senes . 1s . geometric . wit . hr

=

8 9

=

1

-2-1- -98

= -18.

� -15 (�)" = -15�

. and, smce

8 191

Example 6.44. Consider the geometric series

00

=

8 9

2': 1, the senes . d.1verges

where the initial value of n is 1 instead of 0. Then n=l

00

n=l

r + r2

+ r3 + • • •

-1 + ( 1 + r + r2

00

n=O

1 -1+-1-r r 1-r

(ff

+ r3 + ... )

.

212

CHAPTER 6. SEQUENCES AND SERIES

More generally, for any k � l, 00

n=k

-1 - r - r2

+ (1 + r + r2 + • • • + r k -l + r k + r k+l + ... ) - (1 + r + r2 + · · · + r k -1) + (1 + r + r 2 + • • •)

-L k-1

r + n

-r k

-� + rk

For example,

Lr oo

n

n=O

n=O

l

r k -l

- · · · -

l (by Example 6.39) 1_r

l-r

and

Example 6.45. Consider the series 00

00

1

� n(n + 1) = �

(1 1 ) ;, n + l

where the right side is obtained from the left by partial fractions. Then the n th partial sum of the series is Sn

n

= �

(1

1

k- k+1

(1 _

)

1 _ 2:.) + (2:. __ 1 !2 ) + (!2 _ !3 ) + (!3 _ !4 ) +... + () 1 nn n n +l

1 1---. Hence, lim

n-+oo

n+l

Sn

=

l, i.e., the series converges and 1 � n(n + 1) = 1. 00

A series in which such cancellations occur in

Sn

is often called a telescoping series.

6.2. SERIES

213

Example 6.46. Consider a series with only a finite number of nonzero terms, i.e., 00

with an = 0 for n > m � 0. Then the partial sums Sn of the series are given by S1 s2

s3

Sm Sm+1

Sm+2

etc., i.e., Sn

a1

+ a2 a1 + a2 + a3 a1

a1 a1

a1

+ a2 + a3 + · · · + am =

L ak m

k=l

+ a2 + a3 + ... + am + 0 =

k=l

+ a2 + a3 + ... + am + 0 + 0 =

L ak, m

k=l

L ak for all n � m and, hence, lim S L ak. Thus, m

=

L ak m

m

n-+oo

k=l

n

=

k=l

m

oo

"ak = lim Sn = L "ak, L n--+oo k=l

k=l

i.e., the series reduces to an ordinary finite sum. Such a series is called a finite series. The convergence or divergence of a geometric or a telescoping series and, in the case of convergence, the sum of the series, has been established without any great difficulty. For other types of series, these properties cannot, in general, be determined by first principles, and various theorems must be employed. 00

Theorem 6.9. If Lan converges, then lim an = 0. Equivalently, if lim an =/= 0, n--+oo

n=l

00

n--+oo

then Lan diverges. The latter version is often called the nth -term test. n=l

oo

n

Proof. Lets= Lan = lim Sn , where Sn = L "ak is the n th partial sum of the series. n=l

n--+oo

k=l

214

CHAPTER 6. SEQUENCES AND SERIES

Then

Hence,

lim

n---+oo

an

=

lim

n---+oo

Sn

a1

Sn-l

a1

(sn - Sn-1)

+ a2 + · · · + an-1 + an + a2 + · · · + an-1

and

= n---+oo lim Sn - lim Sn-1 = s - s = 0. n---+oo

The second assertion in the theorem is the contrapositive of the first, and the two are logically equivalent. Note that the proof fails if

00

diverges for, then, the numbers does not exist,

Lan n=l

and the term s - s has no meaning.

00 l n Example 6.47. The series� ( ) converges since it is geometric with r

l!I < 1, and lim (!) n->oo 3 3

n

3

= 0,

l-1+1 -1

3 and

00

n=l

with the nth partial sum =

1

as guaranteed by Theorem 6.9.

Example 6.48. Consider the series

Sn

=

+ 1 - 1 + ... + (-l) n -1 =

{

0, if n is even }. 1, if n is odd

Thus, {s n}�=l = {1, 0, 1, 0, 1, 0, · · ·} diverges and, hence, the series diverges. This result can be established much more easily by means of Theorem 6.9. Since 1 1 does not exist, lim ( - 1 =/- 0 and, hence, the series diverges. lim ( - 1

n---+oo

r-

n---+oo

r-

Example 6.49. Consider the series

n Since lim --;::== n->oo ✓n 2 + 1

=

1 =/- 0, the series diverges by the n th-term test (Theorem 6.9).

6.2. SERIES

215

The reader is warned that the converse of Theorem 6.9 is false, i.e., if lim an

=

n-+oo

00

0,

then there is no guarantee that the series Lan converges. It may converge, or it n=l

may diverge, depending upon the particular series. Examples of both will be given shortly.

6.2.1

The Integral Test

Let f be positive, continuous and decreasing for x 2:: 1, and let a n y

=

J(n).

y

a

2

3

II

n-1

X

n

2

3

n-1

n

x

Figure 6.1: Rectangles of lesser area (left) and rectangles of greater area (right). Since every rectangle has unit width, its area is equal to its height. Hence, the areas of the rectangles on the left are a 2, a3, · · · , an , and the areas of the ones on the right are a 1, a 2, · · · , an -J, as depicted in Figure 6.1. Employing the fact that the area n

under each curve for 1 :S: x :S: n is i f ( x) dx, we obtain the inequalities (6.4) 00

Consider the series Lan , and let { sn } n=l be the sequence of n th partial sums. From n=l

216

CHAPTER 6. SEQUENCES AND SERIES

(6.4), we obtain the inequalites a1

Sn Sn

+ (a2 + a3 +···+an)

(a1

=

:S a1

n

(6.5)

+ 1 J(x) dx, n

2: an + 1 J(x) dx.

+ a2 + a3 + · · · + an-1) + an

(6.6)

Since an 2: 0 for n 2: 1, the sequence {s n }�=l is increasing because

If /

00

f(x) dx < oo, then, by (6.5) and the fact that f(x) > 0 for x 2: 1, Sn :S a1

n

+ l J(x) dx :S a1 +

100

J(x) dx < oo

⇒ { s n }�=l is bounded above and, hence, it converges by Theorem 6.6. If

100

J(x) dx = oo, i.e., the improper integral diverges, then, by (6.6), · n

n

S n 2: an + J J(x) dx 2: 1 J(x) dx



⇒ n__.,oo lim S n

n

lim Sn 2: lim J J(x) dx

n->oo

n->oo

1

00

= J f(x) dx = oo 1

00

oo, i.e., the sequence {sn }�=l and, hence, the series"" an , diverges. L,_; n= l In summary, we have derived the following: =

Theorem 6.10. Let f be positive, continuous and decreasing for x 2: 1, and let an = f(n). Then the series 00

converges if and only if

j00 J(x) dx
0 and lim � = lim n-P = oo-/= 0. Hence, the series diverges n-+oo n-+oo nP by the nth-term test. 1 If p = 0, then lim - = lim 1 = 1 -/= 0. Hence, the series diverges by the nth -term n-+oo n-+oo nP test. 1 If p > 0, let f(x) = -. Since f is positive, continuous and decreasing for x �land xP an = f(n), the integral ·test applies. The series diverges if p = l by Example 6.51 and, for p -/= 1, 00 1 x1 -P 1 00 oo, if p < l . -dx= x P dx=-- = { 1 ' if p >l} 1-p i 1 xP p-1 l

J

,

J

,

00

_

Hence, the series converges for p > land diverges for p ::; 1 by Theorem 6.10. Example 6.53. Consider the series 00

1

I::-fo n=l

1 , this is a p-series with p l fo n /2 by Example 6.52. Since

1

=

=

1

2

and, since

1

2 ::;

1, the series diverges

219

6.2. SERIES Example 6.54. Consider the series

1

00

�nvn· 3 2

3 = - and, smce .

S.1nce

l = l -, th.1s 1s · a p-senes · wit · hp r,;;; nyn n31 2 by Example 6.52.

2

. > 1, the senes converges

Example 6.55. Consider the series

1 � nln(n)" 00

1 . Since f is positive, continuous and decreasing for x > 2 and ( ) x 1 nx an = J(n), the integral test applies. Thus, making the substitution u = ln(x), Let J(x) =

1 2

00

1-dx = X ln( X)

-

f00

..!. du= ln 1u1l

J1n(2) U

00

Jn(2)

=

00.

Hence, the series diverges by Theorem 6 10. . Approximations of Series 00

Let Lan be a convergent series. If the sum s of the series cannot be determined, n=l

then it can be approximated. By definition of limit, since oo

s = � an = lim

L

n

n-+oa

n=l

Sn

= lim � ak, n-?OO

L k=l

the terms Sn can be made arbitrarily close to s by taking n sufficiently large. Thus, Sn is approximately equal to s, written Sn � s, and the larger we take n, the better the approximation. Whenever an approximation is made, knowledge of the maximum possible error is essential. The error made in approximating s by Sn is DO

n

00

n=l

k=l

k=n+l

called the remainder of the series.

220

CHAPTER 6. SEQUENCES AND SERIES

00

In the present discussion, we restrict our attention to series Lan where an = f(n) n=l

and f is positive, continuous and decreasing for x 2". 1. In order to determine upper and lower bounds for Rn, we appeal to Figure 6.2. From the graph on the left, we obtain Rn= an +I

+ an+2 + · · · �

Rn= a n+I

+ an +2 + · · ·

and, from the one on the right, we find

y

2':

1

100

n+l

J(x) dx,

(6.9)

J(x) dx.

(6.10)

y

2

3

n

n+l

n+2

2

X

3

n+l

n+2 n+3 X

Figure 6.2: Upper and lower bounds for the remainder Rn . Combining (6.9) and (6.10), we obtain the inequalities

1

00

n +I

f(x) dx�Rn�

Hence, the error m approximating s by

1

00

J(x) dx.

Sn

100 n

J(x) dx.

is at least

100

n +l

(6.11) f (x) dx and at most

221

6.2. SERIES Example 6.56. Consider the p-series

Since p = 3 > 1, the series converges by the integral test (Theorem 6.10). If the sum of the series is approximated by s2 , i.e., n = 2, then

= s=L n=l

and

3 � s2 = n

2

L k23 = 2 + 41 = 2.25, 2

k=l

= = 1 1 1 ⇒ < R2 < -. :S R2 :S -2 2 x 13 x1 1 2 9 - 4 . 1 1 Thus, the minimum error is � 0.111 and the maximum error is = 0.25. If, instead, 4 9 the sum is approximated by s5, i.e., n = 5, then 2 = � -3 ,.:;::; �n =

S

and

1

-x2

=

16

2 =�� k3 5

S5

1

2

2

2



1 1 (X)

� ln(l + L n n=l

+ ¼).

n

(1 + n�)

¼) = � nln(l + ¾) L n=l

n2

Since the logarithmic function is continuous,

=

n

= e, express the given series as

n � ln[(l + ¾) ] . L n2

n=l

226

CHAPTER 6. SEQUENCES AND SERIES

By definition of limit, the quantity ln [ ( 1

+

¾) n]

can be made arbitrarily close to

1 and, hence, less than 2, by taking n sufficiently large. Thus, there exists an integer

N such that

from which we obtain

ln(l

L

L

+ ¾)

n

_ ---�ln[(l + ¾rl 2 < n2

-

n2 '

n � N.

L

ln(l + .!.n ) 1 2 converges, it follows by the comparison test that =2 Since 2 2 n n n n=l n=l n=l converges. 00

00

Example 6.65. Consider the series � [ln(n)] 6 L n2 n=l Since lim

X->00

[ln(x)] P Xq

= 0 for any p > 0, q > 0, by L'Hopital's rule, lim

n->OO

Theorem 6.3. Thus,

[ln 1 6 tl = 0 n->00 n

and lim

and, since

f n=l

nq

= 0 by

[ln(n)] 6 1 nl/2 n3/2 '

[ln(n)] 6 n2 =}

[ln(n)J P

there exists an integer N such that [ln(n)] 6 1 nl/2
1 and lim n- oo � � n=l 5 3 1 1 p > 1 and q > Osuch that p + q = , e.g., p = and q = , and let bn = 5 4. Then 4 4 2 n1 Since

.:.....--'---c._

an bn

(Theorem 6.12).

=0

n

converges, Lan converges by the limit comparison test

00

00

n=l

[ln(n)] 2 n l /4

= nlim -oo

Lb

and, since L = 0 and

= nlim -oo

L

n=l

Example 6. 71. Consider the series 00

00

�an and let bn

L

=

=

1 � Jn+T[ln(n-1)]7'

1 Then n

= -.

a lim ___!1:. n -oo bn

=

n lim ------n -oo Jn+T[ln(n - 1)] 7

=

n 1/2 - = oo lim ------ n[ln(n _ 1)]7 00

00

F+I

00

[ln(n - 1)] 7 because lim .:;____ = 0. Since L = oo and "b "an diverges by L n diverges, L n -oo nl/2 n=l n =3 the limit comparison test (Theorem 6.12). _c._

6.2.3

Alternating Series

Definition 6.8. A series of the form 00

I)-l)nbn = bo - b1 + b2 - b3 + b4 - b5 + · · · , n =O where bn > 0 for all n � 0, is called an alternating series. More generally, for any integers k and m, n =k with bn > 0 for all n � k, is an alternating series.

230

CHAPTER 6. SEQUENCES AND SERIES

Thus, an alternating series is one where the terms are alternately positive and negative. Example 6. 72. The series

1 > 0. is an alternating series with bn = - - > 0 for n -

n+ 1

Example 6. 73. The series 1 1 f (-1r- = 1 __

n=l

y'n

is an alternating series with bn =

y12

+ _1

y'3

1 Jn > 0 for n 2

Example 6. 74. The series

_�

2

+ ...

1.

I)-1r sin(n) n=l

is not an alternating series because there are values of n for which bn = sin(n) f 0. Theorem 6.13. Consider the alternating series 00

If {bn } is decreasing for n 2 k, where k is an integer, and lim bn = 0, then the series n->oo

converges. This theorem is called the alternating series test. Example 6. 75. The alternating series (-1r

1

1

1

= -1+- -- +- - ... L-n 2 3 4 00

n=l

converges by the alternaLing t>eries test (Theorem 6.10) since { �} �= is decreasing 1 for n 2 1 and lim - = 0. n->oo n

l

2 31

6.2. SERIES Example 6. 76. Consider the alternating series

Let J(x)

=

ln(x). X

Then f'(x)

=

ln (n) . I)-1r n n=2 1 - l ( x) � < 0 for x > e

X

⇒ J is decreasing for x 2:: e.

ln(n) ln (n) is decreasing for n 2:: 3 > e, an d, since lim -Hence, ·{ -- } n-+oo n n n=2 converges by the alternating series test (Theorem 6.13). 00

. = 0, the senes

Approximations of Alternating Series

Theorem 6.14. Suppose that b n > 0 and {bn } is decreasing for n 2:: 0, and that lim bn = 0, so that the alternatin g series n-+oo 00

n=O

converges by the alternating series test. If the sum 00

n=O

is approximated by the nth partial sum n

I)-llbk = bo - b1 + b2 - · · · + (-ltb n , k =O then the absolute value of the error is less than the absolute va lue b n+l of the first neglected term (which is ±b n + 1). In other words, Sn

where Rn=

=

L (-l) b 00

k

k=n+l

IR..I

k

= Is - sn l < bn +l,

is the remainder, as defined in Section 6.2.1.

Proof. S - Sn

00

k= n +l

(-1r+ 1 bn+ l + (-l)n +2b n +2 + (-1r+3 b n+3 + (-1r+4 bn+4 (-1r+1 (bn+l - bn+2 + bn+3 - bn+4 + b n +5 - · · ·)

b n +l - bn +2 + bn +3 - b n +4 + bn +5 - · · · bn +l - (bn +2 - bn +3) - (bn +4 - b n +5) - · · · < bn +l

+ ...

232

CHAPTER 6. SEQUENCES AND SERIES

because {bn } is decreasing for n 2: 0 => -(bn+m - bn+m+1 )

< 0 for all m 2: 0.

Example 6. 77. Consider the alternating series 00

Since {bn}

I:

n=l

= { �4 }

(-1r-1 n4

is decreasing and bn

> 0 for n 2: 1, and

J�� b

n

=

0, the series

converges by the alternating series test (Theorem 6.13). Suppose that the sum s of the series is approximated by s3, i.e.,

Then

s

=

L oo

n=l

(-1r-1 � n4

S3

=

L 3

k=l

(-l)k-1 k4

=

1-

1 16

+

1 1 0.003 , = 256 � 9 44 i.e., the error, in absolute value, is at most 0.0039. IR3I = Is - s3I

1 � 0.9498. 81


0 for n 2: 2, and

J�� b

n

= 0, the series

converges by the alternating series test (Theorem 6.13). How large must n be in order that the error in the approximation of s by Sn be less than 0.0001 in absolute value? Since

lerro rl = IRnl = Is - sn l < bn+ l, the required accuracy is achieved by requiring that 1 bn+ l = l '.S 0.0001, ln(n + ) which holds if and only if ln(n + 1) 2: 10,000, i.e.'

e 10 ,ooo - 1 n> '

1 which is an astronomically large number. The reason is that the terms - - - 0 ln n) ( far too slowly compared to, say, -;- , and such a large number of terms is required in n order for the finite sum Sn to become sufficiently close to the infinite sum s.

233

6.2. SERIES

6.2.4

Absolute and Conditional Convergence

Theorem 6.15. If diverges, then

00

00

L la l converges, then

n=l

n

L la l diverges. n=l

00

Lan n=l

00 converges. Equivalently, if Lan n=l

n

00

00

00

If L lan l converges, then L 2lanl = 2 L lanl converges and, by the comparison n =l n=l n=l 00 test, L(an + lan l) converges. Hence, n=l 00

Lan n =l

=

00

L[(an + lan l) - lanl] n=l

converges by Theorem 6.8. The second statement is the contrapositive of the first, and is logically equivalent to it. Example 6.79. Since

f-\n=l n

converges (it is a p-series with p = 2 > 1) and 00 _!_ 00 (-1r 1 2 l= L 2 L n' n n=l

it follows by Theorem 6.15 that

n=l

-1r converges. Of course, this result can also L(-n2 00

n=l

-

be established independently by the alternating series test. Example 6.80. The series

the series

00

L(-1r -converges by the alternating series test, but n n=l

001(-1rl

00 i

I:=I:-n n n=l

diverges (it is a p-series with p = 1 � 1).

n=l

234

00

CHAPTER 6. SEQUENCES AND SERIES 00 Examples 6.79 and 6.80 demonstrate that, for certain series, both La n and

L 00 if L n=l

n=l

L

00 00 lanl converge, whereas for others, La n converges but lanl diverges. Of course,

n=l

n=l

n=l

lanl

00 converges, then La n converges by Theorem 6.15. n=l

00 Definition 6.9. A series La n converges absolutely if both n=l

00

00

n=l

n=l

00 00 00 converge. If La n converges but L lanl diverges, then the series Lan converges conditionally.

n=l

n=l

n=l

If a n 2: 0 for all n 2: 1, then

lanl

00 = a n and, hence, La n converges if and only if n=l

it converges absolutely. Thus, there is a difference between absolute and conditional 00 convergence only for series La n where the a n take both positive and negative values. n=l

00 It must be emphasized that, whether La n converges absolutely or conditionally, n=l

it converges. The distinction between absolute and conditional convergence pertains 00 only to whether or not the series lanl converges.

L

n=l

Example 6.81. By Example 6.79, the series

by Example 6.80, the series

00

00

(-1r L-converges absolutely and, n2 n=l

-

-1r L (-converges conditionally. Note that both series n

n=l

converge. The distinction arises because, in the former, 00 00 (-1r _ 21 n2 l - Ln2

L n=l

converges, whereas in the latter,

n=l

00 (-1r i 1 \ L-=Ln n n=l n=l 00

235

6.2. SERIES diverges. Example 6.82. Consider the series

This series converges by the alternating series test. However,

1

I (-1r I � nln(n) = � nln(n) (X)

(X)

oo

(-l)n diverges by the integral test. Hence,� converges conditionally. nln(n) Example 6.83. Consider the series ( 2r L�00

-

n=O

The series

�J�= "I = t, �: = t, )

or r

oo (L

2 . . converges smce 1t 1s geometnc -. wit . h I r I < 1. Hence, 5n converges abso 1 ute1y. . n=O Example 6.84. The series

(-1r ✓n2 + 1 diverges by the n th-term test because lim - - -- =f. 0 (the limit does not n-->oo n+l exist ). It follows by Theorem 6 15 that .

f

(-1r ✓n2 n+ l n=O

+1 =

f Jn + 2

n=O

n

+l

1

Jn2 + 1 n-->oo n+l

diverges, a result which can be established independently since lim

= 1 =f. 0.

236

CHAPTER 6. SEQUENCES AND SERIES

Theorem 6.16. Let {an } be a sequence and suppose that

L = lim n----> exists. Then: 00

1. If L < 1, then Lan converges absolutely. n =O 00

2. If L > 1, then Lan diverges. n=O 3. If L = 1, then no conclusions can be drawn regarding convergence. As usual, with regard to convergence, the initial value O of n can be replaced by any other integer. This theorem is called the ratio test. Note that, since the ratio test involves lan l and not an, it cannot be employed as a test for conditional convergence. Example 6.85. Consider the series

!

!

1 3 1 3 2n an+l = lim (n + ) 3 = lim (n +3 ) = < 1 n---->oo 2 I an I n---->oo 2n+l n 2 n---->oo n

L = lim 00

⇒ Lan converges absolutely by the ratio test (Theorem 6.16). Thus, both n =O 00

Lan n=O converge.

L 00

=

(-1rn3 and 2n n=O

L=O lanl = L=O ; 00

n

00

n

3

n

6.2. SERIES

237

Example 6.86. Consider the series

L= lim

n--->oo

3n+l n4 + 1 an n4 + 1 - =3 > 1 _ _+_l = lim ----= 3 lim ---n--->oo (n + 1)4 + 1 I an I n--->oo (n + 1)4 + 1 3n

00

=?

Lan diverges by the ratio test (Theorem 6.16). Thus, both n=O

diverge. Example 6.87. For the series

which diverges,

and, for the series

which converges,

n -- = h. m-L= h. m lan+ll =1 n--->oo n + l n--->oo an

an l n I + I = lim = 1. n--->oo a n--->oo ( n + l ) 2

L = lim

2

n

Thus, convergence or divergence cannot be determined by the ratio test if L = l. Example 6.88. Consider the series (X)

00

'

� an = � !!..:_. � n=l

�nn n=l

238

. S1nce

an

> 0,

n

n +l)!

I anI = an= -n! and I an+l I = ( ( + 1 )n+l n

n



n n +l)n (n+ l) nn = l m _(_ _ _ _ _ lim _ _ _ n_ ! _ i n-+oo ( n +l)n+l n-+oo ( n +l) +l n! n 1 1 1 lm n . . -i --- = 1lffi 1 lffi ( -- ) n n-+oo n-+oo (1 + n-+oo ( !l )n n+ 1 e

a_+l_ lim 1- n n-+oo an

L

CHAPTER 6. SEQUENCES AND SERIES

I=

¾t

00

Hence, Lan converges by the ratio test. Since n=l convergence are synonymous.

an

=

n l im n-+oo ( n + l)n

n

< 1.

> 0, convergence and absolute

Example 6.89. Consider the series

(2n+ 2)! 2n(n!)2 2n+ 1 [(n+ 1)!]2 (2n)!



L

+ 1)(2n+ 1) (n

=

(-1-)

n! (2n + 2)(2n+ 1) ] [ 2 (n+ 1)!

2

n+l n+ 1 an l + I = 1.lffi 2--1.lffi I -n-+oo n + 1 n-+oo an

2

2n+ 1 n+l

= 2 > 1.

Hence, Lan diverges by the ratio test. n=O 00

Example 6.90. Consider the series

� _ � 1 ·3 · 5 ···(2n + 1) . L an - L 3nn ! n=O n=O

⇒ 00

L

3nn ! 1·3·5·· · (2n+ 1)(2n + 3) 1·3·5···(2n+l) 3n + 1 (n+l)! a 2n 3 + = � < 1. lim I n+l I = lim n---+oo 3( n + 1) 3 n---+oo an

Hence, Lan converges by the ratio test. n=O

2n+3 3( n + 1)

6.2. SERIES

239

Theorem 6.17. Let { an } be a sequence and suppose that

L

= nlim

---+=

exists. Then:

lan l l /n

1. If L < 1, then Lan converges absolutely. n=O (X)

2. If L > 1, then Lan diverges. n=O

3. If L

=

1, then no conclusions can be drawn regarding convergence.

Again, the initial value O of n can be replaced by any other integer. This theorem is called the root test. As for the ratio test, the root test cannot be employed as a test for conditional convergence. Example 6.91. Consider the series =

Lan

n=l Since

L

=

= [ln(n)] 5n

L

n= l

= nlim lanll/n = lim ---+ n (X)

---+(X)

n3

n

ln n 5 [ ( )] n3

= 0 < 1,

Lan converges by the root test (Theorem 6.17). n=l

Example 6.92. Consider the series

Since

f, = lim lan l l /n = lim (1 + n�oo

n�oo

(X)

Lan diverges by the root test (Theorem 6.17). n= l

.£) n

n

= e > 1,

240

CHAPTER 6. SEQUENCES AND SERIES

Exercises 6. 2 00

1. Write down the n

th

1 partial sum Sn of the series � for n L..t 2n n=O

f�

2. Write down the nth partial sum Sn of the series

n=l n

for n

3. Let { sn }�=l denote the sequence of partial sums of the series the sum s of the series if Sn

=

3n - 1 + for n � l. 2n 1

= 0, l, • • •

=

, 4.

l, 2, · · · , 5.

Lan . Determine

n=l

In Exercises 4-11, determine the sum of the given series. 00

4.

00

1 L5n n=O

5.

oo 2. 3n 6. I: � n=l

L 5. 2 00

8.

3n

n=O

00

1

7. L3n n= 2

3-2n

L 5 (-2)-3 n

n

L5 . 3 00

9.

2n

n= l

2-4n

2 L 4n +8n+3 00

00

10.

2n � (-5) n

11.

n=2

2

n=O

In Exercises 12-23, determine whether the given series converges or diverges. 12.

� 2Jn2 +3 L..t n n=l

13.

00

11.

(-1ri I:[1 n=O 00

1

16.

L n=l

n2 vn

18.

L=l

1 , r>0 r+l n

CX)

n

f

cos(n) 2 n=O 2 + sin (n)

n3 + 2n2 - 1 �-­ L..t n2 +n+l n=O 00

15.

17.

�vn L..t n2

n=l 00

19.



L..t n=l

n

r-1

'

241

6.2. SERIES 20.

1 L2n-l n=l

21. L2ne-n n=l

22.

1 P' p < 1 � n[ln(n)]

23.

00

00

00

24. Consider the series

2

1 p>l � n[ln(n)]P' 00

f

3 . 4 n=l n

(a) Approximate the sums of the series by s 2 and determine lower and upper bounds for the error R2.

(b) How large must n be taken in order that greater that 0.001? 25. Consider the series

Sn

approximate s with error no

f

1 .. r,;:; nyn n=l

(a) Approximate the sums of the series by s 3 and determine lower and upper bounds for the error R3.

(b) How large must n be taken in order that greater that 0.0001?

Sn

approximate s with error no

In Exercises 26-45, determine whether the given series converges or diverges. 1 26. � nyri,+3 00

28.

3n L4n+1 n=O 00

2n + 1 30. L� n=O oo

32. 34.

27.

1 �yn-1 00

3 29. L 2 n=O 2n -3 00

31.

00

L n=O

2n + n en

f

1 L n=l vn+n

33.

n3 +2n-1 3 +2 4 n=O n +n

-{jn2 +1 35. L 2 n=l n +n-l

00

f

vn-1 n=l n+l 00

2 42

CHAPTER 6. SEQUENCES AND SERIES 00

36.

38.

40.



f n=l oo

I:: n=l 00

42.

44.

n 213

1 ln(n)

ln(l + vn

37.

¾)

39.

(-)l n n 2/3

f f

ln(n) n2 n =2

[ln(n)] 3 nfo n=2

(-r 1 41. � + 2 sin(n)

(-1r

00

43.

�Jln(n)

f

(-1r - 1 n n=l

45.

(-r l n � en ln(n)

f

l n 1+(-r 2 n n=l

L (-n1r-1 00

46. Consider the series

2

n=l

(a) Approximate the sums of the series by s 3 and determine an upper bound for IRnl(b) How large must n be taken in order that than 0.0001 in absolute value?

Sn

approximate s with error less

L n(-+r1 1 00

47. Consider the series

n=O

3

(a) Approximate the sums of the series by s 2 and determine an upper bound for IRnl(b) How large must n be taken in order that 1 error of ? 1, 001 ±

Sn

approximates with a maximum

In Exercises 48 5 - 3, determine whether the given series converges absolutely, converges conditionally, or diverges.

243

CHAPTER 6 EXERCISES 48. 50. 52.

(-1r L n -1 n=2 00

f n=O

f

(-l)n vn n+l

(-l)n [ln(n)] 2 n n=2

49.

(-1r+1 � n Jn

51.

(-2r L� n=l

53.

CX)

00

(-2r L 3n _ 1 n=I 00

In Exercises 54 63, determine whether the given series converges or diverges. 54.

L 00

n=l

3n-l n 2 55. � n 1 2 + (n + 1)4

2n(n + 1)3 3nn 2

CX)

n 3 + 2n2 - 1

56 .

L=O

58.

!;

(n!)2 n 2 (2n)!

59.

60.

(2n + 1)! � n !(n + 1)!

61.

62.

(-n r � [ln(n)J 2n

63.

00

n

CX)

57.

n,

00

CX)

2n (n + 1)! n=O

L CX)

nl

CX)

� 3 · 6 · 9 ·. · · (3n)

f n=O

n !(n

+ 3)! (2n + 3)!

(-1rn2n � 2n (n 2 + l)n CX)

Chapter 6 Exercises 1. Prove, by induction, the formula geometric series

Sn

Lrn. CX)

=

1 - rn+l for the n th partial sum of the 1 -r

---

n=O

CX)

2. Prove that Lan converges if and only if n=l m 2': 1. 3. Determine whether or not the series

f

La CX)

n=ni

n

converges for every integer

L onverges. n){ [ln(n)]} c n ln( n=l6

244

CHAPTER 6. SEQUENCES AND SERIES

'°' 00

4. Determine whether or not the series

1 converges. L1+2+3+···+n n=l

'°' 00

5. Determine whether or not the series

n converges. 2 L1+4+9+···+n n=l

6. Prove that every finite series converges absolutely. 7. Consider the Fibonacci sequence a1 = a2 = 1 and an = an -I + an -2 for n 2: 3. The latter relation is called a difference equation, which is the discrete version of a differential equation. Find two solutions in the form an = rn , take a linear combination, and impose the "initial conditions" a 1 = a2 = 1 to determine an explicitly in terms of n for every n 2: 1.

Chapter 7 Taylor Series The subject of Chapter 6 has been the study of sequences and series of numbers. The topic of the present chapter is the study of series which contain a variable x, i.e., series of functions. Thus, the convergence or divergence of the series, as well as its sum, will depend upon the value of x, making the infinite series a function of x.

7.1

Power Series

Definition 7.1. Let a be a real number. A series of the form 00

2:::C n (x - a)11 n=O

=Co+ c1(x - a)+ c2 (x - a) 2 + c3(x - a)3 + · · ·

(7.1)

is called a power series about a or centred at a, and the numbers Cn, n = 0, l, 2, · · · are called the coefficients of the power series. In the special case where a = 0, the series reduces to

L 00

n=O

CnX

n

= Co+ C1X + C2X + C3X + · · · , 2

3

i.e., a power series about 0. Note that, at x = a, the power series (7.1) reduces to the single term Co and, hence, converges absolutely. However, only those power series which converge for all x in an interval arc of any practical use. Example 7.1. The power series 00

Lnnx n n=O

245

246

CHAPTER 7. TAYLOR SERIES

is centred at 0 and has the coefficients Cn series converges absolutely if lim lan l l / n

n---+oo

= n---+oo lim ln

=n n

n

By the root test (Theorem 6.17), the

.

x l / n l

n

= n-+oo lim nlxl
oo I a-l I < 1, n-->oo an Cn i.e.,

and diverges if

Ix - al


1�1,

oo Cn +1

1�1-

Ix - al> lim n -->oo Cn l +

CHAPTER 7. TAYLOR SERIES

248

Hence,

R= hm

. I

-nC

n->oo C n+ 1

is the radius of convergence.

I

Once R has been determined, the series converges absolutely at least for

a-R < x < a+ R. In order to determine convergence or divergence at the two endpoints Ix - al = R, and, hence, the entire interval of convergence, a test other than the ratio test must be employed. Example 7.4. Consider the power series oo

n

Ln:l n=O 1 about O with the coefficients en= - -. The radius of convergence is n+l Cn I . I. n+2 R= hm - = hm --=1. n->oo n + l n->oo Cn+ 1

Hence, the series converges absolutely for lxl At x =l, the series becomes

< 1, i.e., -1 < x < l.

1 L --, which diverges. n+1 00

n=O

(-1r

00

At x = -l, the series becomes�--, which converges conditionally. Thus, the Ln+l n=O interval of convergence is I = [-1, 1). Example 7.5. Consider the power series

1

about 3 with the coefficients Cn = -, n � l. This is a power series with Co = 0. The n radius of convergence is

R = lim

n->oo C n+l 1�

= lim 1

n->oo

n

+ n

1

=1.

7.1. POWER SERIES

249

Hence, the series converges absolutely for lx-31 < 1, i.e., -1 < x-3 < 1, or 2 < x < 4. At x At x

=

2, the series becomes

f

(-)" l , which converges conditionally. n

n=l

= 4, the series becomes

is I= [2, 4).

L=l n-,1 which diverges. Thus, the interval of convergence 00

n

Example 7.6. Consider the power series � (3 - 2x)" L n-1 n=2 Before the point a at which the series is centred, as well as its coefficients Cn, can be determined, the series must be placed in the standard power series form oo (-2)" 3 - 2x)" � = (x - )2 Loo=2 ( n-1 Ln-1 =2 n

n

(n-1

3 from which we find a= - and c11 2 The radius of convergence is

.

= - 2)",-n 2". 2.

. I

R = hm - Cn

n->

oo Cn+l

n

'

This power series has Co

I = hm .

211 n ---n+n->oo n-1 2 l

1

= c1 = 0.

= -. 2

Hence, the series converges absolutely for

I �1 < ! x-

At x

=

2

'2

i.e.

'

- < x -� < 2 2

!

! '2

or

1 n 00

1

J(x) = 1-t = which converges absolutely for Ix -21 series is therefore R = 1.

n=O

=

00

=

I)x - 2r, n=O

ltl < 1. The radius of convergence of the

Example 7.12. Consider the function 1

f(x) = 8-2x In order to express f(x) as a power series about 2, write f(x) as

f(x) = where t =

1 8- 2x

1 8 - 2(x - 2) - 4

1 4-2(x- 2)

1

1

41 -

x-2 2

1 1 41- t'

x-2 Then 2

- -.

2 which converges absolutely for Ix; \

= itl < 1, convergence of the series is therefore R = 2.

i.e., for Ix-21 < 2. The radius of

Example 7.13. Consider the function .

4

J(x) = 4 + x2. In order to express f(x) as a power series about 0, write J(x) as 1 1 4 ( 2 + j X) = 4 x = 1 + x4 = 1- t' .

2

7.2. REPRESENTATIONS OF FUNCTIONS BY POWER SERIES where t

x2

= --



255

Then

The series converges absolutely for

2

I- : 1 =!ti< 1, i.e., for lxl < 2. Thus, the radius

= 2.

of convergence of the series is R

Example 7.14. Consider the function

J(x) With t

1

1

+-=1 +x l - 2x

= -x,

with radius of convergence R 1

=

with radius of convergence R2

= .

1, and, with t

1

2

= 2x,

It follows that

The radius of convergence is 1 . 2' 1 It can be shown, in general, that the 2 radius of convergence R of a linear combination of two power series is at least the smaller of the radii of convergence R 1 and R2 of the two series. which is the smaller of R 1

=

1 and R2

= -.

Theorem 7.2. Suppose that

J(x)

= Lcn(x - at, n=O

Ix - al< R,

256

CHAPTER 7. TAYLOR SERIES

with the radius of convergence R > 0. Then the derivative of the series can be obtained by term-by-term differentiation of the series, i.e.,

with radius of convergence R. Since the first term (n of (7.3) is 0, the series may be expressed as 00

00

n=O

n=l

and, making the change of index k

= 0)

in the series on the right

n - l, we obtain

=

00

00

n=l

k=O

which is the more standard form of a power series. Example 7.15. Consider the function J(x)

1

= (1 - x)2

1 Since f(x) is the derivative of--, we obtain, by Theorem 7.2, l-x 1 (1 - x) 2

where k

=

00

-5!:___1_ = -5!:__ L Xn dx l - X dx n=O

= n - 1,

00

00

00

n=O

n=l

k=O

= I:nxn-l = I:nxn-1 = L(k+ l)x k ,

with radius of convergence R

=

1.

Theorem 7.3. Suppose that J(x)

00

= I:en (x - at, n=O

Ix - al< R,

with the radius of convergence R > 0. Then the integral of the series can be obtained by term-by-term integration of the series, i.e., J

oo oo oo Cn ( X - a)n+l J f(x)dx= f Lcn(x-atdx= [en(x-at]dx= n+l +C, n=O n=O n=O

L

L

(7.4)

257

7.2. REPRESENTATIONS OF FUNCTIONS BY POWER SERIES with radius of convergence R. Making the change of index k the right of (7.4), the series may be expressed as oo

� Cn

� n=O

(x -

a )n+l

= n + l in the series on

oo � Ck-1 (x - a )k C C = + ' + � k k=l

n+ 1

where C is an arbitrary constant of integration. Example 7.16. Consider the function

f(x) = ln(l - x), \x\ < 1.

1 Since J(x) is an antiderivative of---, we obtain, by Theorem 7.3,

l-x

ln(l- x)

= -

J

l

-- dx = l-x oo

00 J � x n dx

� n=O

xk = C-L k'

00

xn+l = - � _ _l + C

�n+ n=O

k=l

where k = n + l. Since both the series and ln(l-x) have the value 0 at x = 0, C = 0 and, hence, 00 1 k ln(l-x) = (7.5) x ,

L� k=l

with radius of convergence R = l. Since the series converges for any x with \xi < 1, and I - x\ = \x\, replacing x by -x in (7.5) gives ln(l + x)

=

oo

oo

k=l

k=l

oo (

L �1 (-xl = L - (�l)k x = L -l)kk-1 x

as the power series representation off (x)

k

= ln(l + x)

k=l

about 0 with R

k

(7.6)

= l.

In all of the preceding examples, the representation of J(x) as a power series has relied upon manipulations of the geometric series, i.e., Equation (7.2). This is not always adequate, and more general methods are required. Theorem 7.4. If

J(x) =

L Gri(x - at,

\x - a\ < R,

n=O

with the radius of convergence R > 0, then the coefficients

Cn

are given by (7.7)

258

CHAPTER 7. TAYLOR SERIES

Proof. By Theorem 7.2, the power series may be differentiated term-by-term. Thus, J(x)

5 4 2 co+ c1(x-a)+ c2 (x - a) + c3 (x - a)3 + c4 (x - a) + c5(x - a) + · · · , 4 2 3 c1 + 2c2 (x - a)+ 3c3 (x-a) + 4c4 (x-a) + 5c5 (x - a) + · · · , 2c2 + 2 · 3c3 (x - a)+ 3 · 4c4 (x - a) 2 + 4 • 5c5 (x - a)3 + • • • ,

J '(x) f"(x) f" '(x) /4l(x)

2 · 3c3 + 2 · 3 · 4c4 (x -a) + 3 · 4 · 5c5 ( x - a) 2 + · · · , 2 · 3 · 4c4 + 2 · 3 · 4 · 5c5 ( x - a) + · · · ,

etc. It follows that

etc., and, hence, Co = J(a) =

J (a) j(0 l(a) f"(a) , C1 = J '(a) = -' -, C2 = ' ! 2! 0 1,

and, in general,

Cn

=

j(n)(a) for all n 2:: 0. n.1

Example 7.17. Let us confirm the formula (7.7) for the geometric series J(x) = about a

= 0,

which has

Cn

1 l-x

= 1 for all n 2::

L n=O

J(0) = 1 = 0!,

1 ⇒ (1-x)2 2 ⇒ (1-x)3 2-3 ⇒ (1-x)4 2-3-4 ⇒ (1 - x)5

f '(x) f"(x) J" '(:r) j(4l(x) etc., and, in general, j( n)(0)

= � xn

0.

1 -⇒ l-x

J(x)

CX)

=

n!. Hence, Cn

Definition 7.3. Given a function

J,

=

j'(0)

=

1 = 1!,

j"(0) = 2 = 2!, J"'(0) = 3!, JC4l(o) = 4!, j(n)(0) = 1, as expected. n.1

the series

7.2. REPRESENTATIONS OF FUNCTIONS BY POWER SERIES

259

is called the Taylor series of f about a or centred at a. In the special case where a= 0, the Taylor series off about 0 is also called the Maclaurin series off. By Theorem 7.4 and Definition 7.3, the power series which represents a function about a point a is its Taylor series about a. A distinction between the two arises, however, due to the fact that there are functions J for which the Taylor series T(x) is not equal to f (x) on an interval Ix - al < R, R > 0, and is therefore not a power series representation of J(x). We shall see later how to prove that T(x) = f(x) for functions for which it is true. It follows by Theorem 7.4 that the power series representation of a function f about a point a is unique, i.e., if

J(x)

00

= :z::::>n(X - ar = n=O

L d (X - ar, 00

n

n=O

Ix - al < R,

R > 0,

j(n)(a) = dn , n 2:: 0. Thus, as observed in Example 7.17, different methods n.1 of computing the series representation of a function must yield identical results. then Cn

=

It follows also by Theorem 7.4 that if

J(x)

= :z::= en(x - ar n=O

= o,

then en = 0 for all n 2:: 0. This fact is essential in the solution of differential equations by infinite series. Example 7.18. Let f(x)

= ex

and a= 0. Then 1 n!

and, hence, the Taylor series of ex about 0, i.e., its Maclaurin series, is

with the radius of convergence R

=

lim

n-+oo 1 � Cn+l

1

= nlim

-+oo

( n + l) ! nl

lim

n-+oo ( n

+ l) = oo.

260

CHAPTER 7. TAYLOR SERIES

Thus, the series converges absolutely for all x. We shall prove later, but assume for the present, that T(x) = f(x) = ex for all x, i.e., 1 ex = � -xn for all x. L..t n! n=O CX)

Example 7.19. Let f(x)

x by 2x gives

Notice that

Cn

e 2n 1 n.

=

=

2

x

= e 2x and

00

L

=

n=O

a= 0. Since ex =

1 (2xr , n.

=

00

2 x L, n. n

n=O

n

L 1n.1 x CX)

n

n=O

for all x, replacing

for all x.

j( n)(o) , in accord with Theorem 7.4. n.1

Example 7.20. Let f(x) = ex and a = 3. Then j(n)(x) = ex for all n ?: 0 and, hence, e3 f(n) (3) oo e3 (x -3 and T ( x) = = Cn = 1 1 n. n. n.1 n=O

L

t

is the Taylor series of e about 3. An alternative is to employ e x

.

and to replace x by x -3 to obtain e =e e x

3 x 3

=

00

1 -(x-3t e L..t n! n=O 3



x

=

1 x L1 n. CX)

n=O

n

for all x

00

e3 L..t n!

= � -(x-3r, n=O

which is the power series representation of e about 3 and, hence, its Taylor series about 3, as above. x

.

Example 7.21. Let f(x)

=

tan- 1 (x), - 1- l n=O



1

1+X

00

00

n=O

n=O

= 2)-xr = I)-1rxn , lxl < 1.

1 Alternatively, since -- = (1 + x)- 1, we may also employ the binomial series with l+x a= -1. Thus, with a= -1, (�) = 1 and, fork 2:'. 1,

(�) = a(a- l)(a - 2)(a � 3) •••(a - k + l) k (-1)(-2)(-3) · · · (-k) kl

=

(-l) kk! kl

= (-l). k

266

CHAPTER 7. TAYLOR SERIES

Hence,

as above. Example 7.27. Consider the function J(x) =

(l

1

_

x)3

= (1 - x )- 3. Its Taylor series

about O (Maclaurin series) is the binomial series with a by -x. Thus, with a= -3, (�)

= 1 and,

=

-3 and with x replaced

fork � 1,

a(a - l)(a - 2)(a - 3) ···(a - k +

k! (-3)(-4)(-5)(-6) · · ·(-k - 2) k! k (-l) · 3 · 4 · 5 · 6 · · · (k + 2) k!

1)

In order to place the last expression into a more concise form, multiply and divide by 2. Then

(-l)k · 3 · 4 · 5 · 6 · · · (k + 2) 2 (-l)k (k

2

k!

+ 2)!

2k! (-l)k (k + l)(k + 2) 2

Hence, 1

(1 -

)

X 3

t, (�) 1+

f

(-x)'

(-l)k (k

k=l



l)(k + 2)

� (k + l)(k + 2) x k , � k=O

since

(k + l)(k + 2) = 1 when k = 0. 2

2

(-1/ xk

lxl < 1,

267

7.2. REPRESENTATIONS OF FUNCTIONS BY POWER SERIES Alternatively, employing the geometric series and Theorem 7.2, 1

l- x

00

L k

Xk

=O

�-1- = � � xk = � kxk -1 � dx l- x dx � (1 - ) k=O k =O � 1 = � � kxk-l =�k(k -l) xk -2 = � k(k -l) xk-2 2 3 � ) � X d x (l x) dx � ::::} (1 k=O k =O k=2 since k(k - 1) = 0 when k = 0 and k = l. Making the change of index n = k -2, we 1

::::}---

X 2

obtain

_ � (n + 2)(n + 1) X n 2 (1 - X )3 � 1

as above.

'

I X I < l,

1

Example 7.28. Consider the function f (x) = y11+x = (1

l+ x

about O (Maclaurin series) is the binomial series with a=

+ x )- 2. Its Taylor series

1 -2"

I

Thus, with a= -

(�) = 1 and, fork :2: 1,

1

2,

a(a - l)(a -2 )(a - 3) · · ·(a -k + l)

k! (-½)(-�)(-�)(-�)... (-¥) k! k · 3 · 5 · 7 · · · (2k- 1) · 1 (-1) k 2 k!

In order to place the last expression into a more concise form, multiply and divide by 2 · 4 · 6 · · · (2k). Then

( -1)k · l · 3 · 5 · 7 · · · (2k- 1) 2 · 4 · 6 · · · (2k) 2 kk! 2 · 4 · 6 · · · (2k) k (-l) (2k) ! 1 k k 2 k! 2 · 1 · 2 · 3 · · · k (-l)k(2k)! 22k(k!)2 ·

Hence,

1

vl1+x

k k =�(a) X k = 1 � (-lk) (2k) !X k = � (-l) (2k)! X k + � 22 (k!)2 � 22k(k!)2 k k=l k =O k =O �

)

lxl < l,

268

CHAPTER 7. TAYLOR SERIES

. (-l)k(2k)! smce 22 k(k!)2 Example 7.29.

= 1 when k = 0. = �- In 8+x

Consider the function J(x)

3

binomial series, express f(x) as X

X

X

2

(

x2

f(x)=-;;-;===2=---2 =- l+8 2 \1/8 + x 2{/1 + x

order to employ the

)-½

8

The Taylor series off about O is then the binomial series of (1 a=

_!,3 multiplied by �-2

Thus, with a=

(�) =

a(a -

_!,3

l)(a - 2)(a

k

(a) = 0



2

+ t)°' with t = �

and

1 and, for k > l,

3)•·•(a - k + l)

(-½)(-½)(-D(-�) ... (-¥)

k! (-l)k · 1 · 4 · 7 · 10··· (3k - 2) Hence,

f--- (a) x

(1 + t) -½ _ - L.., k=O

k

f---. (-l) · 1 · 4· 7· 10··· (3k - 2) t , 3kk!

k _

- l + L..., k=l

k

k

ltl < 1

(x2 )k]

k � l + f---(-l) -1-4-7-10···(3k - 2) [ 3kk! L..., 8 2 k =l x (-l)k · 1 · 4 · 7 · 10··· (3k - 2) 2 k+l + x 2 L..., 23 k+l3k k! k=l

f---

valid for

2

I �

1

=

ltl < 1, i.e., for lxl < 2yl2.

Example 7.30. Consider the function J(x)

= v1f+x = (1 +x)½.

about O is the binomial series with a=�- Thus, (�)=1,

k "?. 2, a(a -

=a=�' and, for

l)(a - 2)(a - 3)···(a - k + 1) k!

(½)(-½)(-�)(-�) ... (-¥) (-1) -

(7)

k 1

·

k! · 1 · 3 5. ··(2k - 3) 2kk!

Its Taylor series

7.2. REPRESENTATIONS OF FUNCTIONS BY POWER SERIES

269

noting that there are k factors in the numerator, only k - 1 of which have minus signs. Multiplying and dividing by 2 · 4 · 6 · · · (2k -2), we obtain (-l)k-l · 1 · 3 · 5 · · · (2k - 3) 2 · 4 · 6 · · · (2k-2) 2 · 4 · 6 · · · (2k - 2) ' 2k k! k 1 1 (-l) - (2k-2)! 2k-l · 1 · 2 · 3 · · · (k. - 1) 2k k! (-l)k-1(2k-2)! k>2 - ) 22 k-lk!(k- 1)! ) which is also correct for k = 1. Hence,

(a) L 00

v1f+x

=

k=O

k

k x =

00

1

+

L k=l

(-1t-1(2k-2)! k lxl < 1. 22k-lk!(k- 1)! x '

Taylor Polynomials and Approximations Definition 7.4. Consider the Taylor series of a function J about a point a, i.e.,

T(x) =

oo

L n=O

f( n) (a ) (x n.I

- ar.

T he nth partial sum Tn (x) of the Taylor series is

L n

f(k) ( )

k=O

k!

a (x - ) k a

J(a) + J'(a)(x

+···+

f "(a)

f"(a) ' (x - a) 3 -(x - a) 2 + - a)+ � 21

f( n l(a) (x -a)'\ n.1

and is a polynomial of degree n, called the nth -degree Taylor polynomial off about a. If the Taylor series of a function converges to the function on an interval, i.e., T(x) = f(x) for Ix - al 0, then J(x) can be approximated by its Taylor polynomial Tn (x), and the error is

11,,-,(x) = f(x) - Tn (x) called the remainder.

=

oo

L

k=n+l

f(k) ( a )

kl

(x

- a)k ,

270

CHAPTER 7. TAYLOR SERIES

The product of two Taylor series, or their quotient, about the same point a, is again a Taylor series about a, but the formula for the coefficients of the product is complicated ( see Exercise 26 in Section 7.1), and none exists for the quotient. In such cases, their Taylor polynomials may still furnish useful information. Example 7.31. Consider the function f(x) = ex cos(x). In order to determine its 4th -degree Taylor polynomial about 0, the Taylor polynomials of suitably high degree of the functions ex and cos(x) must be multiplied. Since

eX

12

l3

14

2

6

24

= 1 + X + -X + -X + -X + · · ·

and

cos(x)

1

1

= 1 - -x2 + -x + · · · ' 24 2 4

and the product is required only up to the fourth power of x, the terms xk for k � 5 are neglected. Thus, ex cos(x)

=

2 (1 + x + !x

1+X Hence, T4 (x)

= 1 + x - !x3 3

-

+ !x3 + 2-x4) 6

24

- -X - -X

+ ··· .

!x4 for f(x)

= ex cos(x)

2

13

1 4

3

6

6

about 0.

Example 7.32. Consider the function J(x) = tan(x). In order to determine its 5th -degree Taylor polynomial about 0, the Taylor polynomial of suitably high degree of the function sin( x) must be divided by that of cos( x). Thus, 1 5 1 3 x x + 120 6

sin(x)

=x-

cos(x)

= 1 - -x + -x + · · · ' 24 2

and

1

1

2

+···

4

and, dividing the former by the latter and neglecting the terms xk for k � 6, we obtain sin(x) cos(x)

1-

X

Hence, T5 (x)

+ _1 120 x 2 4 lx + ....!... 24x 2

3 x - lx 6

+

1 3

-X

3

5

2 +-x5 + · · ·. 15

= x + !x3 + �x5 for tan(x) about 0. 3

15

271

7.2. REPRESENTATIONS OF FUNCTIONS BY POWER SERIES

Example 7.33. Consider the function J(x) = (x2 -1)ln(x). Its Taylor series about 1 can be determined by multiplying the Taylor series of ln(x) about 1 by the Taylor series of x2 - 1 about 1. Since x2 - 1 is a polynomial of degree 2, its Taylor series is also a polynomial of degree 2, and the multiplication can be performed without great difficulty. By Example 7.22, ln(x) =

n =l

Ix -11 < 1.

= x2 - 1 about 1 can be obtained by Equation (7.7). = 2x, g"(x) = 2, and g(n) (x) = 0 for n 2:: 3 ::::}

The Taylor series of g(x)

g'(x)

T hus,

g(n) (l) g"(l) = = 0 for n 2:: 3, = -- = 1, and Cn n.1 2 1 about 1 is T(x) = 2(x - 1)+ (x - 1)2 . As can be

g, (l) = 2, c2

= g(l) = 0,

C1 = and the Taylor series of x2 readily verified, co

00

L ( -1nr-1(x - 1r,

-

T(x) = 2(x - 1)+(x - 1)2 = x2 - 1 for all x.

Then the product is (x - 1)ln(x) 2

=

2 [2(x - 1) + (x - 1) ]

00 (

L - 1nr-1(x -1r

n=l

Making the change of index n - n+l in the first series on the right, we obtain (x2 - l)ln(x)

=

� 2(-lt (x - 1r+ 2 + � (-1r-1 (x - 1r+ 2 2(x -1)2 +L L n+l n n =l n =l � [2(-1)" + (-1r-1] (x - 1r+2 2(x -1)2 + L n+l n n =l n n � [2(-l) - (-l) ] (x - 1r+2 2(x -1)2 + L n+l n n =l - )n ( "+2 l) 2(.r: - 1)2 + [( l n - ] (x - 1) n n + 1) ( n =l

2(x - 1)2 +

f f n=2

[

(-l)n (n - l) +2 ] (x- 1r , Ix - 11 < 1. n(n + 1)

272

CHAPTER 7. TAYLOR SERIES

The last series on the right begins at n = 2 because n - l = 0 when n = l. The series converges absolutely for Ix - 11 < 1 since the Taylor series of ln(x) converges absolutely for Ix - 11 < 1. Truncation of the series by neglecting the terms beyond (x - 1r yields 2(x - 1)2

+

f k=2

( )k (k - 1) [ -l ] (x - 1t+2 k(k + 1)

i.e., the Taylor polynomial Tn (x) of degree n. Example 7.34. Consider the function f (x) of ex converges to ex for all x, we obtain = n ex = � :._ L n! n =O

=}

= e-x

2

. Assuming that the Taylor series

= ( x2 )n = (-l)n x 2n 2 � -e-x = � = L n! L n! n =O n =O

Since this is an alternating series for x =/- 0, if f(x) is approximated by its nth-degree Taylor polynomial Tn (x), the error, in absolute value, will be less than the absolute value of the first neglected term. Thus, with bn (x )

=

IRn(x)I If, for instance, f

(1) = e-¼

and 1 1 e-4 � T2 ( )

2

=

2

L k=O

x

n.

lf(x) -Tn (x)I < bn+1(x) is approximated by T2

(-l) k (-!-2) 2k kl

1

2

2n

= -1 , =

x 2n+2

(n + l)!

(1), l

1

·

then

4

= 1 - (2) + 2 (2) =

25 32

= 0 . 78125.

The value of e-¼ obtained by calculator is approximately 0.7788. The difference between the exact value and the approximate value is -0.00245, and

I - 0.002451 = 0.00245 < 0.0026, as guaranteed by the general theory. The difference between the exact value and the approximate value is negative because the first neglected term is negative and, hence, the approximate value is greater than the exact value.

7.2. REPRESENTATIONS OF FUNCTIONS BY POWER SERIES

273

Example 7.35. Consider the definite integral

r½ e-x dx.

lo

2

Since f(x) = e-x does not have an elementary function as an antiderivative, we must 2 appeal to infinite series. Employing the Taylor series of e-x in Example 7.34 and Theorem 7.3, we obtain 2

r½ e-x2 dx r½

lo

= lo

00



00 00 (-1r (-1rx2n+1 ½ (-1rx2n dx = 1 = 1 2n � 2 + (2n+ l)n!· � (2n+ l)n! 0 n!

The sum s of the alternating series

I)-1 tb 00

n

n=O

on the right is the exact value of the

integral. If the series is approximated by its nth partial sum S n , then 1 IRnl = Is - sn l < bn+l = 2n+3 (2n+ 3)(n+ l)! • 2

If, for instance, an approximation of the integral is required with error less than 0.01 in absolute value, then n must be sufficiently large in order that IRnl < 0.01. With n = l, 1 1 = - � 0.003 < 0.01, IR1I < b2 = 5 2 ·5·2 320 and

1 0

2

1 1 11 e -x2 dx � s1 = - - -= - � 0.4583 ' 3 2 2 .3 24

with the required accuracy.

Example 7.36. Consider the definite integral

=

l / ¼ --;== dx. lo + x6

vl

By Example 7.28 with x replaced by x6, we obtain 00 1 (-l) k (2k)! 6k Vl + x6 = � 22k (k!)2 x ' lxl < 1.

274

CHAPTER 7. TAYLOR SERIES

Hence, by Theorem 7.3,

1

1

1 -dx ¼ Jl + 6 x o

k ¼ � (-l) (2k)! x6kdx 22k (k!)2 0 �

00

=

� �

(-l)k (2k)! � 22k (k!) 2(6k + 1)4 6k+l

(-l)k (2k)!x6k+1 ¼ I 22k (k!)2 (6k + 1) 0

00

( -1)k ( 2k)! � (k!)2(6k + 1 )4 7k+l 00

is an alternating series, and IRkl < bk+l

=

(2k + 2)! . [(k + 1)!]2(6k + 7)4 7 k+8

2 1 1 If k = 0, then IRol < b1 = = 229,3 � 0.00000 4359. Hence, = 8 15 . . 7 2 7 4 76

1

1 dx � s0 ¼ � o V 1 + x6

= -1 = 0.25

with error less than 0.00000 5 in absolute value.

4

Taylor polynomials can be employed in the evaluation of limits, even in cases where L'Hopital's rule is inapplicable or too tedious . Example 7.37. Consider the limit

. (2x2 +3)sin 2 (x) 1 Im--2----. x---->O x cos (x)

Employing the Taylor series of sin (x) and cos(x) about O and retaining only the first two terms, we obtain 2 2 (x) (2x+3)sin --1 11n x---->O x2 cos (

.

x)

2 x3 +-··)2 . (2x +3)(x -l 6 hm x---->O x 2 (1 - ½x2 + • • •) 2 + ... 2 ) . (2x2 +3)x2 (1 -l 6x hm 2 2 x---->O x (1 - ½x + • • •) 2 + x2 + ... )2 . (2x 3)(1 - l 6 hm x---->O ( 1 - ½ x 2 + ... )

7.2. REPRESENTATIONS OF FUNCTIONS BY POWER SERIES since x k ----, 0 for all k � 1 as x ----, 0. Example 7.38. Consider the limit . (ex - x - 1) cos(x) -1 1m--- -x--->O sin2 (x) Since e x = 1

x2 2

x3 6

+ x+ - + - + · · • 1 x2 +... ) (x x +... ) ( 1 - +2

. (ex - x - 1) cos(x) 1Im----=---- -x--->O si1i(x)

lim

2

3

6

2

+... )2 (x _ x 2 ( l2 + £126 +...) ( 1 - l2 x 2 +...) lim x--->O x2 (1 - ½x 2 + • • • )2 (l + £12 + .. ·. )(1 - l2 x2 + · · ·) 6 lim 2 x--->O (1 - ½x2 + ... ) 2 x--->O

½x3

1 2

Example 7.39. Consider the one-sided limit e x sin(x) . 1 1m -;::::===----, x--->O- v'2x2 + x3 cos2 (x) i.e., x----, 0 with x < 0. Then e x sin(x) lim --;::::===--­ x--->O- J2x2 + x3 cos 2 (x)

(1 + x + • • • )(x - l6 x3 +...) . hm x--->O- jxjJ2 + x(1 - ½x 2 +... ) 2 x2 +...) X (1 + X + · · · )(1 - l . 6 hm x--->O- lxl J2 + x(1 _ ½x2 +... )2 lim

x--->O-

1

· · ) ( 1 - ½x 2 + · · · ) J2 + x(1 - ½x2 +... )2

(1 + X + ·

J2 since

vlx2 = lxl = -x for x < 0.

Theorem 7.5. Suppose that the function f has the Taylor series oo

T(x) = � 6

n=O

j(n)( ) a (x at n!

275

276

CHAPTER 7. TAYLOR SERIES

about the point a, and let Tn (x) =

L n

j(k)( a)

k!

k=O

(x - a) k

be its n th -degree Taylor polynomial. T hen the remainder Rn(x) = f(x) - Tn(x) is given by f (n+l)(z ) (x - a)n+1 Rn (x) = (7.9) (n + 1)! where z is a number between a and x. Equation (7.9) is known as Taylor's remainder formula.

Example 7.40. Consider the function J(x) = ex . By Example 7.18, its Taylor series about 0 is

and converges absolutely for all x. If ex is approximated by its n th-degree Taylor polynomial n 1 k Tn(x) = x , k!

L k=O

then, employing Theorem 7.5 and the fact that j( n+l)(z) = ez for all n � 0, we obtain ez Rn (x) = ---!xn+l (n + 1) and, hence,

IRn(x) I = (n

e

2

+ 1)!

lxl n+l ·

Suppose that we wish to approximate ve = e 112 with error less than 0.001 in absolute 1 1 1 value. Since x = and z is between a= 0 and x = , 0 :S z :S < 1. Then

2

if (n + 1)! 2 obtain

n+l

I

Rn

( 21) I

z

1)

e = --- ( ( n + l) ! 2

n+

2

2

l

e

< 0.001 < ( n + l) ! 2n+ 1 -

� lO00e � 2718.28, which is satisfied for n � 4. Taking n = 4, we 4

L

1 1 1 1 1 1+ +8+ � 1.6484· + = 3 k ! -le� k=O k 2 84 48 2 The value of ft obtained by calculator is approximately 1.6487, with a difference of 0.0003 < 0.001, as required.

7.2. REPRESENTATIONS OF FUNCTIONS BY POWER SERIES

277

Theorem 7.6. Suppose that the function f has the Taylor series oo

(nl

J (a) "°""' --(x - at L n!

T(x) =

n=O

about the point a, with radius of convergence R > 0. Let

Tn (x)

=

L J k!( a) (x - al n

(k)

k=O

be its n th -degree Taylor polynomial, and let Rn (x) Then T(x) = J(x) for Ix - al < R if and only if lim Rn(x)

n-+oo

= 0,

= J(x) -Tn (x) be the remainder.

Ix - al < R.

Proof. T(x) = f(x) if and only if J(x) - T(x) = 0 if and only if

Rn (x) smce lim Tn (x) = T(x).

= J(x) - Tn (x) - 0

as n - oo

n-+oo

Example 7.41. In order to prove that the Taylor series of f(x) converges to ex for all x, it must be shown that

ex about 0

By Theorem 7.6, it suffices to show that Rn(x) = f(x) - Tn (x) - 0 as n - oo. By Theorem 7.5, f( n+l) (z) n+l -,------� ez = xn+l x Rn(x) =

(n+l)!

(n+l)!

where z is a number between 0 and x.

Since �

e

x

� (n +l)! I X I

n+ l -

-

'

l x � I I n+ 1 e � (n +l)! X ex

lx/ n+l = 0 by Theorem 6.9. (n + l ) ! Hence, lim /Rn(x)I = 0 by Theorem 6.4 and, by Theorem 6.2, lim Rn(x) = 0.

converges (absolutely) by the ratio test, lim

n-+oo

n-.oo

n__.oo

CHAPTER 7. TAYLOR SERIES

278 Similarly, if x< 0 , then x :::; z :::; 0

⇒ ex :::; e z :::; 1 ⇒

ez 1 n+l < 0< --lxl n+l --t 0 - I.Rn(x)I = --lxl (n + 1)! - (n + 1)! 00

1

⇒ lim Rn(x) = 0. Thus, ex ="' -xn for all x. L n! n---+oo n=O

More generally, for any real number a,

i.e., the Taylor series of ex about any real number a converges to ex for all x. Example 7.42. In order t o prove that the Taylor series of f (x) converges t o sin(x) for all x, it must be shown that OO

(-ll

T(x) = I:--x2k+I l k + )! (2

=

sin(x) about 0

= sin(x) = f(x).

k=O

By Theorem 7.6, it suffices t o show that Rn(x) = f(x) - Tn (x) --t Oas n --too. By Theorem 7.5, f (n+l)( z ) n+l x x Rn ( ) = (n + 1)! where z is a number between O and x. Since j(n+I )(z)

= ±sin(z) or ±cos(z), lf (n+ l)(z)I:::;

1 for all real z and n 2: 0. Hence,

1 l (n+l) (z)I lxln+ l --t 0 0< - IRn(x)I = f(n + 1)! lxln+l < - (n + 1)!

(-l)k 2k+I for all x. x n---+oo L 2 k + 1 ) .1 k=O oo ( l)k 2k x for all x. It follows by Example 7.24 that cos(x) = ; ( k)! ⇒ lim Rn(x)

= 0.

oo

Thus, sin(x) ="'

(

L k=O

Definition 7.5. Let f be a funct ion and a a real number. The funct ion f is analytic at a if f has a Taylor series T(x) about a, with radius of convergence R > 0, and the Taylor series converges t o J(x), i.e., T(x) = f(x), for all x with Ix -al< R. If f does not have a Taylor series about a, or if the Taylor series does not converge to f(x) for all x with Ix - al < R, then the point a is called a singularity or singular point of the function f.

7.2. REPRESENTATIONS OF FUNCTIONS BY POWER SERIES Example 7.43. The function J(x) shown by Example 7.41.

279

= ex is analytic at a for every real number a, as

The functions sin(x) and cos(x) are analytic at 0, as shown by

Example 7.44. Example 7.42.

Example 7.45. The function

is analytic at 0. However, it is not analytic at 1 because J(l) is undefined and, hence, J does not have a Taylor series about 1. Thus, 1 is a singularity off. For any a =I= 1, J is analytic at a since, for Ix - al < 11 - al,

l __ 1 -x

=

1 1 - (x - a) - a

=

1 _1_ -a 1 - a 1 - xl-a

=� � n=O

1 (x - at. ( 1 - a)n+l

Example 7.46. The function J(x) = ,Ix = x 1 ! 2 is not analytic at 0 because 1 J'(x) = ;;:;: => f'(O) is undefined and, hence, f does not have a Taylor series about 2yx 0. Thus, 0 is a singular point of f. Theorem 7. 7. Suppose that J and g are analytic at a. Then: 1. cif ( x)

+ c2g (x) is analytic at a for any constants

c1

and c2.

2. J(x)g(x) is analytic at a. 3.

��:?

is analytic at a provided that g(a) =I= 0.

The concept of analytic function is required in the solution of differential equations with variable coefficients by infinite series, as will be seen in Chapter 9.

Exercises 7. 2 In Exercises l 10, express the given function J cIB a power series about 0 and determine its radius of convergence. l. f (X)

=

1

1

+ 2x

280

CHAPTER 7. TAYLOR SERIES

2. f(x) = - 62-x

3.f(x)=

1

3+x

4. f(x) = � 4-x 5. f(X)

1

= + 6 2x 1

.

6. j(x) = - 2 l-x 1 7. J(x) = --3 2+x 1

1

8· f(x) = - 1 x l - 3x 1 9. f(x) = - 2 +x

10. f(x) =

x2

+1

+

1

3 - 2X

X 2- 1

In Exercises 11 28, find the Taylor series of the given function f about the given point a and determine its radius of convergence.

11. J(x) 12. f(x)

= e-x , a= 0

= e-3x, a=

0

13. J(x)=ex , a=2 14. J(x) = ex, a= -2 15. f(x)

= e- 3x , a= 0 2

16. J(x) = e- 3x, a= l

17. J(x) = sin(2x), a= 0

18. f(x) = cos(3x), a= 0 19. J(x)

= sin(x), a= 1r

20. J(x) = cos(x), a= 1r

7.2. REPRESENTATIONS OF FUNCTIONS BY POWER SERIES 22. f(x)

= ln(x-1),

281

a= 2

23. J(x)

= ln(x),

24. J(x)

= tan- 1 (x - 2),

25. J(x)

=

26. f(x)

=

a= 2

-� < tan-1(x) < �' a= 2

4 ft+x' a= 0 x2

ijf+x

' a= 0

27. f(x)=xJl+x2 , a=0 28. f(x) =

X

, (l-x3)312

a=0

29. Find the 4th-degree Taylor polynomial of J(x)

= ex sin(x) about 0.

. 30. Fmd the 4 th -degree Taylor polynomial of f(x) = 31. Find the 4th -degree Taylor polynomial of f(x)

cos(x) about 0. . 1 + sm( x )

= (x2 + 1) cos(x) about 0.

x ln(x) . 32. Find the 4th -degree Taylor polynomial of j(x) = -- about 1. 2-x 33. Approximate sin (

1)

with error less than 0.001 in absolute value, and confirm

your answer with a calculator.

34. Approximate

11 12

cos( Jx) dx with error less than 0.001 in absolute value.

I

35. Approximate

1

Jl +x8

dx with error less than 0.001 in absolute value.

. x( e 3x - 1)2 cos2 (x) 36. Evaluate hm . 3( ) x--+0

. (x2 37. Eva1uate 1 1m x--+O

Sln

-

X

l)(e-x + x - 1) cos(x) . . X Sln (3X)

. (x+2)3 ln(l-x) 38. Evaluate hm -'--------'--�. x--+O sin(2x)cos(3x)

xJ4x4 - x 5 -) . 39. Evaluate lim - 3- - x--+O- tan (2x) cos2 (3x2

CHAPTER 7. TAYLOR SERIES

282

40. Determine how large n must be taken in order to approximate ffe the Taylor polynomial of

e

x

about O at x

=

=

e 1 13 by

with error less than 0.0001 in

3

absolute value, and give the approximate value. Confirm your answer with a calculator. 41. Determine how large n must be taken in order to approximate ln(0.9) by the Taylor polynomial of ln(l - x) about O at x = O.l with error less than 0.001 in absolute value, and give the approximate value. Confirm your answer with a calculator. 42. Show that f (x) = sin(x) is analytic at every real number a. 43. Show that f (x)

= cos(x)

is analytic at every real number a.

44. Employ power series to prove Euler's identity:

e

ix

= cos(x) + isin(x).

45. (a) Show that, for any two real numbers a and b, ab::::; lallbl. (b) Show that, for any two real numbers a and b, la+ bl ::::; lal + lbl. This inequality is known as the triangle inequality. (c) Show that, if

00

n=O

n=O

have radii of convergence R 1 > 0 and R 2 > 0, respectively, then the radius of convergence R of 00

f(x)

+ g(x) = I)an + bn)Xn

is at least the lesser of R 1 and R2.

n=O

Chapter 8 Fourier Series The topic of Chapter 7 was the repre entation of function by power erie . Variou problem require the repre entation of function by eries which contain trigonometric functions. The content of the pre ent chapter i the tudy of such serie . s

s

s

s

s

s

8.1

s

s

s

s

s

s

s

Fourier Series of Periodic Functions

Definition 8.1. A function f i periodic if there exi t T > 0 uch that s

s

f (x + T)

= f (x)

s

s

for all real x.

Such a function f i said to have periodT (or to be T-periodic). The least (positive) period off i called it fundamental period. s

s

s

Example 8.1. The function J(x) = sin(x) i periodic with period T = 2mr for any integer n 2:'. 1 ince in(x + 2mr) = in(x) for all x. The fundamental period of in(x) i 21r. s

s

s

s

s

s

Let f be periodic with period T by a trigonometric series, i.e.,

J(x)

=

ao



2+L n=l

= 2L, L > 0,

n1rx [a n COS ( L )

and suppo e that f is represented s

+b

11

. sm

(

n1rx )]

(8.1)

L

where a n , n 2:'. 0, and bn , n 2:'. 1, are constants. The fir t term in the erie , which correspond to n = 0, i denoted by �o in tead of a0 merely for convenience, in order that a 0 and a n for n 2:'. 1 can be defined by a single formula in tead of two eparate one . s

s

s

s

s

s

s

s

283

s

284

CHAPTER 8. FOURIER SERIES

Given that Equation (8.1) holds, the coefficients an , n ;::=: 0, and bn, n ;::=: 1, will be determined formally, i.e., without justification of the procedures employed in their derivation. Subsequently, the conditions under which and the values of x for which such a series converges to f ( x) will be given.

.

.

In order to determine the coefficients an , n ;::=: 0, let m ;::=: 0 be any integer, multiply m1rx Equat10n (8.1) by cos ( L) , and mtegrate from -L to L to obtam

L 1L f ( x) cos (m1rx £) dx = 1 a ° cos (m1rx L) dx + -L

-L

1L f--, [ -L

2

m1rx n1rx n1rx m1rx )] L an cos (L) cos ( L) dx + bn sin (L) cos ( L dx.

n=l

Reverse the order of integration and summation to obtain

1L J(x)cos (m1rx) dx = a 1L cos (m1rx) dx + L 2 L L L f--, [a 1 cos (n1rx L) cos (m1rx L) dx] . L) cos (m1rx L) dx + bn 1 sin (n1rx -L



If m

0

n

= 0, then

-L

1

L

-L

m1rx cos ( -- ) d.T

L

=

-L

1L

-L

-L

1 dx

= 2L,

and

Hence,

a J(x) dx = o2L = a0 L, from which it follows that 2 -L L

j

ao = If m

;::=:

1, then

1

L

-L

l

L

m1rx cos ( L) dx

1L f(x) dx.

=

(8.2)

-L

L m1rx sin ( L) m1r

L

= 0. -L

8.1. FOURIER SERIES OF PERIODIC FUNCTIONS

285

Employing the trigonometric identities

cos(a + b) { cos(a - b)

= cos(a) cos(b) - sin(a) sin(b) = cos(a) cos(b) + sin(a) sin(b)

cos(a) cos(b) n1rx with a= Land b

m1rx

= L'

1

£ -L

=

}



1

2 [cos(a - b) + cos(a + b)]

we obtain

m1rx n1rx cos ( L ) cos (L ) dx

12 1_£ { cos [(n - Lm)1rx] + cos [(n + Lm)1rx]} dx £

0, n i- m } { L, n=m

since, for any integer k i- 0,

Employing the trigonometric identities sin(a + b) { sin(a - b)

= sin(a) cos(b) + cos(a) sin(b) = sin(a) cos(b) - cos(a) sin(b)

sin(a) cos(b) n1rx with a= L and b

m1rx

= L'

1

£ -£

=

1 [sin(a - b)

.

}



+ sin(a + b)]

we obtam

n1rx m1rx sin ( L) cos (L ) dx

11: {

0

1rx sin [(n -;i) ]

+ sin

+ ) [(n ;i 1rx]} dx

siuce, for any integer k i- 0,

1L



sin

(brx)

L

k1rx dx = - kn cos (L)

L

L

-L

= 0,

286

CHAPTER 8. FOURIER SERIES

£ brx - O d x = 0. Thus, for any and, if k = 0, then 1- sin ( L ) dx = 1 £

1:

L

L

m x

f(x ) cos ( ; ) dx = amL, i.e., am

=

±

1:

m

� 1,

m x

(8.3)

f(x ) cos ( ; ) dx.

Combining Equations (8.2) and (8.3), we obtain

am

=

±

1:

m x

f(x ) cos ( ; ) dx ,

m

(8.4)

� 0. m x

Similarly (see Exercise 3), multiplication of Equation (8.1) by sin ( ; ) ,

and integration from - L to L yields

bm =

£

--y-x ) dx, L11-L f(x ) sin (mn

m

m

� 1, (8.5)

� I.

Definition 8.2. Suppose that f is periodic with period T = 2L. The trigonometric senes nnx ao � . nnx )] [anCOS ( L ) +bnSIIl ( L + 2 L n=l

with an, n 2". 0, and bn, n 2". 1, defined by Equations (8.4) and (8.5), respectively, is called the Fourier series off. The coefficients an, n 2". 0, are called the Fourier cosine coefficients, and bn, n � 1, the Fourier sine coefficients off. The integral formulas (8.4) and (8.5) which define the coefficients an and bn are called the Euler-Fourier formulas. Since f has period 2£, the above integrals can be evaluated on any interval of length 2L, and one should select the interval on which an explicit formula for f(x ) is available. Example 8.2. Consider the function defined by

f(x ) = x 2 + x, -1 ::; x < 1, f(x + 2) = f(x) for all Then f is periodic with period T = 2 and, with L = the form

Note that the formula f ( x) = x 2

+x

T

2 = 1,

is valid only for

x

x.

its Fourier series takes

in the interval [-1, 1). For

287

8.1. FOURIER SERIES OF PERIODIC FUNCTIONS y 2

3 X

Figure 8.1: The graph of J in Example 8.2. any x outside the interval [-1, 1), the 2-periodicity off must be employed in order to determine the value f(x). The graph of f is depicted by Figure 8.1, where three periods are displayed. The curve from -1 to 1 repeats over the entire real line. By Equations (8.4) and (8.5), the coefficients an , n 2: 0, and bn , n 2: 1, are given by 3 1>x2 +x)dx = (� 1� (x

2

1 -(x 2 n1r 1

+ x) cos(mrx)dx

+ x) sin(mrx)

2(2x

7r

1� (x2

1

1

-1

+ 1) cos(n1rx) 1

2 4(-lr n 2: 1, n2 1r 2 '

n

2

+�)

2

[

3'

1

1 -n1r

1

-1

11

-1

(2x + 1) sin(mrx)dx

1 - 22 n 7r

11

-1

+ x) sin(n1rx)dx

1 --(x 2 n1r

+ x) cos(n1rx) 1

2(- )n n1r

1

-1

1 n1r

+- 1

1

-1

2 cos(n1rx)dx

(2x

+ 1) cos(n1rx)dx

1 11 1 11 2 2 sin(n1rx)dx 2(2x + 1) sin(n1rx) n 1r n 1r _ 1 _1 2 2 2(-lr-l 2 --- + 33 cos(n1rx) n1r n 1r 11_1

- --l + 2(-1r-1

n 2: 1.

288

CHAPTER 8. FOURIER SERIES

Thus, the Fourier series of f is

1

-3

00

+

L n=l

4(-lt

[

n2 1r 2

cos(mrx)

2(-1r- 1 ] + --sin(mrx) n1r

.

The following two definitions are required in order to state a convergence theorem for Fourier series. Definition 8.3. A function f is piecewise continuous on an interval [a, ,BJ if [a, ,BJ can be partitioned into a finite number of subintervals such that f is continuous on each open subinterval and approaches a finite limit as x approaches an endpoint of any of the subintervals. A function is piecewise continuous if it is piecewise continuous on every finite interval. Thus, a piecewise continuous function may have finite jump discontinuities on [a, ,BJ, but no other types of discontinuity. The graph of a typical piecewise continuous function on [a, ,BJ is displayed in Figure 8.2. y

a

X

Figure 8.2: The graph of a piecewise continuous function on [a, ,BJ.

Definition 8.4. The left-hand limit f (x-) of a function f at x is defined by lim J(y). J(x-) = y.-xThe right-hand limit f ( x+) of a function f at x is defined by lim J(y). J(x+) = y.-x+ The average value of f at x is defined by 1

2[J(x+) + f(x- )].

8.1. FOURIER SERIES OF PERIODIC FUNCTIONS

289

If J is continuous at x, then f (x-) = f (x+) = f (x ), and the average value off at x is f(x). The one-sided limits and the average value off at x are useful notions if J has a jump discontinuity at x. Example 8.3. The function f in Example 8.2 is piecewise continuous. It has jump discontinuities at every odd integer x =2n + 1, where n is an integer. At x = l, f(l-)=2, f(l+)=O, and

1 [f(l+)+J(l-)]=1 2

is the average value. Theorem 8.1. Suppose that J is periodic with period T = 2L and that f and f' are piecewise continuous. Then, at any point x, the Fourier series of f converges to

tu(x+) + J(x-)], i.e., the average value off at x. If f is continuous at x, then the Fourier series off at x converges to f ( x). This theorem is called the Fourier convergence theorem. In certain books, a discontinuous function is redefined at a point x of discontinuity 1 by J(x) = f(x +)+ f(x-)], in which case, the Fourier series of J converges to J(x) [ 2 for all x. Example 8.4. Let J(x) = 1 - x for O � x < l and f(x + 1) = f(x) for all x. The graph of f is displayed in F igure 8.3. y

-2

-I

0

2

3

X

F igure 8.3: The graph off in Example 8.4. The function J has finite jump discontinuities at the integers, and no other types of discontinuity. Hence, f is piecewise continuous. Its derivative f' ( x) = -1 everywhere except at the integers, where it is undefined, but lim J'(x) = -1 is finite for every X--->n±

290

CHAPTER 8. FOURIER SERIES

integer n. Hence, f' is piecewise continuous, and the Fourier convergence theorem (Theorem 8.1) applies. Since J has period T

=

1, L

=

T

1

2 2, and the Fourier series off takes the form =

ao

L...,[an cos(2mrx) + bn sm(2mrx)], 2+� n=l and, since an explicit formula for f(x) is available on the interval [0, 1), the integrals in Equations (8.4) and (8.5) must be evaluated from 0 to 1. Thus,

ao

2

an

2

fo\1

1fo

- x) dx = (2x - x2 )

(1 -x) cos(2mrx) dx

1 1 -(1 - x) sin(2mrx) n1r o 1 -- 2 2 cos(2n1rx) 2n 7r 1o 0, n 2: 1, 1

bn

2

fo1

1

+-

I:=

11

n1r o

1,

sin(2mrx) dx

1

(1 - x) sin(2n1rx) dx

11

1 1 cos(2n1rx) dx --(1 -x) cos(2n1rx) - n1r o n1r o 1 n 2: 1. n1r 1

1

)

Hence, for all x except the integers,

f(x) At x

= m,

= - + I:- sin(2n1rx). 00

1

2

n=l

1 n1r

where m is an integer, the series converges to the average value

evident from the series since sin(2n1rm)

= 0 for every n 2: 1.

1-

1

2, as is

The number to which the series converges can be determined for any real number x.

For example, at x =

the 1-periodici ty of

1

2

�'

the series converges to J (�)

J, the series converges to

[!(731-) + f(731+)]

1

=

At x

1

= 2 [!(0-) + f(O+)] = 2,

=

731, employing

291

8.1. FOURIER SERIES OF PERIODIC FUNCTIONS and, at x=731.2, the series converges to f (731.2) = f (0.2) =1 - 0.2=0.8.

{

0 - 1. 2 n1r 1

an

+

1

2 1

=

n1r 2 sin ( ) n1r 2

Thus, the Fourier series of f is the cosine series 00

3 2 n1r n1rx -+ L--sin (-) cos ( --) , 2 2 2 n=l n1r

4 .

n1r

- n1r sm (2)

8.1. FOURIER SERIES OF PERIODIC FUNCTIONS

297

and converges to f ( x) at every x except the odd integers, where it converges to �.

Exercises 8 .1 1. Show that if f is periodic with period T, then f (x integer n.

+ nT) =

f (x) for every

2. Let J(x) = x 2 + 1 for -1::::; x < 2 and f(x + 3) = J(x) for all x. Determine the values f(x) off at x = 5, 7, -4, 783 and -291. 3. Let f be 2L-periodic, and suppose that f is represented by a trigonometric series, i.e.,

f(x) =

ao

2

n1rx � n1rx )] + L [an COS (L ) + bn sin (

11:

L

n=l

The coefficients an have been determined in the text. In a similar way, show (formally) that bn

=

n x

f(x) sin ( � ) dx for every n 2'. 1.

4. Let f(x) = { �: -� � �: � } and f(x+4) = f(x) for all x . Find the Fourier series off. 5. Let J(x) = { series off.

O O_ 0 . n+l ' n n+l For the coefficients an with n even,

n = 0 ⇒ a2 = 2ao, n = 2 ⇒ a4 = 0, n = 4 ⇒ a6 = 0, etc.; hence,

a2k

= 0 for k � 2.

For the coefficients an with n odd,

n=l



n=3



n=5



n=7



n=9



a1 a = 3 2, -a3 - a1 a5 - 4 - 2. 4' -3as 3a1 a7- --6 2-4·6' -3· Sa1 -Sa1 ag=-- = 8 2·4·6·8' 7a9 3 · 5 · 7a1 au = -- = 2-4-6-8-10' 10

etc., and, in general, (-1 )k -l1· 3· 5 · 7··· (2k - 3) a1 2 · 4 · 6· 8 · 10 · · · (2k) k (-1) - 1 1. 3. 5 • 7··· (2k - 3)a 1 2k k! k (-l) -t1 • 3· 5 · 7 · · · (2k - 3) a 1 2· 4 • 6 ·8· 10 ··· (2k - 2) 2·4·6·8· 10 ··· (2k - 2) 2k k! 1 k (-1) - (2k - 2)!a 1 , k � l. 22 k -lk!(k - 1)!

324

CHAPTER 9. SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS

Hence,

is the general solution, where

1 � (-1t- (2k - 2)! 2k +l Y1 = 1 + 2x 2 and Y2 = x + L 2k x 2 -lk!(k - 1)! k=l

Thus, y1 is an elementary function, but y2 is not.

Suppose that the initial conditions y(0) = -3 and y'(0) = 6 are given. Since y(0) = a0 and y'(0) = a 1 , the solution of the initial-value problem is Of course, if y(0) = 0, then a 0 = 0 and y 1 need not be computed, and if y'(0) = 0, then a 1 = 0 and Y2 need not be computed.

By Theorem 9.1, since the only singularities of the equation are x 1 = ±i, and the distance from x0 = 0 to x 1 = ±i is 1, the radius of convergence of the general solution, as well as that of any particular solution, is R 2:: 1. Since y 1 is a polynomial, i.e., a finite Taylor series, its radius of convergence is R1 = oo. In order to determine the precise value of the radius of convergence R2 of y2, we employ

L ck(x),

the ratio test. Thus, with Y2(x) = x + Ck 1 ( x) lim I + I k--+oo Ck (X)

=

k=l

(2k)!x2 k+3 22k -lk!(k - 1)! I k.:.� 2 2 k+ 1 (k + l)!k! (2k - 2)!x 2k+l (2k)(2k - 1) 2 lxl lim 2 k --+oo 2 (k + l)k 2 lxl < 1

r

I

if lxl < 1. Hence, the radius of convergence R2 of y2 is precisely 1.

325

9.1. SOLUTIONS ABOUT ORDINARY POINTS

In equations where the coefficient recursion relation cannot be solved explicitly for all n � 0, approximate solutions of the differential equation can be obtained by retaining a finite number of terms of the series. Example 9. 7. Consider the equation y" + cos(x)y = 0,

=

with p(x) 0 and q(x) solution is given by

=

cos(x). Since p and q are analytic at x0

=

L= (-lt x k = 1 - 2!x + 4!x - 6!x + · · ·

Since cos(x)

k=O

(2k)!

2

2

4

= 0,

the general

6

is an infinite series and not a polynomial, we shall retain only the first six terms in y. Thus, y

y'

y"

ao + a1x + a2 x 2 + a3x.3 + a 4 x4 + a5x5 + · · a1 + 2a2x + 3a3x2 + 4a4 x3 + 5a5x4 + · · · , 2a2 + 6a3x + 12a4x2 + 20a5x3 + · • •

·,

and y is a solution of the equation if and only if

Performing the multiplication and collecting terms, we obtain the condition

which leads to the relations

i.e., a partial set of recursion relations for the coefficients an . It follows that

326

CHAPTER 9. SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS

and ao and a 1 are arbitrary. Thus,

where Y1

=

12

1- x

2

1 4

+ 12x + · ··

and Y2

= x - 13 x +

are two linearly independent solutions.

6

1 5

x +··· 30

If a solution of the equation

y" + p(x)y' + q(x)y = 0

(9.2)

is sought about an ordinary point x0 -=f. 0, then the change of variables

t

= x - x0, y(x) = z(t), p(x) = P(t), q(x) = Q(t),

transforms the equation into the equivalent equation

z" + P(t)z' + Q(t)z = 0,

(9.3)

with t = t 0 = 0 as the ordinary point corresponding to x = x0. Once the solution z(t) of Equation (9.3) is determined, the solution y(x) of Equation (9.2) is obtained by y(x) = z(t) = z(x - xo). If the solution of a differential equation is sought in a neighbourhood of a singular point x 1, then Theorem 9.1 does not apply and the methods employed in the foregoing examples fail. In order to obtain such a solution, a different procedure is required, called the Frobenius method, a topic which is beyond the scope of this book.

Exercises 9 .1 1. Determine all points x0 where the given function is analytic. 3

(a) f(x) = x+2

2 x + 1 (b) J(x) = x 2 -4 (c) f(x) = tan(x)

327

9.1. SOLUTIONS ABOUT ORDINARY POINTS x+l e -1 Jx + 2, x

(d) J(x)

= -;;---

(e)

=

f(x)

2:'. -2

x2 - x + 1 . 2. Compute the Taylor series of f (x) = ---- about xo = 0 by d1v1·d·mg 1 + x

x+l

into 1 - x + x2 .

La x . 00

3. Consider the series

n

n

n= 2

(a) Write down the first four terms. (b) Let n = m + 2 and express the series as one involving the index m. Write down the first four terms. (c) Replace m by n in the series obtained in part (b) and write down the first four terms. (d) Show that step (b) may be avoided simply by replacing n by n + 2 in the original series. 4. Determine an , n

2:'. 0,

L na x + L a x + 00

if

00

n

n

n=O

n

n 2

= 0.

n=O

5. Consider the differential equation y" + xy' + 2y = 0. (a) Find the coefficient recursion relation for the general series solution about Xo = 0. (b) Solve the recursion relation, i.e, find a n explicitly in terms of n, and thereby obtain the general solution and two linearly independent solutions. (c) Write down the first four terms of each one of the two series solutions. (d) Express one of the series solutions as an elementary function. (e) Deduce the radius of convergence of each series solution. (f) Solve the initial-value problem for the given equation if y(0) y'(0) = 2. 6. Consider the differential equation (1

0 and

+ x 2 )y" - 4xy' + 6y = 0.

(a) Find the coefficient recursion relation for the general series solution about Xo = 0. (b) Solve the recursion relation and thereby determine the general solution and two linearly independent solutions. (c) Find y satisfying the given equation if y(0)

=

-3 and y'(0)

=

1.

328

CHAPTER 9. SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS

7. Find two linearly independent solutions of the differential equation

y" + xy' - 2y = 0. 8. Find two linearly independent solutions of the differential equation

(2 + x2 )y" + 2xy' - 2y = 0. 9. Consider the equation (a) Find the 5 th -degree Taylor polynomial T5 (x) of the general solution. (b) Find T5 (x) for each one of two independent solutions. 10. Consider Legendre's equation of order v,

(1 - x2 )y" - 2xy' + v(v + l)y = 0, where v is a constant. (a) Find the coefficient recursion relation for the general series solution about Xo

= 0.

(b) Deduce from the recursion relation that if v = m is a nonnegative integer, then one of the two series solutions is a polynomial Pm (x) of degree m (i.e., show that am+2 = am+4 = am+6 = · · · = 0).

(c) Impose the condition Pm (l) = 1 (which specifies the arbitrary constant within the polynomial solution) and find Pm (x) for 0 :S m :S 3. T he polynomials Pm (x), with Pm(l) = 1, are called Legendre polynomials.

9.2

Periodic Solutions

Consider the nonhomogeneous equation

y" + )..y = j(x),

-00


..y = 0. The solution in that case is not periodic, but contains a term proportional to x. Example 9.8. Let f(x) = x3 - 2x2 + 1 on [O, 2] and f(x + 2) = f(x) for all x. Since f is continuous on [0, 2] and f(0) = 1 = f(2), f is continuous everywhere and f' is piecewise continuous. Hence, -00

< X < 00.

A 2-periodic solution of the equation y" + 3y = f(x) is therefore sought in the form

� + I)cn cos(mrx) + dn sin(mrx)], -oo < x < oo.

y=

Since >..

n=l

= 3 -=f. n2 1r2 for any integer n 2 0, Cn =

and L

=

1,

an , n 2 0, and d n 3 - n27r 2

=

bn , n2 1, 3 - n 27r 2

and it remains to determine the Fourier coefficients of f. Thus,

an

12

2 3' (x3

-

2x2 + 1) cos(n1rx) dx

-(x 3 n1r

-

2x

1

2 + 1) sin(n1rx) 2 - -1 1 2(3x2 2 o n1r1 1o 2 1

-

4x) sin(n1rx) dx

1 (6x - 4) cos(n1rx) dx (3x2 - 4x) cos(n1rx) 1 - 22 n 1r o n 7r o 2 2 �- �(6x - 4)sin(n1rx)\ + 6 [ sin(n1rx)dx n 1r n 1r n 3 1r3 Jo O 2 1 4 4 6 n21, -44cos(n1rx) , 22 n 1r n 1r O n 1r

-2 2

22

331

9.2. PERIODIC SOLUTIONS 2 3 \x - 2x + 1) sin(mrx) dx o f 2 2 1 3 - 2x 2 + 1) cos(mrx) 1 + 1 1 (3x 2 - 4x) cos(mrx) dx --(x nn nn 0 O 2 2 1 1 (3x2 - 4x) sin(nnx) 1 - 22 (6x - 4) sin(nnx) dx 22 nn n n 0 O

bn

1

2 1 6 1 cos(nnx) dx 33(6x - 4) cos(nnx) 1 - 33 nn n n 0 O 12 n "2 1. 2

Hence,

1 y= 9

oo [ 4cos(nnx) L + - - --- +- - --n=l

n 2n2(3 -

n

2n2 )

12sin(nnx) ] n 2n2 )

n3n3(3 -

,

-OO

0 any real number.

Make the substitution x

4.

J

(x2

= a sec(t),

7f

0 S: t
0. Complete the square to obtain

a + bx + x2

= (x + �)

2

- ( � _ a) ,

Partial Fractions Partial fractions are required in order to evaluate integrals of rational functions, i.e.,

j

��:�

dx, where P and Qare polynomials. First, suppose that deg(P) < deg(Q).

1. If Q factors into distinct linear factors, i.e., then there exist constants A 1, A 2,

· · ·,

An , such that

A a2 x+b2

A an x+bn

P(x) 2 n A1 -= --+ --+ ...+ ---, Q(x)

a1x+b1

called the partial fraction decomposition of ��:�. 2. If Q contains a repeated linear factor ( ax + b?, r � 2, then the partial fraction decomposition of ��:� must include the r terms B

B1 2 Br ---+----+···+( ax + b)r ' ax + b ( ax + b)2

where B 1, B 2, ···,Br are constants. These terms must be included in addition to the terms which correspond to the distinct linear factors of Q.

3. If Q contains the distinct, irreducible, quadratic factor ax2 +bx+ c, then the partial fraction decomposition of��:� must include the term Ax+B ax2 +bx+c'

where A and B are constants, in addition to the terms which correspond to the other factors of Q. A quadratic ax2 + bx + c is irreducible if its roots are complex, which occurs if and only if b2 - 4ac < 0.

340

APPENDIX A. FORMULAS AND TECHNIQUES

4. If Q contains the repeated, irreducible, quadratic factor (ax2 +bx+cf, r 2:: 2, then the partial fraction decomposition of��:� must include the terms

Ar x+Br A2 x+B2 A1x+B1 2 2 2 +bx+ 2 ----- c)-- + ... + --cY' (ax +bx+ (axax +bx+ c + --where Ai and Bi, 1 :S i :S r, are constants, in addition to the terms which correspond to the other factors of Q.

If deg(P) 2:: deg(Q), then perform long division to express ��:� as P1 (x) P(x) (x) Q(x) = R + Q(x)' where Ris a polynomial and deg(Pi) < deg(Q).

A.3

Matrix Inversion

Let A be an n x n matrix. Two methods of computing A- 1 are given below. 1. Write down the augmented matrix (Alf), where I is then x n identity matrix, and row reduce the augmented matrix in order to transform A into I. The resulting augmented matrix is then (IIA- 1). 2. Let Mij denote the determinant of the submatrix obtained from A by deleting the i-th row and j-th column, called the ij-minor of A. Let Aj = (-1)i +j Mij , called the ij-cofactor of A. The transpose of the matrix with entries Aj is called the classical adjoint of A, and denoted by adj(A). Then

A_ 1 =

adj(A) det(A)'

i.e.'

In particular, if A = ( : � ) is a 2 x 2 matrix, then A_ 1 =

1 ( d -b ) ad- be -c a

Appendix B Proofs of Selected Theorems B.1

The Comparison Test

Theorem 6.11 Let {an } and {bn } be sequences, and suppose that

for all n 2='. k, where k is an integer. Then:

Lb 00

1. If

n=l

La 00

n

converges, then

00

n

n=l

Lb

also converges.

00

2. If Lan diverges, then n=l

n

also diverges.

n=l

Proof. Without loss of generality, assume O � an � bn for all n 2='. 1. Let S n and t n

n

= I: bk. k=l

n

= I: ak k=l

Since an 2='. 0 and bn 2='. 0, { s n } and { t n } are increasing sequences. In

addition, 0 � Sn � l n because O � an � bn . 00

1. If� L-t bn converges, then lim t n = t exists. Hence, S n � l n � t < oo ⇒ { s n } is n=l

n-+oo

00

bounded above. Since it is increasing, it converges. Thus, Lan converges. n=l

2. If� L-t an diverges, then lim S n n=l

⇒ n-+(X) lim l n = oo, i.e., L bn diverges. 00

00

n-+oo

=

oo

341

n=l

342

APPENDIX B. PROOFS OF SELECTED THEOREMS

B.2

The Limit Comparison Test

Theorem 6.12 Let {an } and {bn } be sequences with an > 0 and bn where k is an integer, and let . a L = 11m -n .

> 0 for n � k,

n-+oo bn

1. If O

00

< L < oo, then either both Lan and

2. If L = 0 and

Lb 00

n =l

3. If L = oo and

n

Lb 00

n =l

n =l

00

Lb 00

n =l

n

converge, or they both diverge.

converges, then Lan converges. n =l 00

n

diverges, then Lan diverges. n =l

Proof. Without loss of generality, assume k 1. If O < L < oo, let E

=

l.

> 0 with E < L. Then there exists an integer N such that n � N ⇒ I:: - LI < E. Hence, for all n > N, -E < :: - L < E, i.e., L - E < :: < L + E. Let a = L - E and /3 = L + E. Then O < a < f3, and

a< : < f3 ⇒ abn < an < /3b11 • By the comparison test, if n

n

00

then Lan converges, and if n=l

00

if La n converges, then n=l diverges.

2. If L

Lb 00

n

n=l

Lb 00

n

n=l

00

Lb 00

n

n=l

converges,

diverges, then Lan diverges. In addition, n=l 00

converges, and if Lan diverges, then n=l

Lb 00

n

n=l

> 0. Then there exists an integer N such that n � N ⇒ < E, i.e., -E < :: < E, or -Ebn < an < Ebn. By the comparison test, if

=

I:: I

Lb 00

n

n=l

0, let E

converges, then

La 00

n =l

n

converges.

b oo, then nlim � = 0. Thus, there exists an integer N such that n � N an I bn I . ->oo bn ⇒ - < E, 1.e., -E < - < E, or -Ean < bn < Ean . By the comparison test, if an an

3. If L

=

Lb 00

n

n=l

00

diverges, then Lan diverges. n =l

343

B.3. THE ALTERNATING SERIES TEST

B.3

The Alternating Series Test

Theorem 6.13 Consider the alternating series

If {bn } is decreasing for n � k, where k is an integer, and lim bn n->oo converges.

= 0, then the series

Proof. Without loss of generality, assume {bn } is decreasing for n � 0. Consider the sequence { sn} with n = 2k even, i.e., 2k I)-l) i bi = b1 - b2

S2k

i=O

+ b3 - b4 + b5 - · · · + b2k-l - b2k

Since bi - bi+ 1 > 0 for all i � 0, the sequence { s2k} is increasing. In addition, 2k I)-l) i bi

S2k

i=O

= bi - (b2 - b3) - (b4 - b5) - · · · - (b2k-2 - b2k-1) - b2k

< b1,

i.e., the sequence { s2k} is bounded above and, hence, it converges. Thus, lim s 2k exists. Next, consider the sequence { sn } with n S2k+1 Thus, for all ls2k+1 - sl
oo

= k->oo lim S2k - lim b2k+1 = s. k->oo

=s

> 0, there exists an integer K such that k � K ⇒ ls2k - sl < E: and E:. It follows that lsn - sl < E: for all n � N = 2K + 1, i.e., lim Sn = s. n->oo

n

converges.

The Ratio Test

Theorem 6.16 Let { an} be a sequence and suppose that L exists. Then:

Then

E

2:) - 1tb 00

Thus,

= S2k - b2k+1 ⇒

= 2k + l odd.

k->oo

I an+ 1 = n->oo lim an

I

344

APPENDIX B. PROOFS OF SELECTED THEOREMS 00

1. If L < 1, then Lan converges absolutely. n =O 00

2. If L > 1, then Lan diverges. n=O

3. If L = 1, then no conclusions can be drawn regarding convergence. Proof. Since lim -n-+oo

I aann+l I = L � 0 exists, there exists an integer k such that an =J. 0

. an+l for all n � k because, otherwise, - - would be undefined. Assume, without loss of an generality, that k = 0. 1. If L < 1, there exists K > 0 such that L < K < 1. Let c there exists an integer N such that n � N

< I a:: 1



11 a:: 1 1- LI
0 such that 1 < K < L. Let there exists an integer N such that

Hence, lan l > K laol n

by the n tl'-term test.

---+

oo

⇒ n___.CX) lim lan l =J. 0 ⇒ lim an n-.oo

E

= L- K >

=J. 0

3. If L = 1, the test is inconclusive, as shown by Example 6.87.



00

0. Then

Lan diverges

n=O

B.5. THE ROOT TEST

345

The Root Test

B.5

Theorem 6.17 Let {an } be a sequence and suppose that L = lim lan l l / n n---->oo exists. Then: 00

1. If L < l, then Lan converges absolutely. n=O 00

2. If L > l, then Lan diverges. n=O 3. If L = l, then no conclusions can be drawn regarding convergence.

Proof. 1. If L < l, there exists K > 0 such that L < K < 1. Let there exists an integer N such that

⇒ Ian I

< K

n



E

= K - L > 0. Then

L Ia I converges by the comparison test since L K 00

00

n=O

n

n=O

00

n

is a

convergent geometric series. Thus, Lan converges absolutely. n=O 2. If L > l, there exists K > 0 such that 1 < K < L. Let there exists an integer N such that



la n l > K

n

-----+

oo

lim lan l ⇒ n-+oo

/: 0

lim an i= 0 ⇒ n--400

E

00 =}

Lan diverges by the n=O

. 3. If L = l, then the test is inconclusive For example, /n

L = lim (-;) l n ---->oo

n

1 = , and

f n=O

= L - K > 0. Then

..!_ diverges and L = lim n n-oo

L 2n1 converges and 00

(..!..)

n=O

n

l

/n

=

1.

346

APPENDIX B. PROOFS OF SELECTED THEOREMS

The Binomial Theorem

B.6

For any real number a, (1 + x)"

(:)x• for !xi< I.

= t,

t (:)x•, lxl

Proof. Since the series converges absolutely for !xi < 1, let f(x)

= t,

(:)x• = 1 +

If f(x) = (1 +x)°', then

< I.

° =;, (1+x)J'(x) = a(l+x)

( ) = a(l+x)° - 1 f' x

( ) f'x Conversely, if (1+x)f'x ( ) = af(x), then - ( ) fx

= - - =;, a l+x

= af(x).

= a ln(l+x)+ c = ln[(l+x)° ] + c =;, f(x) = k(l +x)° and, since (1+x)°' = 1 = f(O) at x = 0, k = 1 and f(x) = (1+x)°' . Thus, we must ln lf(x)I

show that (1+x)f'(x) = af(x). By Theorem 7.2, since f(x) = 1+

J'(x) =



(1+x)J' (x)

=

t

f

k (:) x._,

= t, ( k+1) (

f

k

:

l) x'

because (

7)

k=l

(�)xk ,

k (:)xk -l (k+1) ( : )xk +x 1 k k=l k=O

= t,(k+ l)( : )x• + tk(:)x• k 1

=

f

(7)

+

t

= c, +at

[

(k+

l)c: 1)

+k(:)

l

x'

(:)x• = c, [1 + t (:)x'] = c,j(x),

= a and

(k+l)( : ) +k(:) k l a(a :-- 1) •••(a - k+l)(a - k) a(a - 1) ···(a - k+1) +k = (k+ 1) k! (k+1)! a _ a(a - 1) •••(a - k+1) a(a - 1) ···(a - k+1) = 0( ) +k = (a k) k . k! k!

Appendix C Solutions C.l

Chapter 1 Solutions

Exercises 1.1 - page 7 1. y' + 2x = xex

⇒ y = J xex - 2xdx = xex - ex - x2 + C.

1 2. -y' = 4cos(2x) ⇒ y = J 4xcos(2x)dx = 2xsin(2x) - J 2sin(2x)dx X = 2xsin(2x) + cos(2x)+ c.

J 2xcos(x2 )dx = sin(x2 ) + c. 2 2 y = ± J 2xex dx = ±ex + c. y' = ±2xex (*)2 = 4x2 e2x co1(x ) y' = 2sin(x) =* y = J 2sin(x)cos(x)dx = J sin(2x)dx = -½ cos(2x) + c.

3. * = 2x cos(x2 ) ⇒ y =

4. 5.

6. �

=}

2

=}

2

= ln(l) ⇒ y = f ln(l) dt = t ln(t) - t + c, y(l) = 2 ⇒ c = 3.

7. y' = 4cos2 (x) ⇒ y = y (�) = � =} C = -1.

J 4cos2 (x)dx = J 2 + 2cos(2x)dx = 2x + sin(2x) + c,

y = ¼ J �3 - x!2 dx = ¼[6 ln Ix - 31 - ln Ix+ 21] + c. x 2 9. �! = 12e x sin(3x) ⇒ y = J 12e2x sin(3x)dx = 6e2x sin(3x) - J 18e2x cos(3x)dx = 6e2x sin(3x) - 9e2x cos(3x) - J 27e2x sin(3x)dx =} J 39e2x sin(3x)dx = 6e2x sin(3x) - 9e2x cos(3x) + c1 ⇒ y = 1� [6e2x sin(3x) - 9e2x cos(3x)] + c.

8. y'

10. 1Jt 11. 1Jt

= xL�3_6

=}

= 3y ⇒ � = 3 ⇒ lnlvl = 3t + c1 ⇒ y = ce3t , y(O) = 2 ⇒ c = 2.

= -2y ⇒ � = -2 ⇒ ln IYI = -2t + c1 ⇒ y = ce- 2 t, y(O) = 3 ⇒ c = 3.

12. y' + cos(x)y = 0 =* 'Ji.. = -cos(x) ⇒ lnlyl y y(1r) = -3 ⇒ c= -3. 347

= -sin(x) + c1 ⇒ y = ce-sin(x),

348

APPENDIX C. SOLUTIONS

J

J

J

13. 3y2 y' = 1 ⇒ 3y 2 y' dx = 1 dx ⇒ 3y2 dy = 1 3 = �x + c, y(O) = 2 ⇒ c = 8. y = (x + c) 1

C.2

J 1 dx ⇒

y

= x+c ⇒

3

Chapter 2 Solutions

Exercises 2 .1 - page 14 '

= 3x2 y2 ⇒

'

=



y

2.

y

7⇒y

= 3x2 ⇒

JL y2 3

'

y

_.!y

= x3 + c ⇒ = y

= Jx+1 ⇒ ¼Y4 =

;;!�

3+c' y(O) x--=.L

�(x+1)3 / 2+c 1

= .!2 ⇒ c =

⇒ =

-2. 1 4

3 2 rn(x + 1) / + cJ / _

y

⇒ y3 = ln(x2 + 1) + tan-1(x) + c 3. (x2 + 1)!� = 2;; 1 ⇒ 3y y' = 2 + 1) + tan -1(x) + c] 1 1 3 , y(O) = 3 ⇒ c = 27. y = [ln(x 2

y' X 4. 4x 2 +3 - 4 Y3 x -2 - 5 4 -4y3+1 ::::} (5 Y4 y _ y4 + y = x4 + x3 _ x2 + c. y

+

5

x 5. xy' - ln� )

=

0 ::::} yy'

=

ln�x) ::::} ½Y2

1)YI - 4X 3

=

+ 3X2 - 2X

½[ln(x)] 2 + C1::::}



=}

y

= ± ✓[ln(x)] 2 + c,

⇒ c = 9 and y = - ✓[ln(x)] 2 + 9. ' = (x + l)eY ⇒ e-Y ' = x!l = 1 + � ⇒ -e-Y = x + ln lxl + c ⇒

y(l)

6. xy y= ?

·

=

-3

y

- ln(-x - ln lxl - c).

xy+x =} _]f__ ' = ___3:__ ::::} (1 __l_)y' = 1 + _1_ ::::} y x- 1 x- 1 y+l xy-y y +l l y n IY + 11 = x + ln Ix - 11 + c, y(2) = -2 ⇒ c = -4.

y

'

=

=

8. cos(x) ln(y)�



y sin(x)



In�y) y'

ln(y) = ± 2 ln I sec(x)I + c::::} = J2 Inlsec(x)l+4_ Y e

9.

10.

y

= tan(x) ⇒ ½[ln(y)] 2 = ln I sec(x)I + c 1 ⇒ = e ± J2 Inlsec(x)l+c , (O) = e2 ::::} c = 4 an d y

' = � ⇒ yy' = x2 ⇒ ½y 2 = ½x3 +c1 ⇒ The orthogonal trajectories: = - ;2 ⇒ ln IYI = � + C1 y' = 2

y

--;z::::} �

y

= ±j�x3 + c is the general solution.

⇒ Y = ce l /x_

' = 2x(y2 + 1) ⇒ )'� 1 = 2x ⇒ tan- 1 (y) = x2 + c ⇒ y = tan(x2 + c) is the general solution. The orthogonal trajectories: y' = 2x (;l+ l ⇒ ) (y2 + l)y' = ;; ⇒ ½Y3 + y = -½ ln lxl + c1, or 31n lxl + 6y + 2y3 = c. y

11.

y

12.

y

= x3 + c ⇒ =

kx4



trajectories:

'

y

= 3x2 .

x- 4 y '

y

=

= �:

k

The orthogonal trajectories:



⇒4



y

'

= ;\ ⇒ = y

3� + c.

= ; . The orthogonal ' = -x ⇒ 2y2 = -½x2 + c1 ⇒ = ±Jc - ¼x2 .

-4x-

yy

5

y

+ x-4 y'

=

0

y'

y

C.2. CHAPTER 2 SOLUTIONS 13.

349

= ecx > 0 ⇒ x- 1 ln(y) = c ⇒ -x - 2 ln(y) + x- 1 - 1 ' =0 orthogonal trajecto ries: ' = yl�(y) ⇒ ln( ) '=-x ⇒ ½y 2 ln(y) - ¾Y 2 = -½x 2 + c1, or 2x 2 - y 2 + 2y 2 ln(y) =c. y

y

y

y

y y

y

'= Y 1 :(y). The

y y

Exercises 2. 2 - page 18 1.



y

+ ⇒ u + xu' = 1 + u ' = x X y =1 + '11.X, u = 'Ji. X =x(ln /x/ + c) , y(l) = 2 ⇒ c= 2.



u'= X1



u = ln /x/ + c



t

2. x '= y2 - x 2 ⇒ '= ;; - �, u = ;; ⇒ u +xu' = u - ⇒ uu'=- ¼ ⇒ ½u 2 =- ln/xl +c1 ⇒ u = ±Jc-2 ln/x/ ⇒ = ±xJc-2ln /x/, y(-1) = 2 ⇒ c =4 and y =-xJ4 - 21n Ix/ . y

yy

y

u ⇒ u1 l 2 u'=x.!. ⇒ 3lu 3! 2 =ln/xi+c1 ⇒ ' = ,/x ,/u + ./Y + x'Jl. ' u =x'11. ⇒ u + xu' = _!_ u= (� ln/xi+ c)213 ⇒ y =X ln /xi+ c)213, y( l) =9 ⇒ C=27. + XUt_ + 1/3 1 1/ 2 1 -----,,, ____._ l _ ___,__ '11.+ 'Jl.+ x 4 . y '' U - 'Ji. -----,,, U - U 3� x 3 !Ti" x u x y /Ti + +u x 2 y+�+x V �+y �+l 3 2 +u=ln (u 113 +u112 +1)u'= ¼ ⇒ ¾u413 +�u 1 /x/ +c ⇒ 3 2 3 (;;)'1 1 1 ( +� ;;) +;; =ln /xi + c. ¾



rn

y

+xsec (;;)- x ' =0 ⇒ ' =;; + sec(;;), u=;; ⇒ u +xu'=u +sec(u) ⇒ cos(u)u'= ¾ ⇒ sin(u) =ln /xi+c ⇒ u=sin- 1 (ln /x/ + c) ⇒ y = x sin- 1 (ln /xi+ c), y(l) = f ⇒ c = {!-.

5.

y

6.

y

y

'=

.

2 x+3y 3 X- 2 y

2

= 3 +32 : ) u = XV. -

X

y



u +xu' =

2 + 3u 3 - 2U

⇒ xu'= 32 + 2 u ⇒ 2

()U - ,._

3tan- 1 (u) - ln(l +u2 ) = 21n/xi + c ⇒ 3tan- 1

8.

u'= 'J.X



ln (1 + ;�) =21n/xi+c.

⇒ u +xu' = QU + 1 ⇒ xu' = 5+uU-u2 ⇒ y' = 6x + ⇒ ' = 5yx + 1 ) u ='Ji. X _u_ u' = - 1 ⇒ (1 -1-1-) u' = - 1 ⇒ 1 ln / u-3/+l ln/ u+21 =-ln /x/+c + 5l u+ 2 x 5 u-3 2 x 5 5 u - u-6 ⇒ }lnl;;-31 +glnl;; +21 =-ln/xl +c, y(l) = 4 ⇒ c=gln(6).

y

y

y

Exercises 2.3 - page 23 l. y'+y = e-x, I(x) = efldx =ex y(O) = -2 ⇒ c = -2. 2.

(;;) -

3- 2 � 1 +u

⇒ (ex

y

)'= 1

⇒ ex

y

=x+c



y

=(x+c)e-x ,

' _ 3 =1, J(x) =ef-3dx = e- x3 ⇒ ( e- 3 x y )' = e- x3 ⇒ e-3x = -:}e-3x + C y = ce x3 - ½-

y

y

y

=?

350

APPENDIX C. SOLUTIONS

-

3. exy ' + 2ex y 1 e2Xy = ex + C 4.

y'

+ 2xy =

y = l+ ce-



= y

2x, I(x)

x2





=

=

ef2xdx

ex

2

.

5. xy' + y = � =} xy = ln(x) + c

7. x2y'+xy

0::::} y' + 2y = e-x , I(x) = e-x + ce-2x ) y(O) = -3

=



ef 2dx = e2x C = -4.

⇒ ( )' = ex

2

y

2xex

2



+ � y = x\, I(x) = ef½ dx = e1n( x) = c = 2. y = �[ln(x)+ c], y (l) = 2

y



'

1-1l. =In( x) ::::} y'+ x y = x2 ln( x) '

xy = ln I ln(x)I+ c



y

I(x) = ef ½ dx = �(ln I ln(x)I + c).

::::}

ex

x

= e1n( x) = x



(e2x y)' =ex =} 2

y

=}



=

ex

2

+C



(xy)'

1.X ⇒

(xy)'

1=x ln (x )

3 8. cos(x)y' + sin(x)y = cos4 (x) y' + tan(x)y = cos (x), dx x 1n x tan( ( ) lsec ) I = I sec(x)I = ±sec(x), and I(x) = sec(x) I(x) = ef =e 2 (x) scc(x)y]' = cos [ ⇒ sec(x)y = cos2 (x) dx = ½ l+ cos(2x) dx = ½+ ¼ sin(2x)+ c c = -1r. y = cos(x) [½+ ¼ sin(2x)+ c], y (1r) = �

⇒ ⇒

J

J

9. xy' + 2y = e-x =} y' +� y = e x", I(x) x2y = -xe-x e-x + c (x2y)' = xe-x

-



= x�l ::::} y' - � = xX:1 , and I(x) = ¾ ::::} (¾ y )' = x:i = y = x ( x - ln Ix+ l I+ c).

10. xy'

12.

y

y

I(x)

ef f dx = e2Inlxl = lxl2 x xe-x). y = ;2 (c - e-

=



-



=

1 - x!i

-

ef -½ dx

⇒¾

y

= e-In lxl = l;I = ±�, = x - ln Ix+ 11 + c ::::}

ln(x)y' + Y = x;:_ l ::::} y' + x l;(x) Y = ( x2 12) ln( x) ' dx J(x) = ef xlt�(x) = e1n l ln( x )I = lln(x)I = ±ln(x), and I(x) = ln(x) ln(x)y = ln Ix - 11 - ln(x+ 1)+ c ::::} ::::} (ln(x)y)' = x2� 1 = x�l x!i ::::} Y = ln(x) [ln Ix - 11 - ln(x+ 1)+ c]. X

-



y = u 11 3 13. 3y' - y = y\ is a Bernoulli equation with a = -2. u = y 1-0 = y3 l 213u' u 1 13 = u-2/ 3 u' u-213u' u = l ) which is linear , y' = 31ux x u = cex e- x u = -e-x+ c (e- u)' = e- x 1 ::::} and I(x) = ex 1 13 1 13 e = (c c = 9. 1) , y(O) = 2 y = u



⇒ ⇒ -

-





⇒ -



-

C.2. CHAPTER 2 SOLUTIONS

351

½- u = y 1 - 0 = y 1 1 2 14. y' + 4y = 2e-x .jy is a Bernoulli equati on with a 2 2 x ⇒ y = u ⇒ y ' = 2uu' ⇒ 2uu' + 4u = 2e- u ⇒ u' + 2u = e-x , which is linear , and J(x) = e 2x ⇒ (e2x u)' = ex ⇒ e2x u = ex +C ⇒ U = e-x + ce-2x ⇒ y = u2 = (e-x + ce-2x )2 . 1 0 2 2 3 15. 2x y' - 2xy = -y sin(x) is a Bernoulli equati on with a = 3. u = y - = y2 1 3 2 2 2 2 3 3 2 ⇒ y = u-1/ ⇒ y' = -½u- 1 u' ⇒ -x u- 1 u' - 2xu- ! = -u- 1 sin(x) ⇒ x x2 u' +2xu = sin(x) ⇒ (x2 u)'=sin(x) ⇒ x2 u = - cos (x) + c ⇒ u = c-:i ( ) ⇒ x 2x x y = u- 1 /2 = ± 2 = 7r ⇒ C = _!4 and y = ✓¼-cos(x) = ✓l-4cos(x) · ✓c-cos(x)' y ('!!_)

x x 2 x 16. xyy' + y - xe = 0 ⇒ xy' + y = e is a Bernoulli equati on with a = -1. u = y l -a = y2 ⇒ y = u l /2 ⇒ y' = ½u- 1/2 u' ⇒ �u-1f2u' +u l /2 = xex u-1/2 ⇒ u' + �u = 2ex , I(x) = e2 1nl x l = x2 ⇒ (x2 u)' = 2x2 ex ⇒ x2 u = J 2x2 ex dx = 2x2 ex - 4xex +4ex + c ⇒ u = x\ (2x2 ex - 4xex +4ex + c) ⇒ y = u 112 = ±� ✓2x2 ex - 4xex + 4ex + c.

l

u = y 1 - 0 = y- 1 / 2 ⇒ 17. xln(x)y' +y = 2y.,Jy is a Bernoulli equati on with a= u_ = � = u-2 ⇒ y' = -2u- 3 u' ⇒ -2xln(x)u-3u'+u-2 = 2u-3 ⇒ u' __ y 2x l n(x) x ln(x)' 3 2 x I(x) = ef-2xl�( x ) d = e-½ ln(ln(x)] = [ln(x)J-1/2 ⇒ {[ln(x)J-l/2 u}' = _ (In(xr 1 ⇒ [ln(x)J-1l2 u = 2[ln(x)J-1l2+c ⇒ u = 2+cJ!n(x) ⇒ y = u- 2 = [ 2+cJ!n(x)J-2.

Exercises 2.4 - page 27 1. J(x,y)=x3 - y2 +x2 y 2 +1. (a ) f(l,0) = 2, f(3,1) = 36, f(l,3) = 2, f(2, -1) = 1 2. (b) fx (x,y) = 3x2 +2xy2 , fx (l,0) = 3, fx (3,1) = 33, J�(l,3) = 21, fx (2, -1) = 16. fy (x,y) = -2y +2x2 y, ]y (l,0) = 0, fy (3,1) = 16, fy (l,3) = 0, fy (2,-1) = -6. (c) fxx (x,y) = 6x + 2y2 , fxx (l,0) = 6, fxx (3,1) = 20, fxx (l,3) = 24, fxx (2, -1) = 14, fxy (x,y) = fyx (x,y) = 4xy, fxy (l,0) = 0, fxy (3,1) = 12, fxy (l,3) = 1 2, fxy (2, -1) = -8. fyy (x,y) = 2x2 - 2, f�y (l,0) = 0, fyy (3,1) = 16, fyy (l,3) = 0, fyy (2, -1) = 6. 2. f(x,y) = x2 ln(y) - y 3 ex +x3 - y2 +1. 2 fx = 2xln(y) - y3 ex +3x2 , fy = xy - 3y2 ex - 2y, fxx = 2ln(y) - y3 ex +6x, f�y = fyx = 2: - 3y2 ex , fyy = -:� - 6yex - 2. 3. J(x,y) = x4 + 3x2 y2 +y4 , x(t) = t2 +2t, y(i) = t3 - 2t. 3 ftf(x(t),y(t)) = fx �� +J�� = (4x +6xy2 )(2t +2) +(6x2 y +4y3 )(3l 2 - 2).

APPENDIX C. SOLUTIONS

352

4. f(x,y) = x2 -cos(y) + xsin(y), x(t) = 2t2 + 3t - 1, y(l) =et + 2t. 1tf(x(t),y(t)) = [ 2x + sin(y)](4t + 3) + [sin(y) + xcos(y)](e t + 2).

5. f(x,y) = sin(xy) + �' x(t) =et + 1, y(t) = ln(t).

'Jx] e + [xcos(xy) + 1] ½-

1tf(x(t),y(t)) = [ycos(xy) +

t

2

2

6. J(x,y) = x4 + 3x2 y2 + y4, y(x) = x3 + 6x2 - x. 2 3 2 2 3 fxJ(x,y(x)) = fx + f/!;; = 4x + 6xy + (6x y + 4y )(3x + 1 2x - 1). 7. J(x,y) = eY - x2 + y3 ln(x), y(x) = sin(x). 2 fxJ(x,y(x)) = -2x + Y: + [eY + 3y ln(x)] cos(x).

8. f(x,y) =

x2�y2

d!J(x,y(x))

+ xe- Y, y(x) = sin(2x) .

= (x2-:;_�)2 + e-Y+

4 4 9. f ( X' y) = X - y f(X ' y) = 0 =} Jx

- 1.

+ fY ddxy = 0

10. f(x,y) = x3 + y2 - eY. 0 J(x ' y) = 0 ⇒ fx + fY :EL.= dx

2

[(x2�;2)2

- xe- Y ] [2cos(2x)].

=}

dy dx



_b. :EL.= dx fy

4x = xy3 . = - fyfx = - -4y 3

3

3

= -�. 2y -eY

Exercises 2.5 - page 36 1. 3x2 + y + 1 - (3y2 - x + l)y' = 0, Py = 1 = Qx ⇒ the equation is exact. 3 2 fx = P = 3x + y + 1 ⇒ J(x,y) = x + xy + x + g(y), fy = Q =} X + g'(y) = X - 3y2 - 1 ::::} g'(y) = -3y2 - 1 ⇒ g(y) = -y3 - y + c ⇒ f ( x,y) = x3 + xy + x - y3 - y + c, and the general solution is J(x,y) = c 1, or x3 + xy + x - y3 - y = k, where k = c 1 - c.

:t

= 0, Py = x = Qx ⇒ the equation is exact. 2. xy + x + 1 + (½x2 + y + 1) = ½x2 y + ½x2 + x + g(y), = P = xy + x + 1 ⇒ J(x, y) fx 2 2 fy = Q ⇒ ½x + g'(y) = ½x + y + 1 ⇒ g'(y) = y + 1 ⇒ g(y) = ½Y 2 + y + c ⇒ f(x,y) = ½x2y + ½x2 + x + ½Y 2 + y + c =} ½x2 y + ½x2 + x + ½Y2 + y = k. 3. 3x2 y2 - 2x + (2x3 y + 3y2)y' = 0, Py = 6x2 y = Qx ⇒ the equation is exact. 2 2 fx = P = 3x2 y - 2x ⇒ f(x,y) = x3 y2 - x + g(y), fy = Q ⇒ 2x3 y + g'(y) = 2x3y + 3y2 ⇒ g'(y) = 3y2 ⇒ g(y) = y3 + c ⇒ f(x,y) = x3y2 - x2 + y3 + c ⇒ x3y2 - x2 + y3 = k.

4. y3 -3x2 y2+4x3 +(3xy2 -2x3 y-4y3)y' = 0, Py = 3y2 -6x2 y = Qx ⇒ the equation is exact. fx = P = y3 - 3x2 y2 + 4x3 ⇒ J(x,y) = xy3 - x3 y2 + x4 + g(y), fy = Q = 3xy2 - 2x3 y - 4y3 ⇒ g'(y) = -4y3 ⇒ g(y) = -y4 + c ⇒ 3xy2 - 2x3 y 3+ g'(y) 3 2 ⇒ f(x,y) = xy - x y + x4 - y4 + c ⇒ xy3 - x3 y2 + x4 - y4 = k. y(0) = 1 ⇒ k = -1.

353

C.2. CHAPTER 2 SOLUTIONS

2 2 3 u- xy3 ⇒ 2x -3x 5 · y' = 3xx2uy2--2x 3 +x y y + y + (3x y - 2x y + x)y' = 0 is exact because y Py = 6xy2 - 6x2 y + 1 = Qx . fx = P = 2xy -3x y + y =} J(x,y) = x 2 y3 - x 3y2 + xy + g(y), fy = Q ⇒ 3x2y - 2x3 y + x + g'(y) = 3x y - 2x y + x ⇒ g'(y) = 0 ⇒ g(y) = c ⇒ J(x,y) = x y - x y + xy + c ⇒ x y - x y + xy = k. y( 2) = 1 ⇒ k = -2. 2

3

2

3

2

2

2

2

3

2

2

2

6.

3

3

2

3

2

2

/}x + y + ( 21 + x) �� = 0, P

2

3

4)xu

3

2

⇒ the equation is exact. fx = P = /'}x +y ⇒ f(x,y) = ft;y+xy + g(y), fy = Q ⇒ + g'(y) = 21 +x ⇒ g'(y) = 0 ⇒ g(y) = c ⇒ f(x,y) = ft;y+xy +c 21 +x y

=

+ 1 = Qx

= 4 ⇒ k = 6.

⇒ ft;y + xy = k. y( 1)

7. (6xy + eY)y' + xex + 3y2 = 0, Py = 6y = Qx ⇒ the equation is exact. 2 x x 2 x fx = P = xe + 3y ::::} J(x,y) = xe - e + 3xy + g(y), fy = Q =} 6xy + g'(y) = 6xy + eY ⇒ g'(y) = e Y ⇒ g(y) = e Y + c x x 2 x x 2 =} J(x, y) = xe -e + 3xy + eY + c ⇒ xe - e + 3xy + eY = k. - 2y) � = 0, Py = 4xye x +Y = Qx ⇒ the equation 2 2 2 2 2xex +Y + 4x3 ::::} J(x,y) = ex +Y + x4 + g(y), ] = Q

8. 2xe x +Y + 4x + ( 2yex 2

2

3

2

+Y

2

2

2

is exact. fx = P = 2 2 2 2 ::::} 2yex +y + g'(y) = 2yex +y - 2y ::::} g'(y) = -2y ::::} g(y) 2 2 2 2 J(x,y) = ex +y + x4 - y2 + c::::} e x +y + x4 - y2 = k.

=

-y2

y

+ c ::::}

9. ex+y +x+y+cos(x)+(ex+y +x-y)y' = 0, Py = ex+y +l = Qx ⇒ the equation is exact. fx = P = ex+y +x+y+cos(x) ⇒ J(x,y) = ex+y +½x2+xy+sin(x)+g(y), x fy = Q =} e +y +x +g'(y) = ex+y +x- y ⇒ g'(y) = - y =} g(y) = -½y2 +c::::} x +y J(x,y) = e +½x2 +xy+ sin(x)- ½y2 +c ⇒ ex+y + ½x2 +xy+sin(x)-½y2 = k.

-2xy 10 · x2 x2 1 - 2X· + ( x2 Y1P l + 2Y ) ddxy - (x2+y2 - Qx =} the equaf10n - 0 , Fsy + 1) 2 + + +y + 2 2 2 is exact. fx = P = x2 +:2 + 1 - 2x ::::} J(x,y) = ½ ln(x + y + 1) - x + g(y), 2 fy = Q =} x2 +y2 + 1 + g'(y) = x2 y2 +l + 2y ::::} g'(y) = 2y ::::} g(y) = y + C::::} + J(x,y) = ½ ln(x2 + y2 + 1) - x2 + y2 + c ⇒ ½ ln(x2 + y2 + 1) - x2 + y2 = k. y(O) = 0 ⇒ k = 0.

-(2x+ 2 )(2y-3) 3 2 11 · x2 +y22+x2+x y 1 = 1 + �' u = � u =ln JxJ+c ⇒ y = x(ln JxJ + c).

=;,

y

1

= - yx� ⇒

y y 2

1

= -x2 ⇒

u + X'U1 = 1 + u ==> u' = � ==>

(b) x + y- xy1 =0 =} y1 - �y = 1, I(x)= ef-fdx = e-ln lxl = JxJ- 1, J(x) = � 1 ⇒ (�y) = �⇒ �y=lnJxJ+c⇒ y=x(ln JxJ+c).

357

C.2. CHAPTER 2 SOLUTIONS

- 1.y' = Q (c) X + y- xy' = Q ' Py-QQx = -2 ⇒ II('(xx)) = -2 ⇒ J(x) = ..l.. ⇒ 1. + JL x x is exact. fx = �+ ⇒ J(x, y) = ln lxl - �+g(y), fy = -�+g'(y) = -� ⇒ g(y) = c ⇒ lnlxl- � = k ⇒ y = x (ln lxl - k). u = � ⇒ (d) The orthogonal trajectories are given by y' = x�xy = 1 -l-u-u -l u + xu' = __=._!_ u = -ln Ix! + c. ⇒ xu' = l+u ⇒ J -2'c±l._ u +u+l ' dx = J x dx l+u X

2

X

X

2

X

;2

�¥,

2

2

l 1 1 -2'c±1._ _ (u+2l+2 _ .! 2 (u+2) 2 2 2 2 (u+l) +;l u +u+l - (u+l) +;1 2 4 2 4

= 2 1n [ ( u+2 J u u+ +u+l du

�J

2 [(u+.!) +4 2

.2! ln 4.

1

l

2

X

+

__.!._

J3

1 3 1 2 ) +4 ] + v'3

-1 +__1._ v'3 tan

2

73

2 u+ I ) 2 +1 ( 73 73

t an -1

(



2

1 ) ⇒ + v'3

v1:3U

(2.u v'3x+__1._ v'3 ) = -ln !xi+c.

= -� lnlxl+c ⇒ y = kx- 2!3_ ⇒ xu' = -Q3 u ⇒ (b) 3xy' + 2y = 0 ⇒ y' = - l3 1lx ' u = 1l ⇒ u + xu' = - lu 3 5 3 !{ = _Q.!. ⇒ lnlul = _Q lnlxl+c ⇒ u = kx- / ⇒ y = kx- 2 13. 3 3 (c) 3xy'+ 2y = 0 ⇒ y'+3�y = 0 ⇒ I(x) = ef txdx = ei 1 nl x l = lxl2/3 = x213 ⇒ (x 2f3y)' = o ⇒ x 2f3y = c ⇒ y = cx-2/3_ I'( x ) I -1 -1/3 ⇒ (d) 3xy' + 2y = 0 ' Py-QQx = -1 3x ⇒ I( x ) = 3x ⇒ (x) = x 3 2 3 3 1 1 2x- 1 y+3x 1 y' = 0 is exact. fx = 2x- 1 y ⇒ f(x, y) = 3x213y+g(y), fy = 3x 2 13 + g'(y) = 3x 213 ⇒ g(y) = c ⇒ 3x2 13y = k ⇒ y = k 1 x- 2 l3 _ (e) The orthogonal trajectories are given by y' = �: ⇒ 2yy' = 3x ⇒ y 2 = �x 2 + c ⇒ 3x 2 - 2y 2 = k. (a) 3xy'+ 2y = 0 ⇒ 1:;- = -lx ⇒ ln IYI

X

U

X

5. 2x - 2e- x - y 2

=

k ⇒ 2+2e- x - 2yy' = 0 ⇒ y' = l+e x . The orthogonal y trajectories are given by y' = 1 .;/-x ⇒ = 1 �1 x = e��1 ⇒ ln IYI = -ln(ex+1)+C ⇒ Y = ex: 1 •

f

6. y = 2x lnlxl+ex ⇒ � - 2 ln lxl = c ⇒ -;2+� - � = 0 ⇒ y' = �+ 2. The - 12 ⇒ orthogonal trajectories arc given by y' = 11.-+12, u = 1lX ⇒ u+xu' = u+ x 1 _ -l- 2u-u2 l + 1 d _ -1 d ⇒ ⇒ u+2 1 _ -1 ⇒ XU -

u+2

lnju+11- u�l 7. J(x, y)

=

(u+l)2 U -

x

= -lnlxl+c ⇒

J u+l

(u+l) 2

ln[� + 1[- x : y

U -

Jx

X

= -lnlxl+c.

0 defines y as a function of x. Hence, fx+fy ��

=

= - � :­ : = - �:,

0 ⇒ :

J(x, y) = 0 also defines x as a function of y. Hence, fx :+fy = 0 ⇒ -..l... and y'= 2 v-3 = ll_ 2 ' u = Jf..x => u + xu' = 2u-2x 2 X 2 y +4u-3 => u-2 u = -1 =? -1- _1_ 2 =? => xu' = -u u(u-1 + u-3 ) u' = x 2 u2 -4u+3 ' x lnlu-ll+lnlu-31 = -2ln lxl+c => (u-l)(u-3) = x\ => u2 -4u+3- x\ = 0 .J__

=>u=2±J1 +;2 =>y=2x±

9.

✓x2 +k.

(a) xex - xln(x) + y + xy' = 0 is exact since Py = l = Qx , fv = x => f(x, y)=xy+g(x), fx=y+g'(x)=xex -xln(x)+y => g'(x) = xex -x ln(x) => g(x) � xex -ex -½x2 ln(x) +¼x2 +c=> xy+xex -ex -½x2 ln(x)+¼x2 = k x 1 =>y = !:..... X - e + .!x 2 ln(x) - 4 x + !E. X

(b) xex -xln(x)+y+xy' = 0 => xy'+y=xln(x)-xe =>(xy)' = xln(x)-xex =>xy = ½x2 ln(x) - ¼x2 - xex +ex + c =>y = e: - ex + ½xln(x) - ¼x + �-

10. (a) g(y)y'= f(x) => f(x) - g(y)y' = 0 and [f(x)Jv = 0 = [- g(y)] x .

n y) (b) y+ xln(x)ln(y)y'= 0 => l � y' = x l�(x) => ½[ ln(y)]2 = -lnlln(x)I + c => ± ✓k-2 lnlln(x)I_ y=e 11=> + ln(yy) y'= 0 is exact. fx = (c) ln(yy) y' = --=.!._ x ln( x ) => xln( x) x ln(x) 1 Y) 2 f(x, y)=ln I ln(x)I + g(y), fv = g'(y)= "� => g(y) = ½[ln(y)] + c => ln I ln(x)I + ½[ln(y)] 2=k.

11. (a) y' = J(ax +by +c), b-/- 0, u(x) =ax +by +c => u'=a +by'= a +bf(u) => a+t; (u) =1, which is separable. (b) y' = (x + y + 1)2 , u= x + y + l =>u' = 1 + y'= 1 + u2 => 1��2 = 1 => tan-1(u)=x+ c =>u= tan(x + c) => y = u - x - l= tan(x + c) - x - l. y'u ( ) 1 1 _ y'u+l _ C yI = Jx+iF°2' U=x +y - 2 ⇒ uI=1 +yI=1+ Ju - .ju ⇒ y'u+luI - 1 => J 1 du = J 1 dx = x + c, and v= fa =>



J }£_1 du = J -:;;h 2v dv = 2 J v - l + :;¼i- dv = v 2 - 2v + 2 ln(v + 1) =>

x + y - 2 - 2 ✓x + y - 2 + 2ln( ✓x + y - 2 + 1)=x+ c.

(d) y' = (x + y) ln(x + y) - 1, u = x+ y => u' = 1 + y' = u ln(u) => ul:(u)u' = 1 x => ln I ln(u)I = x + c => I ln(u)I = ex+c => ln(u) = kex => u = e ke => X y = u - X = e ke - X. 12.

(a) a(y) + [b(y)x - c(y)]y' = 0, Qx;Py = I (y) exists.

b(y�c:;(y)

is independent of x; hence,

(b) a(y) + [b(y)x - c(y)]y' = 0 => a(y) + [b(y)x - c(y)];;, = 0 => a(y)x' + b(y)x = c(y), which is linear for x(y). 13.

P (a) y + [(2y2 + l)x - 2y]y' = 0, Qx; y = 2y => �(�i = 2y => I(y) = eY => 2 2 2 2 ye Y +[(2y2 +1)x-2y]eY y'=0 is exact. fx=yeY => f(x, y)=xyeY +g(y), 2

359

C.2. CHAPTER 2 SOLUTIONS

(b)

2 2 fy = (2y2 + l)xeY + g'(y) = [(2y2 + l)x - 2 y]eY ⇒ g'(y) 2 2 g(y) = -eY + c ⇒ (x y - l)eY = k.

(a)

2 -2 yeY ⇒

2 [(2y +l)x - 2y]y' = 0 ⇒ y + [(2y2 +l)x - 2y]� = 0 ⇒ 2 x = 2. The integrating factor is yx' + (2y + l)x = 2 y ⇒ x' + ( 2 y + 2 Y l Y dY = eY +111 n Y = ± y eY2, and I(y) = yeY2 ⇒ ( yeY2x)' = 2yeY2 J(y) = eJ2+

y+

t)

⇒ y eY

14.

=

2

x

= eY2 + c ⇒

(xy - l)eY

2

= c.

+(3xy+2)y' = 0, Qx; y =; = �(�i ⇒ I(y) = y ⇒ y3+(3xy2+2 y)y' = 0 is exact. fx = y3 ⇒ J(x, y) = xy3 + g(y), fy = 3xy2 + g'(y) = 3xy2 + 2y 2 2 3 ⇒ g(y) = y + c ⇒ xy + y = k. y

P

2

(b) y2 + (3xy + 2)y' = 0 ⇒ y2 + (3xy + 2)� = 0 d x' + �x = ;'1, J(y) = ef f y = e3 Inlyl = ± y3 , I(x) 2 3 2 3 y x = -y + c ⇒ x y + y = c.

⇒ y

=y

3

2

= -2 ⇒ x)' = -2y ⇒

x' + 3yx



(y

3



1 5. 1 + (xy - x- 3 y3)y' = 0 ⇒ 1 + (xy - x-3 y3 = 0 ⇒ x'+ yx = y3 x-3, which is a Bernoulli equation for x(y) with a= -3. u = x4 ⇒ x = u 114 ⇒ x' = ¼u- 314 u' 2 2 4 l 4 3 3 3 ⇒ ¼u- f4u' + yu / = y3u- /4 ⇒ u' + 4 yu = 4y , I(y) = e f y dy = e y ⇒ 2 2 2 2 2 2 ' ( e2Y u) = 4y3 e2Y ⇒ e2Y u = J y2 • 4ye2Y dy = y2 e2Y - ½e2Y + c (by parts) ⇒ /4 u = y2 - ½ + c e-2y2 ⇒ x = (y2 - ½ + ce-2y2)1 . 16.

(a)

= 2/y=l ⇒

y'

(b) The function

y

.J'i=r = 1

2

= 1 satisfies

⇒ y'

jy=1

= x + c ⇒ y = (x + c)2 +1.

= 0 = jy=I and is a singular solution.

(c) At every point (x 0, 1) on the line y = 1, the line is tangent to the parabola y = (x - x0)2 + 1 at its vertex. Hence, the line is the envelope of the one-parameter family of parabolas.

17.



(b) y

= l-y2y

y ' =1 ⇒ =f ✓l - y2 ±� y J(x, y, c) = (x + c) 2 + y2 -1 = 0.

(a) (y')2

2



= X +C ⇒

(x + c)2 + y2

=1

= ±1 are two singular solutions.

(c) f(x, y,c) = (x+c)2 +y2 -l ⇒ y = ±1.

= 0 and fc (x, y,c) = 2(x+c) = 0 ⇒ y2 -1 = 0

(d) The general solution defines a one-parameter family of circles of radius 1 with centres at (-c, 0) on the x-axis. At every point (x 0, ±1) on the lines y = ±1, the lines are tangent to the circle centred at (x0, 0). Hence, each line is an envelope of the family of circles. 18.

(a) y - mx ±

✓m2 + 1 ⇒ y' -- m ⇒

(y') 2 +1

( y- xy')2 -

m2 +1 ( mx ±� m + -xm)2 1.

360

APPENDIX C. SOLUTIONS (b) y = mx ± Jm2 + 1 ⇒ y - mx = ±Jm 2 + 1 ⇒ (y - mx)2 = m2 + 1 ⇒ f(x,y,m) = (y-mx) 2 -m2 - 1 = 0. fm (x,y,m) =-2x(y-mx)-2m = 0 m2 = 2 ⇒ x(mx-y) =m ⇒ x2 = (mx-y )2 m�+1 · (c) x(mx - y) = m::::;, y = m(x;-l) ::::;, m 2(x2-1 ) 2 2 2 1_ _ = (m 2 + 1) (---=L Y = m 2+1 ) = m 2+1 · x2

(d) x2 + Y2 = m�:1 solution.

+ m/+1 = 1



g(x,y) = x2

+ y2 - 1 = 0 is a singular

(e) The graph of the singular solution is the unit circle centred at the origin. At every point on the circle, the circle is tangent to one of the lines defined by the general solution. This can be confirmed as follows: x2 + y2 = 1 ⇒ 2x + 2yy' = 0 ⇒ y' = -�. Thus, the slope of the tangent to the circle at

+ 1 goes through (x0, y0) if and only if Yo = mxo ± Jm + 1. Then (y0 - mx0) 2 = m 2 + 1 and, solving

(xo ,Yo) is m1 = -�- The line y = mx ± Jm 2 2

for m, we obtain m =-xo = m1 . Thus, the line is tangent to the circle at YO (xo ,Yo)-

19. (a) (b) 20.

⇒ 'Ji.. = .!. ⇒ y = kx. y = kx ⇒ y(0) = 0 for all values of k.

y' = Ji.X

y

X

Hence, y(0) = 1 is impossible. The reason is that f(x,y) = � is not continuous at x = 0 and, hence, it is not continuous in a rectangular region containing (0, 1).

⇒ y- 1 13y' = 3x ⇒ �y2!3 = �x2 + c ⇒ y= (x2 + k) 312. y =(x2 + k) 3 1 2 and y(0) = 0 ⇒ k = 0 ⇒ y =x 3 .

(a) y' = 3xy 1 ! 3 (b)

(c) y

= 0 is a singular solution.

=

(d) Both y= x3 and y 0 are solutions of the initial-value problem. Hence, the solution is not unique. The reason is that, with j(x,y) = 3xy 1 1 3, 2 fv(x,y) = xy- ! 3 is not continuous at y = 0 and, hence, it is not continuous in a rectangular region containing (0, 0).

C.3

Chapter 3 Solutions

Exercises 3 .1 - page 46 1. y" + 6x2 + sin(x) = 2 ⇒ y" = 2 - 6x2 - sin(x) ⇒ y' = 2x - 2x3 + cos(x) + c1

⇒ y = x2 - ½x4 + sin(x) + c1x + c2.

x 2. xy" = ln(x) ⇒ y" = In� ) ::::;, y' = ½[ln(x)] 2 + c1 ::::;, y = ½ f[ln(x)] 2dx + C1X + C2. f[ln(x)] 2 dx = 1 · [ln(x)] 2 dx = x[ln(x)] 2 - 2 ln(x)dx = x[ln(x)] 2 - 2x ln(x) + J 2dx = x[ln(x)] 2 - 2x ln(x) + 2x ⇒ y = ½x[ln(x)] 2 - x ln(x) + x + c1x + c2 = ½x[ln(x)] 2 - x ln(x) + c3 x + c2.

J

J

C.3. CHAPTER 3 SOLUTIONS

361

3. y"-4 cos2(x) = 2 ⇒ y" = 4 cos2(x)+2 = 4+2cos(2x) ⇒ y' = 4x+sin(2x)+c1 ⇒ y = 2x2 - ½ cos(2x) + c1x + c2,y(0) = 0 ⇒ c2 =½,and y'(0) = 1 ⇒ c 1 = 1 ⇒ y = 2x2 - ½ cos(2x) + x + ½4.

⇒ y' = xex - ex + C 1 ⇒ y = xex - 2ex + C1 X + C2,y(0) = 1 ⇒ C2 = 3, and y'(0) = 1 ⇒ C 1 = 2 ⇒ y = xex - 2ex + 2x + 3.

x y" = xe

5. (a) xy" - y' + 2y = 0 is linear and homogeneous. (b) 2y"+ ln(x)y = sin(x) is linear and nonhomogeneous. (c) y" + y' + (x2 + y 2) = 0 is nonlinear. (d) y"+ xy' = y 3 is nonlinear. (f)

⇒ x2y"+ xy' - 2y = 0 is linear and homogeneous. y" + y' + y + x = 0 ⇒ y" + y' + y = -x is linear and nonhomogeneous.

(g)

y"+ yy' + y = 0

(e) x2y"+ xy' = 2y

is nonlinear.

6. If y 1 = er i x and y2 = er2x are linearly dependent, then there exist constants C 1 and c2, not both zero, such that c 1 erix + c2er2 x = 0 for all x. Then x = 0 ⇒ c 1 + C 2 = 0 ⇒ c1(er , x - er2x ) = 0 for all x,and Ci -/- 0 (otherwise C2 = -Ci = 0) ⇒ eri x = er2x ⇒ r1 x = r2x,and x = l ⇒ ri = r2. Conversely,if ri = r2,then Yi = Y2 ⇒ Yi and Y2 are linearly dependent. 7. If y1 = xa and y2 = xf3 are linearly dependent, then there exist constants c 1 and c2,not both zero,such that cixa + c2 xf3 = 0 for all x in J. Then x = l ⇒ ci +c2 = 0 ⇒ c1(xa -xf3) = 0 for all x in J, and c 1 -/- 0 (otherwise c2 = -ci = 0) ⇒ xa = xf3,and x = 2 ⇒ 2a = 2/3 ⇒ a= (3. Conversely,if a= /3, then Y 1 = Y2 ⇒ Yi and Y2 are linearly dependent. 8. If Yi = xr and Y2 = esx are linearly dependent, then there exist constants c 1 and c2,not both zero,such that c 1 xr + c2esx = 0 for all x in J. Then x = l ⇒ C1 + C2e5 = 0,X = 2 ⇒ 2r c 1 + C2e 25 = 0,and X = 3 ⇒ 3r ci + C2e35 = 0. These three equations give s = r ln(2) and 2s = r ln(3), which is impossible unless r = s = 0. Conversely,if r = s = 0,then Yi = Y2 = 1 ⇒ Y 1 and Y2 are linearly dependent.

9. If Yi,Y2, · · ·,Yn are linearly dependent,then there exist constants

ci, c2, · · ·, Cn , not all zero, such that C 1 Yi + C 2Y2 + · · · + CnYn = 0. If c1 -/- 0, where' of course' the term - 1 y - · · ·- fn · is omitted yi - £?.. then yJ· = -!:l Cj Cj 2 Cj yn' Cj yJ from the right-hand side. Conversely, if y1 = C 1 Yi + C2Y2 + · · ·+ Cn Yn , then C1Y 1 + C2Y2 + · · · + Cn Yn - l · y1 = 0, with Cj =-1 -/- 0.

10. The functions y 1 = x,y2 = ex and y3 = e-x are linearly independent. Suppose that c 1 x + c2 e + c3 e-x = 0 for all x. Then x = 0 ⇒ c2 + c3 = 0, x = l ⇒ c 1 + c2(e - e-i) = 0,and x = -l ⇒ -ci + c2(e-i - e) = 0. The unique solution of this system of three equations is ci = c2 = c3 = 0.

362

APPENDIX C. SOLUTIONS

1 1. The functions Y1 =x, y2 =e x and y3 =2x are not linearly independent because 2 · Y1 + 0 · Y2 + (-1) · y3 =0 for all x, with c1=2 -/- 0.

=

12. If y 0, then 1 · y = 0 for all x, with c1 = 1 -/- 0. Hence, y is linearly dependent. If y ¢. 0, then there exists x0 such that y(x0) -/- 0. If cy =0 for all x, then cy(x0) = 0 ⇒ c=0, i.e.,y is linearly independent. Thus, y is linearly independent if and only if y ¢. 0.

Exercises 3.2 - page 55 1.

y

11

- y' - 2 y =0,y=e rx=}r2 - r-2=(r- 2)(r+l)=0

⇒ y=c1e 2x + c2e-x .

2.

=e rx=}r2 +r- 6=(r-2)(r+3)=0=}y=c1 e 2x +c2 e- 3x _ 2x -e-3x),y'=c1(2e 2x +3e-3x), and y'(0)= 10 y (0) =0 =} C2 =-Ci=} y= c1(e =} C1=2=}y =2(e 2x -e-3x). Then y(l)=2(e 2 -e-3) and y'(l) =2(2e 2 +3e-3).

3.

=e rx =} r2 +5r+4=(r+l)(r+4) =0 =} y =c1e-x +c2 e-4x _ x 4x x 4x y (0) = 0 =} C2 = -c1 =} y = c1(e- - e- ), y' = c1(-e- + 4e- ), and ) y'(0 =-6 =} C1 =-2=} y =-2(e-x - e-4x ).

y"+ y' -6 y=0, y

y"+5 y' +4 y=0, y

4. 4y" + y' -3 y = 0, y= e rx=} 4r2 + r - 3 = 0 =} r= -1/7 =} r1= ¾, r2 = -1 3 =} Y=c1e 4 x + c2e-x . 5. y" -3 y' + y = 0, � Y2 = e 2 x 6.

y=e

rx

3

..rsx

=} r2 - 3r + l = 0=} r = 3 ±2v'5 =} Y1 = e \

=e rx =} r2 - 1 =0 =} r = ±1 =} y =C1e x +c2 e- x , y' =C1e x - C2 e-x , C2 = -½ =} y (0) = 1 =} C1 + C2 = 1, y' ( 0) = 2 =} C1 - C2 = 2 =} C1 = 3 e - 1... x x � .1.. = ) - .2!.e- · Then y (l)= 2 - 2· and y (-1 Y= 1e 2e 2e 2 y"- y=0, y

t

7. 9y" - 24y' + 16y =0, y = e rx =} 9r2 - 24r + 16 =(3r - 4)2 = 0=} r= ix 3 3 Y1 =e ix , Y2 =xe 8.

and

rx =} r2 - lOr + 25 =(r - 5)2 =0 y" - l0 y' + 25 y=0, y=e 5x 5x y =e (c1 + c2 x). y(0)=l=}C1 = 1 =} y =e (l + c2 x), 5x 5x y'=5e (l + C2 X) + C2 e , y'(0) = 1 =} C2 =-4=} y=e 5x (l

9. 4y" - 4y' + y = 0, y =e½ x (c1 + c2x).

y

! =}

=} r=5=} - 4x).

= e rx =} 4r2 - 4r + 1 = (2r - 1)2 = 0 =} r

1

2

=}

10.

0, y = e rx =} r2 = 0 =} r = 0 =} Y1 = e 0 = l, Y2 = XY1 = x, and y=c1 + c2 x. Alternatively,y"=0 ⇒ y'=k1 =} y=k1x + k2 .

1 1.

y" + y'

y" =

+ y=0, y =e rx =} r2 + r + l =0 =} r = -1±2,/=3 =

y=e-½ x [c1 cos

(-1,;x) + C sin (-1,;x)]. 2

-½ ± i °';' =}

363

C.3. CHAPTER 3 SOLUTIONS 12. y" - 2y' + 2y = 0, y = e rx ⇒ r2 - 2r + 2 = 0 ⇒ r = 2±t=4" = 1 ± i ⇒ y = e x [c 1 cos(x) + C2 sin(x)]. y(0) = 0 ⇒ c 1 = 0 ⇒ y = c2 ex sin(x), y' = c2 [e x sin(x) + ex cos(x)], and y'(0) = 1 ⇒ C2 = 1 ⇒ y = ex sin(x).

⇒ 2r2 - 4r + 3 = 0 ⇒ r = 4±� = 1 ± i4 ⇒ y = ex [C1 cos ( 1 x) + C2 sin ( 1 x) ] .

13. 2y" - 4y' + 3y

16. 17.

0, y

=

erx

⇒ r2 + 4 = 0 ⇒ r = ±2i ⇒ y = c1 cos(2x) + C2 sin(2x). (a) y" + y = 0, y = erx ⇒ r2 + 1 = 0 ⇒ r = ±i ⇒ y = C1 cos(x) + C2 sin(x). (b) y(0) = 1 and y'(0) = 0 ⇒ y = cos(x). (c) y(0) = 0 and y'(0) = 1 ⇒ y = sin(x). x2 y"+2xy'-6y = 0. For x > 0, y = xr ⇒ r(r-1)+2r-6 = 0 ⇒ r2 +r-6 = 0 ⇒ (r + 3)(r - 2) = 0 ⇒ y = c1lxl-3 + c2lxl2 = c3x-3 + c2x2 for x =/- 0. 6x2 y" +7xy' -2y = 0. For x > 0, y = xr ⇒ 6r(r -1) + 7r -2 = 6r2 + r -2 = 0 ⇒ r = -�t7 = ½, -i ⇒ y = c1lxl 1 /2 + c2x-2 l3 for x =/- 0.

14. y" + 4y 15.

=

= 0,

y = e rx

18. x2 y" -7xy'+ 16y = 0. For x > 0, y = xr ⇒ r(r-1)-7r+16 = r2 -8r+16 = 0 ⇒ (r - 4)2 = 0 ⇒ y = x4 [c1 + c2 ln lxl] for x =/- 0. y(l) = 0 ⇒ c 1 = 0 ⇒ y = c2x4 ln lxl, y' = c2 [4x3 ln lxl + x3 ], and y'(l) = 2 ⇒ c2 = 2 ⇒ y = 2x4 ln lxl. Then y (-1) = 0 and y ( e) = 2e 4.

19. 9x2 y"-3xy'+4y = 0. For x > 0, y = xr ⇒ 9r(r-1)-3r+4 ⇒ (3r - 2)2 = 0 ⇒ y = x213 [c1 + c2 ln lxl] for x =/- 0.

= 9r2 -12r+4 = 0

20. x2 y"+xy' = 0. (1) For x > 0, y = xr ⇒ r(r-l)+r = r2 = 0 ⇒ y = Ci +c2 ln lxl for x =/- 0. (2) For x =/- 0, x2 y" + xy' = 0 ⇒ xy" + y' = 0. Integrate by parts to obtain xy' - y' dx + y' dx = k 1 ⇒ y' = � ⇒ y = k1 ln lxl + k2. (3) For ⇒ ln IY'I = - ln lxl + C1 ⇒ y' = � ⇒ y = k1 ln lxl + k2. x =/- 0, =

7



J

J

21. -2x2 y" + 2xy' - 5y = 0 ⇒ 2x2 y" - 2xy' + 5y = 0. For x > 0, y 2r(r -1) -2r + 5 = 2r2 - 4r + 5 = 0 ⇒ r = 4±ij24 = 1 ± i 'I,; ⇒ y = x [c 1 cos (

4 ln lxl) + c sin ( 'I,; ln lxl)] for x =/- 0.

=

xr



2

22. x2 y"-3xy'+ 13y = 0. For x > 0, y = xr ⇒ r(r-1) -3r+ 13 = r2 -4r+13 = 0 ⇒ r = 2 ± 3i ⇒ y = x2 [c1 cos(3 ln Ix!)+ c2 sin(3 ln lxl)] for x =/- 0. Then y(l) = 0 ⇒ c1 = 0 ⇒ y = c2 x2 sin(3 ln lxl), y' = c2[2x sin(3 ln lxl) + 3x cos(3 ln lxl)], and y'(l) = 6 ⇒ c2 = 2 ⇒ y = 2x2 sin(3 ln \x\). 23. x2 y" + xy' + 4y = 0. For x > 0, y = xr ⇒ r(r - 1) + r + 4 r = ±2i ⇒ y = c 1 cos(2 ln lxl) + c2 sin(2 ln lxl) for x =/- 0. ,.I

=

r2 + 4

=

0



364

APPENDIX C. SOLUTIONS

Exercises 3. 3 - page 7 4 l.

+ 4y' = -5e x . For y" + 4y' = 0, y = e rx ⇒ r2 + 4r = r(r + 4) = 0 ⇒ y1 = 1 and Y2 = e-4x _ YP = Ae x ⇒ y� = Ae x , yi = Ae x , and yi + 4y� = -5e x ⇒ A= -1 ⇒ YP = -ex . The general solution is y =-e x + c 1 + c2 e-4x_

2.

- y = 6e 2x , y(0) = 2, y'(0) = 4. For y" - y = 0, y = e rx ⇒ r2 - 1 = 0 ⇒ Yl = e x and Y2 = e-x _ YP = Ae 2x ⇒ y� = 2Ae 2x , Yi= 4Ae 2x , and Yi- yp = 6e 2x ⇒ A = 2 ⇒ YP = 2e2x . The general solution is y = 2e 2x + c1ex + c2 e-x. 2x y(0) = 2 +C1 +C2 = 2 and y'(0) = 4 + C1 - C2 = 4 ⇒ c 1 = C2 = 0 ⇒ y = 2e .

3.

5y' + 6y = 4e- 2x . For y" - 5y' + 6y = 0, y = e rx ⇒ r - 5r + 6 = (r - 2)(r - 3) = 0 ⇒ Y1 = e 2x and Y2 = e 3x . YP = Ae-2x ⇒ YP, = -2Ae-2x ' yP" = 4Ae-2x ' and yP" -5yP' +6yP = 4e-2x ⇒ A=.!.5 ⇒ yP = l e-2x. 5 The general solution is y = ½e- 2x + c 1 e 2x + c3 e 3x .

4.

+ 3y ' - 4y = -lOe x . For y " + 3y ' - 4y = 0, y = e rx ⇒ r +3r-4 = (r-l)(r+4) = 0 ⇒ y1 = e x and Y2 = e-4x _ Since e x is a solution of the homogeneous equation , YP = Axe x . Then y� = Ae x(l+x), yi = Aex(2+x), and yi + 3y� - 4yp = - lOe x ⇒ A = -2 ⇒ YP = -2xe x . The general solution is Y = -2xe x + C1ex + C2 e-4x _

5.

+ 6y' + 9y = 6e-3x _ For y " + 6y' + 9y = 0, y = e rx ⇒ r2 + 6r + 9 = (r + 3)2 = 0 ⇒ Y1 = e- 3x and Y2 = xe- 3x _ Since e-3x and xe- 3x are solutions of the homogeneous equation , YP = Ax2 e-3x _ Then 3x 2 2 3x 3x ⇒ y� = Ae- (2x - 3x ), yi = Ae- (2- l2x + 9x ), and yi + 6y� +9yp = 6e3 3x 2 2 3x 3x A= 3 ⇒ YP = 3x e- _ The general solution is y = 3x e- + c1 e- + c2 xe- x_

y"

y"

y" 2

y" 2

y"

6. 6y " + y ' - 2y = 65sin(x). For 6y" + y' - 2y = 0, y = e rx ⇒ 6r2 + r - 2 = 0 ⇒ x x Y1 = e½ and Y2 = e-� . YP = Acos(x) +B sin(x) ⇒ y� = -Asin(x) +Bcos(x), yi = -Acos(x) - Bsin(x), and 6yi +y� - 2yp = 65sin(x) ⇒ A + 8B = -65 . } ⇒ A= -1 and B = -8 ⇒ YP = -cos(x) - 8sm(x). { -SA + B = 0 The general solution is y = -cos(x) - 8sin(x) + c1e½ x +c2 e-� x . 7.

+ 2y' + y = 50cos(3x), y(0) = 0, y '(0) = 2. For y" + 2y' + y = 0, y = e rx ⇒ r +2r+l = (r+l)2 = 0 ⇒ Y1 = e- x and Y2 = xe-x . YP = Acos(3x)+Bsin(3x) ⇒ y� = -3Asin(3x) + 3Bcos(3x), yi = -9Acos(3x) - 9 Bsin(3x), and Yi + 2y� + YP = (-8A + 6B) cos(3x) +(-6A - 8B)sin(3x) = 50cos(3x) ⇒ -8A + 6B = 50 . } ⇒ A = -4 and B = 3 ⇒ YP = 3sm(3x) - 4cos(3x). { _6A _ SB = 0 The general solution is y = 3sin(3x) - 4cos(3x) +c1e-x + c2 xe-x . y(0) = 0 ⇒ c1 = 4, y'(0) = 2 ⇒ C2 = -3 ⇒ y = 3sin(3x) - 4cos(3x) + 4e-x - 3xe-x .

y" 2

C.3. CHAPTER 3 SOLUTIONS

365

8. y" + y = 4sin(x). For y" + y = 0, y = e rx ⇒ r 2 + 1 = 0 ⇒ y1 = cos(x) and y2 = sin(x). Since cos(x) and sin(x) are solutions of the homogeneous equation, YP = x[Acos(x) + Bsin(x)]. Then y� = [Acos(x) + Bsin(x)] + x[-Asin(x) +Bcos(x)], yi = [-2A sin(x) + 2B cos(x)]+x[-Acos(x) - Bsin(x)], and yi +YP = 4sin(x) ⇒ A= -2 and B = 0 ⇒ YP = -2xcos(x). The general solution is y = - 2xcos(X) + C1 cos(X) + C2 sin(X).

9. y" + 4y = 8cos(2x) + 12sin(2x). For y" + 4y = 0, y = e rx ⇒ r 2 + 4 = 0 ⇒ y1 = cos(2x) and y2 = sin(2x). Since cos(2x) and sin(2x) are solutions of the homogeneous equation, YP = x[Acos(2x) + Bsin(2x)]. Then y� = [Acos(2x) + B sin(2x)] + x[-2Asin(2x) + 2B cos(2x)], yi = [-4A sin(2x) + 4B cos(2x)] + x[-4A cos(2x) - 4B sin(2x)], and yi + 4yP = 8cos(2x) + 12sin(2x) ⇒ A= -3 and B = 2 ⇒ YP = x[2sin(2x) - 3cos(2x)]. The general solution is y = x[2sin(2x) - 3cos(2x)] + c1cos(2x) + c2 sin(2x).

10. y" + 4y' + 4y = 4sin(x) - 3cos(x) . For y" + 4y' + 4y = 0, y = e rx ⇒ r 2+4r+4 = (r+2) 2 = 0 ⇒ Y1 = e- 2x and Y2 = xe- 2x. YP = Acos(x)+Bsin(x) ⇒ y� = -Asin(x) +Bcos(x), yi = -Acos(x) - Bsin(x), and yi +4y� +4yp = (3A+4B) cos(x) +(-4A+3B) sin(x) = 4sin(x) - 3cos(x) ⇒ 3A + 4B = -3 { } ⇒ A= -1 and B = 0 ⇒ yP = - cos(x) . The general _4A+ 3B = 4 solution is y = - cos(x) + c 1 e- 2x + c2 xe- 2x. 11. y" - y = 4. For y" - y = 0, y = e rx :::::} r 2 - 1 = (r - l)(r + 1) = 0 :::::} Y1 = e x and Y2 = e-x. YP = A ⇒ y� = yi = 0, and yi - YP = 4 ⇒ A = -4 ⇒ YP = -4. The general solution is y = -4 + c 1 e x + c2 e- x . 12. y" - y' - 12y = 3 + 12x. For y" - y' - 12y = 0, y = e rx ⇒ r 2 - r - 12 = (r - 4)(r + 3) = 0 ⇒ Y1 = e4x and Y2 = e- 3x_ YP = A+ Bx ⇒ y� = B, yi = 0, and yi-y� - 12yp = (-12A- B) - 12Bx = 3+ 12x ⇒ A=-½ and B = -1 ⇒ YP = -½-x. The general solution is y = -½-x+c1 e 4x +c2 e-3x _ 13. y" + 2y' + 3y = -9x. For y" + 2y' + 3y = 0, y = e rx ⇒ r 2 + 2r + 3 = 0 ⇒ r = -l±v'2i ⇒ Y1 = e- x cos(v'2x) and Y2 = e-xsin(v'2x). YP = A+Bx ⇒ y� = B, yi = 0, and yi + 2y� + 3yp = (2B + 3A) + 3Bx = -9x ⇒ A= 2 and B = -3 ⇒ YP = 2 - 3x. The general solution is y = 2 - 3x + e-x[c1cos(v'2x)+ C2 sin(v'2x)] . 14. y" - 9y = 2 - 3x+ x2. For y" - 9y = 0, y = e rx ⇒ r 2 - 9 = (r - 3)(r + 3) = 0 ⇒ Y1 = e 3x and Y2 = e- 3x_ YP = A+Bx+Cx2 ⇒ y� = B+2Cx, vi= 2C, and 20 YP" - 9yP = (2C - 9A) - 9Bx - 9 Cx2 = 2 - 3x + x2 ⇒ A = - 81' B = .3!. and 20 2 C = _.!.9 ⇒ yP = -81 + lx 3 - lx 9 . The general solution is 20 3x 2 _ .!. _ l Y = 81 + 3 x 9 x + c1e + c2 e-3x

366

APPENDIX C. SOLUTIONS

15. y"+ y ' = 1 + x, y(0) = 1, y '(0) = 1. For y"+ y ' = 0, y = erx =} r2 + r = r(r + 1) = 0 ⇒ y1 = 1 and y2 = e-x . Since 1 is a solution of the homogeneous equation, yP = x(A+Bx)= Ax+ Bx2. Then y�=A+ 2Bx, y; = 2B, and y ; + y� = 1 + x⇒ A=0 and B = ½ ⇒ YP = ½x2 . The general solution is y = ½x2 + c1 + c2e- x . y(0) = 1 ⇒ c1 + c2 = 1, and y'(0) = 1 ⇒ C2 = -1 ⇒ C1 = 2 ::::} y = ½x2 + 2 - e- x . 16. y" - y' = 4x3 . For y" - y' = 0, y = erx ⇒r2 - r = r(r - 1) = 0⇒ Y1 = 1 and y2 = ex . Since 1 is a solution of the homogeneous equation, YP=x(A+Bx+ Cx2 + Dx3 )=Ax+ Bx2+ Cx3+ Dx4 . Then Y,p =A+ 2Bx+ 3Cx2 + 4Dx3 ' y"p = 2B + 6Cx+ 12Dx2 ' and yp" - y'p = 4x3 ⇒ (2B - A)+ (6C - 2B)x+ (12D - 3C)x2 - 4Dx3 = 4x3 ⇒A=-24, B = -12, C = -4 and D = - l⇒ YP = -24x - 12x2 - 4x3 - x4. The general solution is y = -24x - 12x2 - 4x3 - x4 + c 1 + c2 ex . 17. y" - y = -2x + 4ex + 2 sin(x), y (0) = 0, y'(0) = 9. For y" - y = 0, y = erx ⇒ r2 - 1 = (r - l)(r + 1) = 0 ⇒ y1 = ex and Y2 = e- x . For y" - y = -2x, Yp 1 =A+Bx⇒A=0 and B = 2 ⇒YP i = 2x. For y " - y = 4ex , Yp2 = Axex ⇒ A = 2 ⇒ Yp2 = 2xex . For y" - y = 2 sin(x), yp3 = A cos(x) + Bsin(x) ⇒ A = 0 and B = -l ⇒ yp3 = - sin(x). Hence, YP = 2x + 2xex - sin(x). The general solution is y = 2x + 2xex - sin(x) + c1ex + c2 e-x . y(0) = 0 ⇒ c 1 + c2 = 0, and y'(0) = 9 ⇒ 3 + c1 - c2 = 9 ⇒ c1 = 3 and c2 = -3 ⇒ y = 2x+ 2xex - sin(x)+ 3ex - 3e- x .

:=�:

18. y"+y'-2y

= 9ex . For y"+y'-2y = 0, y = erx⇒r2+r-2 = (r-l)(r+2) = 0⇒ Y1 = ex and Y2 = e-2x, with W(x) = I :: _ I = -3e-x and f(x) = 9ex . 2 Then u 1 = J 3dx = 3x, u 2 = J-3e3x dx = -e3x , and YP = 3xex - ex , or YP = 3xex . The general solution is y = 3xex + c1ex + c2e- 2x.

19. y" -4y'+4y

= 6xe2x.

For y" -4y'+4y

= 0, y = erx⇒r2 -4r+4 = (r- 2)2 = 0 e2x xe2x 4x . 2x and W ( x ) = I 2x 2x and Y2 = xe2x, with =} y1 = e 2x I = e e + 2xe 2e f(x) = 6xe2x . Then u 1 = J-6x2 dx = -2x3 , u 2 = J 6xdx = 3x2 , and YP = -2x3 e2x+3x3 e2x = x3 e2x . The general solution is y = x3 e2x+c1e2x+c2 xe2x.

20. y"+y' = 1, y(0) = 2, y'(0)

= 4.

For y"+y'

= 0, y = erx⇒r2+r = r(r+ 1) = 0 ⇒y 1 = 1 and Y2 = e- x , with W(x) = _: x = -e-x and J(x) = 1. Then --x I 0 1 1 u 1 = f ldx = x, u 2 = J-ex dx = -ex , and YP = x - l, or YP = x. The general solution is y = x+ c1 + c2e-x. y (0) = 2 ⇒c1 + c2 = 2, y'(0) = 4 ⇒1 - C2 = 4 ⇒ c2 = -3, C1 = 5, and y = x+ 5 - 3e-x .

367

C.3. CHAPTER 3 SOLUTIONS

21. y" + 9y = 6 s in( 3x). For y" + 9y = 0, y = erx =} r2 + 9 = 0 =} y1 = cos( 3x) and cos( 3x) sin( 3x) = 3 and J(x) = 6 sm( 3x). y2 = sm( 3x), with W(x) = _ 3 s in(3x) 3 cos( 3x) Then u1 = - J 2 sin2( 3x) dx = - J l - cos(6x) dx = -x + ¾ sin( 6x), u2 = J2 sin( 3x) cos( 3x) dx = J sin( 6x) dx = cos(6x), and YP = [-x + ¼ sin( 6x]) cos( 3x) - ¼ cos( 6x) sin( 3x) = -x cos(3x) + ¼ sin( 3x), or YP = -x cos( 3x). The general solution isy = -x cos( 3x) +c1 cos(3x) +c2 sin(3x).

.

.

I

.

I



22. y" + y = sec(x). For y" + y = 0, y = erx ⇒ r2 + 1 = 0 ⇒ Y1 = cos(x) and I c?s( x) sm(x) I = 1 and J(x) = sec( x). Then y2 = sin( x), with W(x) = COS( X ) -Sm ( X ) u1 = - J tan( x) dx = - lnI sec(x)I, u2 = J l dx = x, and YP = -cos(x) InI sec( x)I + xsin( x). The general solution is y = xsin( x) - cos(x) InI sec( x)I + c1 cos(x) + c2 sin( x).

23. y"+6y'+l3y = l6e- 3 x cos( 2x). For y"+6y'+l3y = 0, y = erx =} r2 +6r+13 = 0 ⇒ r =-3 ± 2i ⇒ Y1 = e- 3x cos(2x) and Y2 = e- 3x sin(2x), with W(x) = e- 3x sin( 2x) e- 3x cos( 2x) = 2e-6x and . . . 3x 3x 3x 3x I cos -3ecos 2x) 2x) + 2e-3esm 2x) ( 2x) 2esm ( ( ( I J(x) = l6e-3 x cos( 2x). Then u1 = -8 J sin( 2x) cos(2x) dx =-4 J sin( 4x) dx = cos( 4x), u2 = 8 J cos2(2x) dx = 4f l + cos( 4x) dx = 4x + sin( 4x), and YP = cos( 4x)e- 3 x cos(2x) + [4x + sin( 4x)Je- 3x sin( 2x) = 4xe- 3x sin(2x) + e-3x cos( 2x), or YP = 4xe-3x sin( 2x). The general solution is y = 4xe- 3x sin( 2x) + c 1 e- 3x cos( 2x) + c2e-3x sin( 2x). 2 rx 24. y" - 4y' + l3y = �. cos(3x) For y"- 4y' + l3y = 0 ' y = e =} r - 4r + 13 = 0 =} r = 2 ± 3i ⇒ Y1 = e2x cos( 3x) and Y2 = e2x sin( 3x), with e2x cos(3x) e2x sin( 3x) W(x ) _ , 2x - 3 e4x and 2e cos(3x)- 3e2x s in( 3x) 2e2x sin( 3x) + 3e2x cos(3x) I J(x) = c!se(;x )· Then U1 = -3J tan(3x)dx =- lnlsec(3x)I, U2 = 3f ldx = 3x, and YP = -e2x cos( 3x) lnI sec( 3x)I + 3xe 2x sin( 3x). The general solution is y = 3xe2x sin(3x) - e2x cos( 3x) lnI sec(3x)I + c 1e2x cos( 3x) + c2e2x sin(3x).

25. y"- 2y' + y = ex ln( x), x > 0. For y"- 2y' + y = 0, y = erx r2- 2r + 1 = (r- 1) 2 = 0 ⇒ Y1 = ex and Y2 = xex, with W(x)

= I ee:

xex / = e2x and f(x) e +xex 2 x ln( x) dx = -; ln( x) + x:, u2 x

= ex ln(x). Then

=}

u1 = - J = J ln( x) dx = x ln( x)- x, and 2 YP = [ x: - x2 ln( x)] ex + [xln( x) - x]xex = ex [½x2 ln( x)-¾x2]. The general solution is y = ex [½x2 ln(x)-¾x2] + c1ex + c2 xex .

368 26.

APPENDIX C. SOLUTIONS " + 3y' + 2y = sin(ex ). For y" + 3y' + 2y = 0, y = e rx ⇒ r2 + 3r + 2 = (r + l)(r + 2) = 0::::} y1 = e -x and Y2 = e - 2x , with e e =: =:: I = -e -3x and f(x) = sin(ex ). Then W(x) = I -e - 2e U1 = Jex sin(ex ) dx and u2 = - J e 2x sin(ex )dx. The substitution u = ex , :'. = ex , then gives u1 = Jsin(u)du = - cos(u) = - cos(ex ) and, by integration by parts, u2 = - Ju sin(u)du = ucos(u) - sin(u) = ex cos(ex ) - sin(ex ). Then 2x x x x x x YP = - e - cos(e ) + e - [ e cos(ex ) - sin(e )] = - e -2x sin(e ), and the general solution is y = -e-2x sin(ex ) + c 1 e-x + c2 e-2x . y

27. x2 y" - 6y = l0x5. For x2 y " - 6y = 0 and x > 0, y = xr =} r2 - r - 6 = (r - 3)(r + 2) = 0 ⇒ y1 = x3 and y2 = x-2 . These solutions are defined for all x =I= 0, with

I = -5 and J(x) = l0x3. Then u1 = 2Jxdx = x2, 3;: _2;=: 5 7 7 and yP = x2 x3 - �x x-2 = Qx u2 = -2 Jx6 dx = -�x 7 ' 7 7 . The general solution is y = �x5 + C1X 3 + c2x- 2 for X =I= 0. W(x)

= /

28. x2 y" - 4xy' + 6y = x4 ex . For x2 y " - 4xy' + 6y = 0 and x

= xr ⇒

= (r - 2)(r -3) = 0 ⇒ y1 = x2 and 2 = x3 , defined for all real x, with W(x) = I ;; 3;: I = x4 and f(x) = x2 x . Then U1 = - Jx x dx = x - x x , u2 = J x dx = x , and 2 2 3 YP = ( x - xex)x2 + x3 x = x2 x . The general solution is = x x + c 1 x + c2x > 0,

y

r2 - 5r + 6

y

e

e

e

e

e

for all real x.

e

e

e

y

e

e

29. x2 y " + 5xy' + 3y = 4xex , y (l) = e , y'(l) = e . For x2 y" + 5xy'+3y = 0 and x > 0, y = xr ⇒ r2 +4r+3 = (r+ l)(r+3) = 0 ⇒ x x y1 = x-1 and y2 = x-3, defined for all x f. 0, with W(x) = I =: I = =: -X -3X 2

-2x-5 and f(x) = 4 e; • Then u1 = 2Jxex dx = ex and U2 = -2Jx3 ex dx. Integration by parts gives 2 2 2 2 2 u2 = - Jx2 · 2xex dx = -x2 ex + J2xex dx = -x2 ex + ex . Hence, 1 + ex \1 - x2 )x-3 = x-3ex 2 . The general solution for x f. 0 is x2 YP = e x2 = x-3 ex +c1x-1+c2 x- 3. y(l) = e ⇒ C 1 +c2 = 0 ⇒ y = x-3ex2 +c1(x-1 -x- 3) y 4ex2+2x-2 ex2 +c (-x- 2+3x- 4), and y'(l) = e ⇒ c1 = e ⇒ c2 = -e ⇒ y' = -3x1 2 ⇒ y = x-3 ex + e(x_, - x-3 ). 2

30. x2 y" - 2xy' + 2y = x3 cos(x). For x2 y" - 2xy ' + 2y = 0 and x

2

2

2

= (r - l)(r - 2) = 0 ⇒ y1 = x and 2 = x2 , defined for all real x, with W(x) = I I I = x2 and f(x) = xcos(x). Then u 1 = - Jxcos(x)dx = -x sin(x) - cos(x), u2 = y

> 0,

y

= xr

=}

r2 - 3r + 2

;:

C.3. CHAPTER 3 SOLUTIONS

369

J cos(x) dx = sin(x), and Y

= -[ xsin(x) +cos(x)]x +[sin(x)]x2 = -xcos(x).

The general solution is y = c 1 x + c 2 x2 - x cos(x) for all real x. P

31 . x 2 y" +xy' - y = x2 sin(x). For x2y" +xy' - y = 0 and x

> 0, y = xr



and y2 = x- 1 , defined for all x =/: 0, with W(x) =

½J



_:=:

r2 - 1 = 0 ⇒ r = ±l

I�

I

-½ J

⇒ Y1

=x

= -2x-1 and

f(x) = sin(x). Then u 1 = x2 sin(x) dx = sin(x) dx = cos(x), u 2 = 2 2 ½x cos(x) - x cos(x) dx = ½x cos(x) - x sin(x) - cos(x ), and x . YP = cos(x) +x- 1 [½x2 cos(x) - sin(x) - cos(x)] = - sin(x) - co� ) The general solution is y = - sin(x) - co�x ) + c1 x + c 2 x-1 for x =/: 0.

-½x

J

x

32. x2 y" + 3xy' + y = ln� ), X > 0. For x2 y" +3xy' +y = 0, y = xr

r 2 +2r + 1 = (r + 1 )2 = 0 ⇒ Y1 = x- 1 and 1 -3 x-11n(x) . W ( x) = I x-_ 2 _ 2 I = x and Y 2 = x-1 ln ( x) , with _ 2 -x -x 1 n(x ) +x 2 x 1 2 11; ) dx = ½[ln(x)] , f(x) = 1,�C:). Then U1 = - [ln�)] dx = -½[ln(x)]3 , u 2 = 1 1 3 1 3 3 and YP = -½ x- [ln(x)] +½x- [ln(x)] = ½x- [ln(x)] . The general solution for x > 0 is y = ¼x-1[ln(x)]3 + c 1 x- 1 + c 2 x-1 ln(x). x



J

J

33. x2 y" - 3xy' +4y = x2 ln(x), x > 0. For x2 y" - 3xy' +4y = 0, y = xr ⇒ r2 - 4r +4 = (r - 2)2 = 0 ⇒ y 1 = x2 and x2 x2 ln(x) I = x3 and f(x) = ln(x). Then Y 2 = x2 ln(x), with W(x) = I 2x 2xln(x) +x 1 x 2 U 1 = - [In�)]2 dx = -½[ln(x)]3 , u 2 = 11; ) dx = ½[ln(x)] , and yP = -½ x2[ln(x)]3 +½x2[ln(x)]3 = ½x2[ln(x)]3 . The general solution for x > 0 is y = ¼x2[ln(x)]3 + c1 x2 + c 2 x2 ln(x).

J

J

34. x2y" - 3xy' + l3y = 9x2 . For x 2 y" - 3xy' +l3y = 0 and x > 0, y = x,. ⇒ r2 - 4r +13 = 0 ⇒ r = 2 ± 3i ⇒ Yi = x2 cos[3ln(x)] and Y 2 = x2 sin[3ln(x)], with W(x) = x2 cos[3ln(x)] x2 sin[3ln(x)] I - 3x ' and 2xsin[3ln(x)] +3xcos[3ln(x)] _ I 2xcos[3ln(x)] - 3xsin[3ln(x)] 3 n 3 dx = cos[3ln(x)], J(x) = 9. Then U1 = -3 si [ !n(x)] 3 n u 2 = 3f cos[ � (x)] dx = sin[3ln(x)], and YP = x2 cos 2[3ln(x)] +x2 sin 2[3ln(x)] = x2 . Since yP is a particular solution of the nonhomogeneous equation for any real x, the general solution for x =/: 0 is y = x2[1 +c1 cos(3ln lxl)+c 2 sin(3ln lxl)].

J

35. x2y" + 7xy' + l3y = X3 COS212111 (X )l, x > 0. For x2 y" +7xy' + l3y = 0, y = x1· ⇒ r2 +6r + 13 = 0 ⇒ r = -3 ± 2i ⇒ Yi = x- 3 cos[2ln(x)] and Y 2 = x- 3 sin[2ln(x)], with W(x) = x- 3 cos[2ln(x)] x- 3 sin[2ln(x)] I -3x- 4 cos[2ln(x)] - 2x- 4 sin[2ln(x)] -3x- 4 sin[2ln(x)] +2x- 4 cos[2ln(x)] I

370

APPENDIX C. SOLUTIONS = 2x- 7 and f(x) = xs cos212111 x ]' The substitution u = 2ln(x), �� = �, then ( ) gives s in( u) l_ 1 __ u1 = -2J x scoin[s22[12nln(x(x)] ] dx - -J cos 2 (u) du cos(u) - cos[ 2 ln(x)]' ) U 2 = 2 x cos[i ln(x) ] dx = sec(u) du= ln I sec(u) +tan(u)I = ln I sec[2ln(x)] +tan[2ln(x)]I, and 3 3 YP = -x- + x- sin[2ln(x)]ln I sec[2ln(x)] +tan[2ln(x)]I. The general solution for x > 0 is y = x- 3 {sin[2ln(x)]ln I sec[2ln(x)]+tan[2ln(x)] l-1+c1 cos[2ln(x)]+c 2 sin[2ln(x)]} .

J

J

36. x2 y" - 5xy' + l3y =8x3 sin[2ln(x)], x > 0. For x 2 y" - 5xy' + l3y = 0, y = xr ⇒ r 2 - 6r +13 = 0 ⇒ r = 3 ± 2i ⇒ 3 3 Y 1 =x cos[2ln(x)] and y 2 =x sin[2ln(x)], with 3 x3 sin[2ln(x)] x cos[2ln(x)] x ( 2 2 2 W )-1 - 3x cos[2ln(x)] - 2x sin[2ln(x)] 3x sin[2ln(x)] + 2x2 cos[2ln(x)] I = 2x5 and f(x) = 8xsin[2ln(x)]. The substitution u = 2ln(x), �� =�'then gives n2 u1 = -4 si [ 2�n(x)] dx = -2 sin 2(u) du= cos(2u) -1 du= ½sin(2u) - u = ½ sin[4ln(x)] - 2ln(x), u2 = 4 J sin[ 2 In (x)]xcos[ 2 In(x)] dx = 2 J sin(u) cos(u) du= J sin(2u) du=-½ cos(2u) = -½ cos[4ln(x)], and 3 3 YP = {½ sin[4ln(x)] - 2ln(x)}x cos[2ln(x)] - ½ cos[4 ln(x)]x sin[2ln(x)] 3 = ½x {sin[4ln(x)]cos[2ln(x)] - cos[4ln(x)]sin[2ln(x)]} - 2x3 ln(x)cos[2ln(x)] = ½x3 sin[2ln(x)] - 2x3 ln(x)cos[2ln(x)], or YP = -2x3 ln(x)cos[2ln(x)]. The general solution is y = x3 { c1 cos[2ln(x)] + c 2 sin[2ln(x)] - 2ln(x) cos[2ln(x)]}, X > 0.

J

J

J

37. 4x 2 y" +4xy' - y = x2 . For 4x 2 y" +4xy' -y = 0 and x > 0, y = xr ⇒ 4r(r -1) + 4r -1 = 4r 2 - 1 = 0 1 2 1 2 ⇒ y1 = xt /2 and Y2 = x- 1 12, with W(x) = I 1 x_/112 1 x-_31/ 2 I = -x-1 and

J

2X

J

-2X

512 J(x) = .4!. · Then ul = .4!. x1 12 dx = l6 x31 2 ' u 2 = _.!.4 x312 dx = -l 10 x ' and 2 YP = /5 x . Since YP is a particular solution of the nonhomogeneous equation for all real x, the general solution for x =/- 0 is y = 1�x2 + c 1 lxl1 / 2 + c 2 lxl-112 .

Exercises 3.4 - page 83 1. /� 2 = �+3x2 and z(x) =y'(x) ⇒ z�2 = �+3x2 ⇒ ln lz- 21 = 21n lxl+x3 +c1 ⇒ z - 2 =k 1 x2 ex3 ⇒ y =f k1x2 ex3 + 2 dx = ½k1ex3 + 2x + k 2 .

2. x2 y" = (y') 2 and z(x) = y'(x) ⇒ x2 z' = z 2 ⇒ ;� = ;2 ⇒ -� = -� + c ⇒ x _ -x -. 1- dx = -"'c _ l.. J l - 1-cx -J. 0. ln 11-cxl +k ' if c 1 z =_ c c2 1-cx =} y = J 1-cx dx = _.!. 2 x If c = 0, then z = X =} y = 2 + k 1.

C.3. CHAPTER 3 SOLUTIONS

371

3. y" -x(y')3 =0 and z(x) = y'(x) ⇒ z' -xz3 = 0 ⇒ z- 3 z' =x ⇒ -½z- 2 = ½x2 +c ⇒ z2 = k �x2 ⇒ z = ✓:�x2 and is real if and only if k = µ2 > 0. Then

(.=£)

1 x = µsin(t) ⇒ y = ±J .� µ,2-x2 dx = ±f ldt = ±l+c1 = ±sin - µ, +c1.

4. y'y" + x =0 and z(x) = y'(x) ⇒ zz' + x = 0 ⇒ ½z2 = -½x2 + c ⇒ z = ±Jk - x2 and is real if and only if k = µ2 > 0. Then x = µsin(t) y = ±J µ2 - x2 dx = ±µ2 J cos 2(t)dl = ±� J 1 + cos (2t)dl = ¥ [t+ ½ sin(2t)] +c1 = ¥[t+sin(l) cos(t)] +c1 =

J

± � [sin - 1



(!) + x7] + c

1.

5. xy" = y'ln(y') -y'ln(x) and z(x) = y'(x) ⇒ xz' =zln(z) - zln(x) ⇒ z' = �ln( �), which is homogeneous, and u = � ⇒ u + xu' = u ln( u) ⇒ xu' = u[ln(u) - 1] :::} u[ln(�)-l] =;; :::} ln I ln(u) - 11 =ln(x) +c:::} ln(u) -1 = kx ⇒ u = e kx+l ⇒ z =xekx+1, and k = ±ec -=J 0. Hence, Y =J xekx+ l dx = lk xekx+ l - l.... ekx+ l + kl · k2 6. y" + 2y' = 2e-x # and z(x) = y'(x) ⇒ z' +2z = 2e-xy'z, which is a Bernoulli equation with a=½- Then u = z1l2 ⇒ z = u2 ⇒ z' = 2uu' ⇒ 2uu' + 2u2 = 2e- x u ⇒ u' +u = e- x , which is linear, with the integrating factor I(x) = ex . Then (ex u)' = 1:::} ex u = x+c ⇒ u = (x+c)e-x ⇒ z =(x+c)2e- 2x ⇒ y =J(x +c)2e-2xdx = -½(x+c)2e-2x + f(x +c)e-2xdx = -½(x +c)2e- 2x - ½(x+c)e- 2x + ½ J e- 2x dx = -½(x+c)2e- 2x - ½(x+c)e- 2x - ¼e- 2x + k.

7. x4 +(y')2 - xy'y" =0 and z(x) = y'(x) ⇒ x4 + z2 - xzz' = 0, with x) 3 I = _;!x = I'( P(x ' z) = x4 + z2 ' Q(x ' z) = -xz ' Pz-Qx Q I(x) :::} (x) = x- is an integrating factor. Then x + x- 3 z2 - x- 2 zz' = 0 is exact. fx = x + x-3 z2 ⇒ J(x, z) =½x2 - ½x- 2 z2 +g(z), and fz = -x- 2 z ⇒ f(x, z) = ½x2 - ½x-2z2 + c and ½x2 - ½x- 2 z2 =c1, or x2 - x- 2 z2 =k, and z = ±xJx2 - k. Then y = ±J xJx2 - kdx = ±½(x2 - k)31 2 + k1 . 8. y" = (2x - y' + 3 )2 + 2 and z(x) = y'(x) ⇒ z' = (2x -z + 3 )2 + 2. Let u = 2x - z + 3. Then u' = 2 - z' = 2 -(u2 + 2 ) = -u2 , which is separable. 1 1 1 :::} 2x - z+ 3 = -+ :::} z= 2x+ 3 - -+ Then - Uu� = 1 :::} Ul =x+c ⇒ u = -+ X C X C X C ⇒ y =J 2x+3 - x�c dx = x2 + 3x - ln Ix + cl +k. 9. y" +(2 -6y +12y2) (y')3 = 0 and v(y) = y'(x) ⇒ vv' +(2 - 6y + 12y2 )v3 = 0 . If v = 0, then y = c. If v -=J 0, then - �� = 2 -6y +12y 2 ⇒ � = 2y-3y2 +4y3 +c1 :::} (2y - 3y2 + 4y3 + c1)y' = 1:::} y2 - y3 +y4 + C1Y = X + C2. 10. y" = (y')2 tan(y) and v(y) = y'(x) ⇒ vv' = v2 tan(y). Ifv = 0, then y = c. If v -=J 0, then�= tan(y) ⇒ ln lvl = ln I sec(y)l+c1 ⇒ v = ksec(y) ⇒ cos(y)y' = k

372

APPENDIX C. SOLUTIONS

11. yy" =y'(y' -1) ln(y' -1) and v(y) =y'(x) ⇒ yvv' = v(v-l) ln(v-1), with v-/= 0 since ln(-1) is undefined. Then (v-l)f�(v-l) =t ⇒ ln I ln(v -1)1 = ln IYI+ c ⇒ 'y =l:::::} e kYy' = land k= ±ec ...J. 0 y l:::::} kY ln(v - 1) =ky ⇒ v = ek + l+ckY r · l e + + k y k k + Then -¾ ln(l +e- ) =x+k1 :::::} 1+e- y =e (x ki) :::::} -ky =ln[e- k(x ki ) -1] + :::::} y = -¾ ln[e- k(x k1 ) - 1]. 12. sin(y)y" + cos(y)(y') 2 = (y')3 and v(y) = y'(x) ⇒ sin(y)vv' + cos(y)v2 = v3. If v = 0, then y = c. If v -/= 0, then sin(y)v' + cos(y)v = v2, which is a Bernoulli equation with a = 2. Then u = v- 1 ⇒ v =u- 1 ⇒ v' = -u- 2 u' ⇒ - sin(y)u-2 u1 + cos(y)u-1 = u-2 , or -sin(y)u' + cos(y)u = 1, which is linear. The standard form is u'-cot(y)u= - csc(y), with the integrating factor y)I =lcsc(y) , and with I(y) = csc(y), the equation y y =e1n lc sc( J(y)= e-fcot()d I becomes csc(y)u' -csc(y) cot(y)u =- csc2 (y), i.e., [csc(y)u]'= - csc 2 (y). Then csc(y)u = cot(y)+ k ⇒ u= cos(y)+ ksin(y):::::} v= s )ksin (y) co (y) [cos(y)+ ksin(y)]y'= 1 ⇒ sin(y) - kcos(y)= x + k1 .

*

13. y" =2yy' and v(y) = y'(x) ⇒ vv' =2yv. If v =0, then y = c. If v -/= 0, then v' = 2y ⇒ v = y2 + k ⇒ /�k = l⇒ Y2�k dy = x + c1. If k =0, then _.!.y = x + c1 ⇒ y = X-+1C1 . If k =µ2 > 0, then y =µ tan(t) ⇒ 1 lµ dl = lµ = .l tan-1 µ ⇒ y = µtan[µ(x+c1)]. If k = -µ2 < 0, 2 + 2 dy = µ y µ 1 ln v-µ I = -1- - _1 dy = 1 (ln I 1 then dy =_ I +µ Y- µI -ln IY+µI)= _ +y µ _ 2µ 2µ J y µ 2µ y y -µ e 2µx 2 µx 2 µx :::::} y(l - k1 e )= µ(k1 e +1) :::::} Y= µ�1__ :;:2t;. X + C1 :::::} � = k1 e

J

J

J�

(u)

J

1 14. y" - (y') 2 = 1 and z(x) = y'(x) ⇒ z' - z2 =l⇒ l�1 = 1 ⇒ tan- (z) = x+ c ⇒ z =tan(x+ c) ⇒ y = tan(x+ c) dx =ln I sec(x+ c)I + k.

J

15. v1f+Tl y" = land z(x) = y'(x) ⇒ v'1+z z' =l⇒ �(1 + z)31 2 =x + c 2 3 (1+ z)3 1 2 = �x+ c1 ⇒ z= (�x+ c1) 1 - 1



⇒ y =J (�x+ c1) 213 - 1 dx= � (�x+ c1) 513 - x+ c2.

16. e Y'y" =land z(x) = y'(x) ⇒ e z z' = l ⇒ e z =x+c ⇒ z = ln(x+ c) ⇒ y = ln(x+ c) dx. Let u =x+ c and integrate by parts to obtain l- ln(u) du= u ln(u) - u+ c1. Hence, y =(x+ c) ln(x+ c) - ln(x+ c)+ C1-

J

J

17. cos(y')y" =1 and z(x) =y'(x) ⇒ cos(z)z' = 1 ⇒ sin(z) = x+ c ⇒ z= sin-1(x+ c) ⇒ y = sin- 1 (x+ c) dx. Let u = x+ c and integrate by parts to obtain l -sin-1(u) du= u sin- 1 (u)- �du= u sin- 1 (u)+✓1 - u2+c 1.

J

J

J

Hence, y = (x+ c) sin- (x+ c)+ J1 - (x+ c)2 + c 1. 1

C.3. CHAPTER 3 SOLUTIONS

373

18. y" = y'jl - (y' ) 2 and v(y) = y'(x) => vv' = vvT"=v2. If v = 0, then y = c. If v =/= 0, then ✓iv�v2 = 1 => sin- 1(v) = y+c1 => v = sin(y+c1) => csc(y+c1)y' = 1 => - ln I csc(y + c,) + cot(y + c1)1 = x +k => csc(y + c1) +cot(y + c1) = k 1 e- . x

Chapter 3 Exercises - page 84 1. The standard form of the equation is y" + ¼Y' +(1 - 4;2 )y = 0, with p(x) = ¼ J ¾ dx = co => u'(x) = �e-4-- X Xl = sec 2(x) and Y1 = cosr) yX Y1 s c) 0 s ) 2 u u = y (x) dx = tan(x) => y => = sec 1 = tan(xt 1x) = i� . 2

J

2 . y" +xy' +y = 0, p(x) = x, Y1 = e-x /2 ⇒ u'(x) = -tr e -f x dx 2 2 2 =? U = J e ! dx =? Y2 = UY1 = e - x / J ex / dx. 2

x

2

2

2

=

e

x

2 e

_ x2 /

2

=e

x

2

/

2

1 3. The standard form of the equation is y" + (1 + 2x )y' - 2; y = 0, with p(x) = 1 / -x dx l+t;, J � u => x) x e ( ' = e = and 1 + .1... = xe 2xe - x -½ ln(x) = x1 / e => Y1 � � u = J x1 / 2ex dx => Y2 = uy1 = x- 1 / e -x J x1 / e dx. 2

2

2

2

2

x

x

4. y" + y = cos(x). For y" + y = 0, y1 = cos(x) and Y

2

= sin(x).

(a ) YP = x[Acos(x) + Bsin(x)] => y; = [Acos(x) + B sin(x)] + x[-A sin(x) + B cos(x)], y� = [-2Asin(x) +2Bcos(x)] + x[-Acos(x) - Bsin(x)], 1 y� + YP = -2Asin(x) + 2Bcos(x) = cos(x) => A = 0 and B 2* YP = ½xsin(x) => y = ½xsin(x) + c1 cos(x) + c2 sin(x). (b) W(x) = 1, f(x) = cos(x), J sin(2x) dx = ¼cos(2x), u 1 = - J sin(x)cos(x) dx = 2 u = cos (x) dx = ½ 1 +cos(2x) dx = ½x + ¼sin(2x), YP = ¼cos(2x)cos(x) + [½x + ¼sin(2x)]sin(x) = ½xsin(x) + ¼cos(x), or YP = ½xsin(x) => y = ½xsin(x) + c1 cos(x) + c2 sin(x). 2

J

J



5. y" - y' - 2y = 8x2 . For y" - y' - 2y = 0, Y1

=

e

-x and Y

2

=

e

2x

.

374

APPENDIX C. SOLUTIONS

6. v" -V = cos(x) - 5sin(2x). For v" - v = 0, Vi= ex and v2 = e-x _

(a) For v" - v = cos(x), VPi =Acos(x)+Bsin(x), vi1 =-Acos(x)- Bsin(x), vi1 - VPi = -2Acos(x) - 2Bsin(x) = cos(x) =} A = -½, B = O ⇒ Vp 1 =-½cos(x). For v" - v =-5sin(2x), Vp2 = Acos(2x) + Bsin(2x), vi2 =-4Acos(2x) - 4B sin(2x), vi2 - Vp2 = -5Acos(2x) - 5B sin(2x) =-5sin(2x) ⇒ A = 0, B = l ⇒ Vp2 =sin(2x) =}VP -½cos(x) +sin(2x) =} V = -½cos(x)+ sin(2x) +c 1 ex + c2 e-x . =

(b)

7.

W(x) = -2, f(x) =cos(x) - 5sin(2x), u1 =½Je-x [cos(x) - 5sin(2x)] dx =½Je-x cos(x) dx- � Je-x sin(2x) dx, u2 = -½Jex [cos(x) - 5sin(2x)] dx =-½ Jex cos(x) dx+� Jex sin(2x) dx. J e-xcos(x) dx = -e-x cos(x) - J e-x sin(x) dx =-e-x cos(x) + e-x sin(x) - J e-xcos(x) x x =} J e- cos(x) dx=½e- [sin(x) -cos(x)], Jexcos(x) dx =ex cos(x) +Jex sin(x) dx = ex cos(x) + ex sin(x) -Jex cos(x) =}Jex cos(x) dx =½ex [cos(x)+sin(x)], Je-x sin(2x) dx =-e- x sin(2x) +2Je- x cos(2x) dx = -e-x sin(2x) - 2e-x cos(2x) - 4 Je-x sin(2x) x x =} J e- sin(2x) dx =-ge- [sin(2x) + 2cos(2x)], x sin(2x) dx Je = ex sin(2x) - 2Jex cos(2x) dx=ex sin(2x) - 2ex cos(2x) - 4 Jex sin(2x) x x =} Je sin(2x) dx = ge [sin(2x) - 2cos(2x)], u1 = ¼e-x [sin(x) -cos(x)] +½e- x [sin(2x) + 2cos(2x)], u2 = -¼ex [cos(x) +sin(x)]+ ½ex [sin(2x) - 2cos(2x)], VP=-½cos(x) +sin(2x), and v=-½cos(x) + sin(2x) + c1ex + c2e-x .

v" - 2v' - Sv = 24(x2 + l)e4x .

For v" - 2v' - 8v = 0, V1=e- 2x and V2 = e4x .

(a) VP =x(A +Bx+Cx2 )e4x=(Ax+ Bx2 + Cx3 )e4x , v� =[ A+ (4A+ 2 B)x+ (4B + 3C)x2 +4Cx3]e4x , vi= [(SA+2B) +(16A +16B+6C)x +(l6B +24C)x2 + l6Cx3]e4x, vi - 2v; - 8vP= [(6A+ 2B) + (12B+ 6C)x+ 18Cx2 ]e4x = 24(x2 +l)e4x 38 =} =} C = :! B =-�, A VP = ( 398x _ �3 x2 + :!3 x3 ) e4x 3' 3 9 4x 3 2 =} V = (3 8X - �X + 1x ) e + C1e-2x +C2 e4x _ 9 =

(b) W(x)=l : �: ::: 1=6e 2x ,u1 = -4f(x2 +l)e6x dx -2 4 =-�(x2 + l)e6x + Jxe6x dx = -�(x2 + l)e6x + ixe6x - Je6x dx 19 ) e6x _e6x = -�3 x2 + �9 x _ 27 =-�3 (x2 + l)e6x + �9 xe6x _ _!27 ' t 2 2 3 u2 =4Jx + 1 dx = 1x + 4x, VP = (- �x + ix - ��) + ( 1x3 +4x)] e4x,

1

i

375

C.3. CHAPTER 3 SOLUTIONS

8.

(a) a(x - x0) 2 y" + b(x - x0)y' + cy = 0, l = x - x0 and y(x) = z(t) ⇒ rJ:1!. dz dl dz i!JJ. d2 z dt d2 z. dx = dl dx = dl and dx2 = dl 2 dx = dt 2 Hence ' the equation for y(x) transforms into the Euler equation ai 2 z" + btz' + cz = 0 for z(t).

(b) t = x - 3 and y(x) = z(t) transforms 2(x - 3)2 y" + 3(x - 3)y' + y = 0 into 2l 2 z"+3tz'+z

r = -l�iv'7 =

fort -I- 0 ⇒ y for X-=/- 3. 9.

= 0.

-¼ ±i ./} ⇒ z = =

⇒ 2r(r-1)+3r+l = 2r2 +r+l = 0 ⇒ ltl-1/4 [C 1 cos ( ./} ln ILi) + c2 sin ( ./} ln ltl)]

For t> 0, z

= tr

Ix - 31- 1 /4 [c 1 cos ( ./} ln Ix - 31) + c2 sin ( ./} ln Ix - 31)]

�1: �1 ;,

= (a) ax2 y" +bxy' +cy = 0, x> 0, l = ln(x) and y(x) = z(l) ⇒ � = 2 �:¾ = �:�(;) + �1(�}). Hence, the Euler equation for y(x) transforms into az" + (b - a)z' + cz = 0, which has constant coefficients. (b) z = erl ⇒ ar2 + (b - a)r + c = 0, and y = xr ⇒ ar(r - 1) + br + c = 0 ⇒ + (b - a)r + c = 0. r e = er ln(x) = e'n(x ) = xr ⇒ z = C 1 er 1 t + C2er2 t = C 1 Xr1 + C2 Xr2 = Y, z = ert (c1 + c2 t) = xr [c 1 + c2 ln(x)] = y, and z = eat [c1 cos(,Bl) + c2 sin(,Bt)] = x0 { c1 cos[,8 ln(x)] + C2 sin[,8 ln(x)]} = y. ar2

(c)

10. x2 y11

rt

-

xy' + y

=

l n(x) ,

X>

0.

(a) For x2 y" - xy' + y = 0, y 1 = x and Y2 = x ln(x). ( W(x) = I � l:(�) �l / = x,J(x) =x ,;(x)' u 1 = -J;dx = -ln(x), u 2 = J xl;(x) dx = ln I ln(x)I, YP = -x ln(x) + x ln(x) ln I ln(x)I, or YP = x ln(x) ln I ln(x)I ⇒ y = x ln(x) ln I ln(x)I + c1x + C2 X ln(x).

(b)

(c)

= ln(x) and



= z(l) ⇒ z" - 2z' + z = For z" - 2z' + z = 0, z = ert ⇒ r2 - 2r + 1 = (r - 1)2 = 0 ⇒ z1 = et e Let and z2 = leL . W(t) = I : t I = e2t , J(t) = u 1 = - J 1 dt = -t, e e + teL t

y(x)

u 2 = Ji dt = ln ltl, zp = -tet + tet ln Ill, or Zp ⇒ z = let ln ltl + c1et + c2teL .

(d) y(x)

t,

= tet ln ltl

= z(l) = z(ln(x)) = x ln(x) ln I ln(x)I + c 1 x + c2x ln(x),

as in part (a).

11. 4x2 y" + 4xy' - y = x2lxl 1 1 2.

(a) For 4x2 y" + 4xy' - y = 0, x > 0, y = xr ⇒ 4r(r - 1) + 4r - 1 = O � 4r 2 - 1 = 0 ⇒ Y1 = x 1 /2 and y2 = x-1/2 . I x� 1 /2 x- 1 /32 I = -x-1, (x) = x�12' W(x) = J - 1 /2 - x- /2 ½ ½

376

APPENDIX C. SOLUTIONS 2 -4l X 2dX - - x1 2 , U1 - .4!. Xd X - x8, U2 2 3 5 2 5 2 YP = x8 x l /2 _ x12 X-1/2 = x241 ' y = x241 + C 1 xl/2 + C2 X- 1/2 ·

J

J

3

12. y" + y'=2x.

(a) For y" + y' = 0, Y1 = 1 and Y2 = e-x . YP = x (A + Bx) = Ax + Bx2, y� = A+2Bx, vi=2B, vi+y� = (A+2B)+2Bx = 2x ⇒ B = 1, A=-2 ⇒ YP =-2x + x2 ⇒ y=x2 - 2x + C1 + c2e-x .

(b) (c)

e-x x 2 0 -e-x = -e- , J(x) = 2x, u1 = 2J xdx = x , U2 = -2 xexdx = -2xex +2ex , Yp = x2 + (-2xex +2ex )e-x = x2 -2x+2, or YP = x2 - 2x, and y=x2 - 2x + c1 + c2e-x. W(x) =

1

J

1

I

z(x) = y'(x) =} z' + z = 2x, J(x) = ef ldx =ex =} (exz)' = 2xex ⇒ exz = 2 xexdx = 2xex - 2ex + c ⇒ z = 2x - 2 + ce-x ⇒ y = x2 - 2x - ce-x + k.

J

13. x2y" + xy' = 1, x > 0. (a) For x2y" + xy'

(b)

=

0, Y 1

J

=

1 and Y2

= ln(x).

W(x)

J

=

I�

ln ) t

I

J(x) = }2 , U1 = - Ini ) dx = -½[ln(x)]2, u 2 = ¾ dx = ln(x), YP = -½[ln(x)] 2 + [ln(x)] 2 = ½[ln(x)] 2 =} y = ½[ln(x)] 2 + c 1 + c2ln(x). x

=

¾,

z(x) = y'(x) =} x2z'+xz = 1 =} z'+¾z = ;2 , I(x) = ef½dx = eln(x) = x =} x (xz)' = ¾ =} xz = ln(x) + c =} z = Ini ) + � =} y = ½[ln(x)] 2 + cln(x) + k.

14. 3y'y" = 1.

(a) z(x) = y'(x) ⇒ 3zz' = 1 =} �z2 = x+c1 =} z2 1 2 3 2 ⇒ y = J (�x + c2) 1 dx = (�x + c2) 1 + c.

(b) v(y) = y'(x) ⇒ 3v2v' = 1 ⇒ v3



� (y + k 1 )2 l3 =x + k2

= y+k1

= �x+c2 ⇒ z =

(�x + c2 )

⇒ v = (y+k1)1 l 3 ⇒ (y+k� ) 113

⇒ y= (�x + k3)3

12

- k1.

112

=

1

377

C.4. CHAPTER 4 SOLUTIONS 15.

+ p(x)y� + q(x)y2 = 0 and y� + p(x)y� + q(x)y1 = 0 ==> yi[y� + p(x)y� + q(x)y2] = 0 and Y2 [Y� + p(x)y� + q(x)y 1] = 0 ==> (Y1Y� - y�y2 ) + p(x)(Y1Y� - Y�Y2 ) = 0 ==> W'(x) + p(x)W(x) = 0. (b) I(x) = efp(x)dx ⇒ [efp(x)dxw(x)]' = 0 ⇒ efp(x)dxw(x) = c ==> W(x) = ce-fp(x)dx_ (a)

y�

=

(c) If c = 0, then W (x) 0 on I, and if c-/- 0, then W(x)-/- 0 for any x in I since e-fp(x)dx > 0 for any continuous function p(x).

(d) Suppose that Y1 and Y2 are linearly dependent on I. Then there exist constants c1 and c2 , not both 0, such that c1y1 + c2 y 2 = 0 on I. If c1 -/- 0, then Y1 = - :Y2 ==> y� = -:y� ==> W (x) = Y1Y� - Y�Y2 = 0. Similarly

=

if c2 -/- 0. Conversely, if W(x) = y1y� - y�y2 0 on I, then Y2 Y2 = Yi ==> YI ln IY2I = ln IY1I + c ==> Y2 = ky1 , i.e., Y1 and Y2 are linearly dependent on I.

=

0 on (e) Suppose that Y1 and Y2 are linearly independent on I. If W (x) I, then, by part (d), y 1 and Y2 are linearly dependent on I, which is a contradiction. Hence, W(x) -I- 0 on I. Conversely, suppose that W(x) -I- 0 on I. If y 1 and y 2 are linearly dependent on I, then, by part (d), W(x) 0 on I, which is a contradiction. Hence, y 1 and y 2 are linearly independent on I.

=

16. x2y" - xy'

= 0 on the interval I= (-1, 1).

(a) For x > 0,

(b)

y

= xr

==> r 2 - 2r

= 0 ==>

y1

= 1 and Y2 = x2 , valid for all x.

W ( x) = I � ;; I = 2x.

(c) At x0 = 0, W(0) = 0.

(d) The result of part (c) does not contradict the result of part (e) in Exercise 15 because the standard form of the equation is y" -¾y' = 0, and p(x) = is discontinuous at x = 0, hence not continuous on I.



C.4

Chapter 4 Solutions

Section 4.1 Exercises - page 101 1. y"' - 6y" + lly' - 6y = 0, y = erx ==> P(r) = r3 - 6r 2 + llr - 6 = 0 ==> (r - l)(r2 - 5r + 6) = (r - l)(r - 2)(r - 3) = 0 ==> y = c1ex + c2e2x + C3e3x . 2.

2 - 2y" - y' + 2y = 0, y = erx ==> P(r) = r3 - 2r - r + 2 = 0 ==> 2 (r - l)(r - r - 2) = (r - l)(r + l)(r - 2) = 0 ==> y = c1ex + c2 e- x + c3e2x ⇒ y' = C1ex - C2e-x + 2c3e2x , y" = C1ex + C 2e-x + 4c3e2x , y (0) = 6, y'(0) = 0,

y"'

378

APPENDIX C. SOLUTIONS C1 + C2 + C3 y " (0) = 0 =} { C1 - C 2 + 2c3 C1 + C2 + 4c3 x x = 6e + 2e2e2x . y

3. 9y111 + 9y" - y' - y = 0, (r + 1)(9r2 - 1)

=

= = =

= erx

� } a> c1

= 6,

P(r) = 9r3 (r + 1)(3r -1)(3r + 1) = 0 y

=}

c,

= 2,

c,

=

-2, and

+ 9r2 - r - 1 = 0 =} x x x =} y = c1e- + c 2 e½ + c3 e-½ .

= 0, y = erx =} P(r) = r3 - 5r2 + 8r - 4 = 0 =} = (r - l)(r - 2)2 = 0 =} y = c1ex + c2e2x + c3 xe2x.

4.

- 5y " + 8y ' - 4y (r - l)(r2 - 4r + 4)

5.

- 6y " + 12y' - 8y = 0, y = erx =} P(r) = r3 - 6r2 + 12r - 8 = 0 (r - 2)(r2 - 4r + 4) = (r - 2)3 = 0 =} y = c1e2x + c2 xe2x + C3X2 e2x .

6.

+ y " - y' - y = 0, y = erx =} P(r) = r3 + r2 - r - 1 = 0 =} (r - l)(r2 + 2r + 1) = (r - l)(r + 1)2 = 0 =} y = c1ex + c2 e- x + C3 Xe-x =} x x x x C1ex + 2e x + c3(-2e- x + xe-x), y ' = C1e - C 2 e- + C 3(e- - xe- ) y

y "'

y '"

y "'

� � � ; c: 1 y(0) = 0, y'(0) = 3, y "(0) = 2 =} { C1 - C2 + C3 = 3 } C1 + C 2 - 2c3 = 2 C3 = -1, and y = 2ex - 2e-x - xe-x. 7.

=}

=}

C1

= 2,

- 5y " + 17y' - 13y = 0, y = erx =} P(r) = r3 - 5r 2 + 17r -13 = 0 (r - l)(r 2 - 4r + 13) = 0 ⇒ r 1 = l and r2 , r2 = 4±f36 = 2 ± 3i =} y = c1ex + e2x[c 2 cos(3x) + c3 sin(3x)]. y "'

C2

= -2,

=}

8. y(4) - 5y " + 4 y = 0, y = erx =} P(r) = r4 - 5r2 + 4 = (r2 - l)(r 2 - 4) (r - l)(r + l)(r - 2)(r + 2) = 0 =} y = C1ex + c 2 e- x + C3 e2x + C4e-2x.

=0

10. y(4) _ 7y "' + 17y" -17y' +6y = 0, y = erx =} P(r) = r4 -7r3 +17r2 -17r+6 2 2 =} (r - l)(r - 2) (r - 4r + 3) = (r - 1) (r - 2)(r - 3) = 0 x x 2x 3x =} Y = C1e + C2Xe + C3 e + C4e .

=}

=0

11. y(4) - y "' - 3y " + 5y' - 2y = 0, y = erx =} P(r) = r4 - r3 - 3r2 + 5r - 2 = 0 =} (r-l)(r+2)(r 2 -2r+l) = (r-1)3 (r+2) = 0 =} y = c1ex +c2 xex+c3 x2 ex +c4e-2x .

C.4. CHAPTER 4 SOLUTIONS

379

) - 4y "' + 6y " - 4y ' + y = 0, y = erx ⇒ P(r) = r4 - 4r3 + 6r2 - 4r + 1 = 0. Since r1 = 1 is a root of both P(r) and P'(r) = 4r3 - 12r2 + 12r - 4, (r - 1)2 is a factor of P(r). Hence, P(r) = (r - 1)2 (r2 - 2r + 1) = (r - 1)4 = 0 2 x x x 3 x =} y= C1e + C2Xe + C3X e + C4 X e . 4

12.

y(

13.

y(

14.

y(

15.

y(

16.

y(

17.

y(

18.

y(

) _ 6y111+ 13y" -12y'+4y = 0, y = erx ⇒ P(r) = r4 - 6r3 + 13r2 - 12r+4 ⇒ (r - l)(r - 2)(r2 - 3r + 2)= (r - 1)2(r - 2)2 = 0 2x x x + C4Xe2x . =} Y = C1 e + C2Xe + C3 e 4

=0

) + 2y'" + 4y " - 2y' - 5y = 0, y = erx ⇒ P(r) = r4 + 2r3 + 4r2 - 2r - 5 = 0 ⇒ (r - l)(r + l)(r2 + 2r + 5) = 0 ⇒ r1 = 1, r2 = -1, r3 , r3 = - 2±F15 = -1 ± 2i ⇒ y = c,ex + c2e-x + C3 e-x cos(2x) + C4 e- x sin(2x). 4

) + lOy '" + 41y" + 76y ' + 52y = 0, y = erx =} P(r) = r4 + 10r3 + 41r2 + 76r + 52 = 0. Since r 1 = -2 is a root of both P(r) and P'(r) = 4r3 + 30r2 + 82r + 76, (r + 2)2 is a factor of P(r). Hence, P(r) = (r + 2)2 (r2 + 6r + 13) = 0 ⇒ r1 = -2, r2 , r2 = -B±F15 = -3 ± 2i ⇒ 2x + c2xe-2x + C3 e-3x cos(2x) + C4 e-3x sin(2x). y = c1e4

) + 5y " + 4y = 0, y = erx ⇒ P(r) = r4 + 5r2 + 4 = (r2 + l)(r2 + 4) = 0 ⇒ r1, r1 = ±i, r2 , r2 = ±2i ⇒ y = c 1 cos(x) + c2 sin(x) + C3 cos(2x) + C4 sin(2x). 4

) + 4y = 0, y = erx ⇒ P(r) = r4 + 4 = 0. P(r) = (r2 + ar + 2)(r2 + br + 2) ⇒a= ±2, b = =F2 ⇒ P(r) = (r2 + 2r + 2)(r2 - 2r + 2) = 0 ⇒ r 1 , r 1 = -1 ± i, r2 , r2= 1 ± i ⇒ y= c 1 e-x cos(x) + c2 e-x sin(x) + c3ex cos(x) + C4ex sin(x). 4

-8y '"+26y "-40y '+25y = 0, y = erx ⇒ P(r) = r4 -8r3 +26r2 -40r+25 = 0. P(r) = (r2 +ar+ 5)(r2 +br+ 5) ⇒a= b = -4 ⇒ P(r) = (r2 - 4r + 5)2 = 0 ⇒ r1, r1= 2 ± i ⇒ y= c1e2x cos(x) + c2 e2x sin(x) + c3 xe2x cos(x) + C4Xe2x sin(x). 4l

19. x3y 111 + x2y " - 2xy ' + 2y = 0. For x > 0, y = xr ⇒ P(r) = r(r - 1)(r - 2) + r(r - 1) - 2r + 2 = ( r - 1)[r(r - 2) + r - 2] = (r - l)[r2 - r - 2 ] = (r - l)(r - 2)(r + 1) = 0 ⇒ y = c 1 x + c2x2 + c3 x- 1, x-/= 0. 20. x3y '" - 2x2 y" + 3xy ' - 3y= 0. For x > 0, y= xr ⇒ P(r) = r(r - l)(r - 2) - 2r(r - 1) + 3r - 3 = (r - l)[r(r - 2) - 2r + 3] (r - 1)2(r - 3) = 0 ⇒ y = c1x + c2x ln lxl + c3 x3, x-/= 0.

=

21. x3y 111 + xy ' - y = 0. For x > 0, y = xr ⇒ P(r) = r(r - l)(r - 2) + r - 1 = 0 ⇒ (r - l)[r(r - 2) + 1] = (r - 1)3 = 0 ⇒ y = c 1 x + c2x ln lxl + C3X(ln lxl)2, x-/= 0.

22. x3y111 + 8x2y " + 13xy ' - 13y = 0. For x > 0, y = xr ⇒ P(r) = r(r - l)(r - 2) + 8r(r - 1) + 13r - 13 = (r - l)[r(r - 2) + 8r + 13] = (r - l)(r2 + 6r + 13) = 0 ⇒ r1 = 1 and r2, r2 = - 5 ±F15 = -3 ± 2i ⇒ y = c 1 x + c2 x-3 cos[2 ln(x)] + c3 x-3 sin[2 ln(x)], 4 4 y ' = c 1 + (2c3 - 3c2 )x- cos[2 ln(x)] - (2c2 + 3c3 )x- sin[2 ln(x)],

380

APPENDIX C. SOLUTIONS " = [-4(2c3 - 3c2 ) - 2(2c2 + 3c3)Jx-5 cos[2 ln(x)] + [4(2c2 + 3c3) - 2(2c3 - 3c2)Jx-5 sin[2 ln(x)], or 5 5 y " = (8c2 - 14c3 )x- cos [2 ln (x)] + (14c2 + 8c3 )x- sin[2 ln(x)]. Then y (l) = 0, y'(l) = 6 and y"(l) = -2 ⇒ c1 +c2 = 0, c1-3c2 +2c3 = 6 and 8c2- 14c3 = -2 ⇒ c1 = 2, c2 = -2 and c3 = -1. Hence, y = 2x - 2x-3 cos[2 ln(x)J- x-3 sin[2 ln(x)]. y

23. x4 y( 4 ) + x3 y + x2 y" - 2xy ' + 2y = 0. For x > 0, y = xr ⇒ P(r) = r(r - l)(r - 2)(r - 3) + r(r - l)(r - 2) + r(r - l) - 2r + 2 = 0 ⇒ (r - l)[r(r - 2)(r - 3) + r(r - 2) + r - 2] = (r - l)(r - 2)[r(r - 3) + r + 1] = (r - l)3(r - 2) = 0 ⇒ y = c1x + C2X In lxl + c3 x(ln lxl) 2 + c4 x2 , x =/= 0. m

24. x4y( 4 ) - x3 y + 5x2 y " - lO xy ' + lOy = 0. For x > 0, y = xr ⇒ P(r) = r(r - l)(r - 2)(r - 3) - r(r - l)(r - 2) + 5r(r - l) - lOr + 10 = 0 ⇒ (r - l)[r(r - 2)(r - 3) - r(r - 2) + 5r - 10] = (r - l)(r - 2)[r(r - 3) - r + 5] = (r - l)(r - 2)(r2 - 4r + 5) = 0 ⇒ r1 = 1, r2 = 2 and r3, r3 = 4±'(-4 = 2 ± i ⇒ y = c1x + c2x2 + c3x2 cos(ln lxl) + C4X2 sin(ln lxl), x =/= 0. m

25. x4 y(4 ) + 6x3 y"' + 6x2 y" - 2xy' + 2y = 0. For x > 0, y = xr ⇒ P ( r) = r(r - l)(r - 2)(r - 3) + 6r(r - l)(r - 2) + 6r(r - l) - 2r + 2 = 0 (r - l)[r(r - 2)(r - 3) + 6r(r - 2) + 6r - 2] = (r - l)(r3 + r2 - 2) = (r - 1)2 (r2 + 2r + 2) = 0 ⇒ r1 = 1 and r2 , r2 = - 2±2vGI = -1 ± i ⇒ y = c1x + c2x ln lxl + c3x-1 cos(ln lxl) + c4x-1 sin(ln lxl), x =/= 0.



26. x4 y (4 ) + 2x3 y "' + 3x2 y" - 3xy ' + 4y = 0. For x > 0, y = xr ⇒ P(r) = r(r - l)(r - 2)(r - 3) + 2r(r - l)(r - 2) + 3r(r - 1) - 3r + 4 = 0. P(r) = r4 - 4r3 + 8r2 - 8r + 4 = (r2 + ar + 2)(r2 + br + 2) ⇒ a= b = -2 ⇒ r4 - 4r3 + 8r2 - 8r + 4 = (r2 - 2r + 2)2 = 0 ⇒ r1, r1 = 2±-(-4 = 1 ± i ⇒ y = c1x cos(ln lxl) + c2 x sin(ln lxl) + c3x ln lxl cos(ln lxl) + C4X ln lxl sin(ln lxl)-

Section 4.2 Exercises - page 118 l.

"' - 3y " - y ' + 3y = 8e 2x . For y - 3y" - y' + 3y = 0, y = erx =} P(r) = (r2 -l)(r -3) = 0 ⇒ Y1 = ex , Y2 = e -x and y3 = e3x . With J(x) = 8e 2x , x 3x e -x e e x x 3x = ex [-9e 2x - 3e2x ] - e -x[9e4x - 3e4x ] + e3x[2] = W(x) = e -e - 3e x 9 3x x e ee 3x 6 e -l ' x 3x ee 0 x 0 -e- 3e3x = J(x)[3e2x + e 2x ] = 4e 2x J(x) = 32e4x , �1(x) = x 9 3x ee f(x) 3x e 0 0 3e 3x = -f(x)[3e4x_e4x ] = -2e4xf(x) = -l6e6x , x e f(x) 9e3x y

m

381

C.4. CHAPTER 4 SOLUTIONS

ex e-x 0 x � 3(x)= e -e-x O =f(x)[-1 - 1]= -2f(x)= -16e2x , e-x f(x) ex x U1 = -2Je dx = -2ex , U2 = Je3x dx= ½e3x , U3 =Je-x dx=-e-x Yp = -!e2x ⇒ y =-!e2x + C1ex + C2 e-x + C3 e3x _



2. y"' - 3y" + 2y' =8xe 2x . For y'" - 3y" + 2y' =0, y = erx ⇒ P(r) = r(r - l)(r- 2)= 0 ⇒ Y1 = 1, Y2 =ex and y3 =e2x . With J(x)=8xe2x , 1 ex e2x W(x)= 0 ex 2e2x = 2e3x,� 1 (x)= e3x f(x)= 8xe5x , 0 ex 4e2x � 2 (x)= -2e2x f(x)=-16xe4x ,� 3(x) = ex f(x)= 8xe3x , UJ =4Jxe2x dx = 2xe2x - e2x , U2 = -8Jxex dx=-8xex + 8ex , u3 = 4Jx dx = 2x2 ⇒ YP = (2xe2x - e2x ) + (8ex - 8xex )ex + 2x2 e2x = 7e2x - 6xe2X + 2x2 e2x , or YP = (2x2 - 6x)e2x ⇒ y =(2x2 - 6x)e2x + C1 + c2 ex + c3 e 2x.

3. y'" - 2y" + y' =-2ex cos(x). For y'" - 2y" + y' =0, y =erx ⇒ P(r) =r(r-1)2 = 0 ⇒ Y1 = 1, Y2 = ex and y3 =xex . With J(x)= -2ex cos(x), xe x 1 ex W(x)= 0 ex ex + xex =e2x ,� 1 (x)=e2x f(x)=-2e3x cos(x), 0 ex 2ex + xex � 2 (x) =-(1 + x)ex f(x) = 2(1 + x)e2x cos(x),� 3 (x)= ex J(x)=-2e2x cos(x), u1 =-2Jex cos(x) dx= -2ex cos(x) - 2Jex sin(x) dx = -2ex cos(x) - 2ex sin(x) + 2Jex cos(x) dx ⇒ -4J ex cos(x) dx = -2ex [cos(x) +sin(x)] ⇒ u1 = -ex [cos(x) + sin(x)], u2 =2J(1 + x) cos(x) dx = 2(1 + x)sin(x) + 2 cos(x), u3 =-2Jcos(x) dx= -2sin(x) ⇒ YP= -ex [cos(x) + sin(x)] + [2(1 + x)sin(x) + 2 cos(x)]ex - 2xex sin(x) = ex [cos(x) + sin(x)] ⇒ y =ex [cos(x) +sin(x)] + c 1 + c2 ex + c3xex .

4. y'" - y" + 2y = l0ex . For y'" - y" + 2y= 0, y = erx ⇒ P(r) = (r + l)(r-2 - 2r + 2) = 0 ⇒ r- 1 = -1 and r = 1 ± i ⇒ Y1 = e-x , Y2 = ex cos(x) and Y2 =ex sin(x). With J(x)= l0ex , ex sin(x) e-x ex cos(x) x x x x W(x)= -e- e cos(x) - e sin(x) e sin(x) + ex cos(x) e-x 2ex cos(x) -2ex sin(x) =e-x [2e2x ] + e-x [2e2x ] + e-x [e2x]=5ex ,� 1 (x)= e2x J(x)= 10e3x , � 2 (x)=-[2sin(x) + cos(x)]f(x) =-10ex [2sin(x) + cos(x)], � 3(x) = [2 cos(x) -sin(x)]J(x) = 10ex [2 cos(x) -sin(x)], u1 = 2Je2x dx= e2"', u2 =-2J2sin(x) + cos(x) dx =4cos(x) - 2sin(x), u3 =2J2 cos(x) -sin(x) dx=4sin(x) + 2 cos(x) ⇒ YP = e2x [e-x ]+[4 cos(x)-2sin(x)][ex cos(x)]+[4sin(x)+2 cos(x)][exsin(x)] = 5ex ⇒ y = 5ex + c1e-x + Czex cos(x) + c3 ex sin(x).

382

APPENDIX C. SOLUTIONS

5. y"'- y" + y' - y = 4sin(x) + 4cos(x). For y"' - y" + y' - y = 0, y = erx =} P(r) = (r-l)(r2 + 1) =} y1 = ex , Y2 = cos(x) and y3 = sin(x). With ex cos(x) sin(x) x J(x) = 4[sin(x) + cos(x)], W(x) = e -sin(x) cos(x) = 2ex , x e - cos(x) -sin(x) �1(x) = f(x) = 4[sin(x) + cos(x)], � 2 (x) = -ex [cos(x)-sin(x)]f(x) = -4ex [cos(x) -sin(x)l[sin(x) + cos(x)] = -4ex [cos2 (x) -sin2 (x)] = -4ex cos(2x), � 3 (x) = -ex [sin(x) + cos(x)]f(x) = -4ex [sin(x) + cos(x)] 2 = -4ex [l + sin(2x)], u1 = 2 J e- x [sin(x) + cos(x)] dx = -2e- x cos(x), u2 = -2 J cos(2x) dx = - sin(2x), u 3 = -2 J l + sin(2x1 dx = -2x + cos(2x) ⇒ YP = -2cos(x) -sin(2x) cos(x) + [-2x + cos(2x)] sin(x) = -2cos(x)- 2xsin(x) + [sin(x)cos(2x)-cos(x)sin(2x)] = -2cos(x) - 2xsin(x)-sin(x), or YP = -2xsin(x) ⇒ y = -2xsin(x) + c1 ex + c2 cos(x) + c3 sin(x). P(r) = r2 (r - 1) = 0 =} l X ex x 3 Y1 = 1, Y2 = x and y3 = e . With J(x) = 20x , W(x) = 0 1 ex = ex , 0 0 ex x x 4 3 x x 3 x �1 (x) = (xe - e )f(x) = 20(x - x )e , � 2 (x) = -e f(x) = -20x e , 3 �3(x) = f(x) = 20x , u1 = 20 J x4 - x3 dx = 4x5 - 5x4 , u2 = -20 J x3 dx = -5x4 , u 3 = 20 J x3 e-x dx = -20x3 e- x + 60 J x2 e- x dx = -20x3 e- x - 60x2 e- x + 120 f xe- x dx = -20x3 e-x - 60x2 e-x - l20xe- x-l20e- x ⇒ YP = (4x5 - 5x4 ) - 5x5 - [ 20x3 + 60x2 + 120x + 120] = -x5 - 5x4 - 20x3 - 60x2- 120x- 120, or YP = -(x5 + 5x4 + 20x3 + 60x2 ) ⇒ y = -(x5 + 5x4 + 20x3 + 60x2 ) + c1 + c2 x + c3 ex is the general soluti on. Then y' = -(5x4 + 20x3 + 60x2 + 120x) + C2 + c3 ex , y" = -(20x3 + 60x2 + 120x + 120) + c3 ex , and y(0) = 20, y'(0) = 0 and y"(0) = -20 ⇒ c1 + c3 = 20, c2 + c3 = 0 and -120 + C3 = -20 ⇒ c1 = -80, c 2 = -100 and c3 = 100 =} y = -(x5 + 5x4 + 20x3 + 60x2 ) - 80 - lO0x + lO0ex .

6. y"' - y" = 20x3 . For y"' - y" = 0, y = erx

=}

7. y"'-2y" = 5sin(x). For y"'-2y" = 0, y = erx

P(r) = r2 (r-2) = 0 =} Y1 = 1,

Y2 = x and y3 =

e2x

=}

. With J(x) = 5sin(x), W (x) =

l

X

e 2x

0 1 2e2x 0 0 4e2x

= 4e2x ,

� 1 (x) = [ 2xe2x - e2x ]f(x) = 5(2x -l)e2x sin(x), �2 (x) = -2e2xf(x) = -10 e2x sin(x), � 3 (x) = f(x) = 5sin(x), u1 = ¾ (2x - 1)sin(x) dx = -¾(2x- 1) cos(x) + � sin(x), u2 = -� sin(x) = � cos(x), u3 = ¾ e- 2x sin(x) dx = -¾e- 2x cos(x) - � e- 2x cos(x) dx = -¾e- 2x cos(x) - �e- 2x sin(x) - 5 e- 2x sin(x) dx =} 245 e- 2x sin(x) dx =

J J J

J

J

J

383

C.4. CHAPTER 4 SOLUTIONS - e-2 x [�sin(x) + ¾ cos(x)] ⇒ u3 = -e-2x [½sin(x) + ¼cos(x)] ⇒ YP = -¾(2x- l)cos(x) + �sin(x) + �xcos(x)- [½sin(x) + ¼cos(x)] = 2sin(x) + cos(x) ⇒ y = 2sin(x) + cos(x) + c1 + c2 x + c3 e2x .

P(r) = r2 (r2 - 1) = 0 ⇒ Y 1 = 1, e-x l X ex x O l e -e-x Y2 = x, y3 = ex and y4 = e- x ⇒ W(x) = = -2, and with e-x 0 0 ex 0 0 ex -e-x J(x) = 2, 61(x) = -2xf(x) = -4x, 62 (x) = 2f(x) = 4, 63 (x) = -e- x f(x) = -2e-x , 64 (x) = ex f(x) = 2ex , U 1 = 2J xdx = x2 , U2 = - J 2dx = -2x, U3 = Je- x dx = -e-x , U4 = - Jex dx = -ex ⇒ YP = -x2 - 2, or YP = -x2 ⇒ y = -x2 + C 1 + C2 X + c3 ex + C4e-x.

8. y(4)

- y" =

4 2. For y( )

- y" = 0,

y

= erx ⇒

9. y(4) + y" = 6x. For y( 4) + y" = 0, y = erx =? P(r) = r 2 (r2 + 1) = 0 =? Y1 = 1, Y2 = x, y3 = cos(x) and y4 = sin(x) ⇒ W(x) = sin(x) l x cos(x) . cos(x) 0 1 - sin(x) . -sm (x)[-sm(x)] -cos(x)[-cos(x)] = 1 = . ( ) ( ) 0 0 - COS X - Slll X 0 0 sin(x) -cos(x) (along the fourth row) and, with f(x) = 6x, 6 1 (x) = -f(x)[x - O] = -6x2 , 62 (x) = f(x)[l] = 6x, 63 (x) = -f(x)[-sin(x)] = 6xsin(x), 64(x) = f(x)[-cos(x)] = -6xcos(x), u1 = -6J x2 dx = -2x3 , u2 = 6J xdx = 3x2 , u3 = 6J xsin(x)dx = -6xcos(x) + 6sin(x), U4 = -6J xcos(x)dx = -6xsin(x) - 6cos(x) ⇒ YP = x3 + [6sin(x) - 6x cos(x)] cos(x) - [6cos(x) + 6x sin(x)] sin(x) = x3 - 6x, or YP = x3 ⇒ y = x3 + c 1 + c2 x + c3 cos(x) + C4 sin(x).

10. x3 y111 - x2 y" + 2xy' - 2y = x. For x3 y"' - x2 y" + 2xy' - 2y = 0, x > 0, y = xr ⇒ P ( r) = ( r - l )2 (r - 2) = 0 ⇒ Y 1 = x, Y2 = x ln(x) and y3 = x2 . With f ( x) = x\ , x x ln(x) x2 W(x) = 1 ln(x) + 1 2x = x[ 2ln(x)] - [ 2xln(x) - x] = x, l 0 2 X 2 6 1 (x) = f(x)[x ln(x) - x2 ] = ln(x) - 1, 62 (x) = - f(x)[ x2 ] 63 (x) = J(x)[x] = �, u 1 = J In(�-l dx = ½[ln(x)] 2 - ln(x), u2 = - J � dx = -ln(x), U3 = J ;2 dx = -� ⇒

= -1,

YP = {½[ln(x)] 2 - ln(x)}x - ln(x)[xln(x)] - �[x2 ] = - ½x[ln(x)] 2 - xln(x) - x, or YP = -½x[ln(x)] 2 ⇒ y = -½x[ln(x)] 2 + c1x + c2 xln(x) + c3 x2 .

r 11. x3 y111 + 5x2 y" + 2xy' - 2y = �- For x3 y"' + 5x2 y" + 2xy' - 2y = 0, x > 0, y = x 2 1 ⇒ P(r) = (r - l)(r + l)(r + 2) = 0 ⇒ Yi = x, Y2 = x- and y3 = x- . With x x-2 x-1 J(x) = 2\, W(x) = 1 -x-2 -2x-3 = x[-2x-6] - [4x-5] = -6x- 5, 6x- 4 0 2x- 3

384

APPENDIX C. SOLUTIONS

61(x) = f(x)[-x-4] = -x- 8, 62 (x) = -f(x)[-3x- 2 ] = 3x-6, 63 (x) = f(x)[-2x-1] = -2x- 5, u1 = ¼ J x-3 dx = -/2 x- 2 , u2 = -½ J x-1 dx = -½ ln(x), u3 = J ½ dx = ½x 1 1 1 1 =} Yv = -/2 x- - ½ ln(x)[x- ] + ½x- , or Yv = -½x- ln(x) 1 1 2 =} Y = -½x- ln(x) + C1X + C 2 x- + C3 x- . 12. x3 Y1

y

111

+2x2 " = 1. For x3 y

y

111

+2x2 " = 0, x > 0, y

= 1, Y2 = ln(x) and 3 = x. With f(x) = }3 y

= xr =;,- P(r) = r 2 (r-l) = 0 =;,-

y

,

W(x)

1 ln(x) x 1. 1 - ;2 , xl 0 - x2 0

= 0

1 6 1 (x) = f(x)[ln(x)- 1] = ln(:1- , 6 2 (x) = -f(x)[l] = - ;3 63 (x) = f(x)[¾] = ;4, u1 = J In(�-l dx = ½[ln(x)] 2 - ln(x), u 2 = - J ¾ dx = - In(x), U3 = J x\ dx = -¼ 2 2 2 =} Yv = ½[ln(x)] - ln(x)- [ln(x)] - 1, or Yv = -½[ln(x)] 2 =;,- y = -½[ ln(x)] + c1 + C 2 ln(x) + c3x is the genera l so luti on. Then y' = -� ln(x) + c2 ¾ + c3 , y" = x\ ln(x) - ;2 - c2 ;2 , and y(l) = 5, y'(l) = 1 and y" ( 1) = 0 =} C1 + C3 = 5, C 2 + C 3 = 1, -1 - C 2 = 0 =} C1 = 3, C 2 = -1 and C 3 = 2 =;,- y = -½[ln(x)] 2 + 3 - ln(x) + 2x. ,

r 13. x3 + x2 " - x ' = 4x2 . For x3 + x2 " - x ' = 0, x > 0, = x 2 2 P(r) = r (r - 2) =} Y1 = 1, Y2 = ln(x) and 3 = x . With f(x) = 1 ln(x) x2 W(x) = 0 2x = �, '6. 1 (x) = f(x)[2xln(x) - x] = 8ln (x) - 4, 1. xl 2 0 x2 2x (x) (x) = = - f [ ] -8, 63(x) = f(x)[¼] = ;2, 62 u1 = J 2x ln(x) - x dx = x2 ln(x) - x2 , u 2 = - J 2x dx = -x2 , 2 2 2 U3 = J;; dx = ln(x) =} Yv = x ln (x)- x , or YP = x ln(x) =} y = x2 ln(x) + c1 + c2 ln(x) + c3 x2 . y

111

y

y

y

111

y

!,

y

y

y

=;,­

14. x3 "' + 2x2 y" - 2x ' = x3 cos(x). For x3 "' + 2x2 y" - 2x ' = 0, x > 0, = xr =;,- P(r) = r(r + l)(r - 2) = 0 =;,- Y1 = 1, Y2 = x-t and 3 = x2 . With y

f(x)

y

y

1

x- 1

x2

y

y

y

= cos(x), W(x) = 0 -x- 2 2x = -6x- 2 , '6.1(x) = f(x)[3] = 3cos(x),

0 2x- 3 2 2 2 i6. 2 (x) = - f(x)[2x] = -2xcos(x), i6. 3 (x) = J(x)[-x- ] = -x- cos(x), 2 2 u 1 = -½ J x cos(x) dx = -½x sin(x) + J xsin(x) dx = -½x2 sin(x)- x cos(x) + sin(x), u 2 = ½ J x3 cos(x) dx = ½x3 sin(x) - J x2 sin(x) dx = ½x3 sin(x) + x2 cos(x) - 2 J xcos(x) dx = ½x3 sin(x) + x2 cos(x) - 2xsin(x) - 2cos(x), u 3 = ½ J cos(x) dx = ½sin(x) =} Yv = [-½x2 sin(x) - xcos(x) + sin(x)] + [½x3 sin(x) + x2 cos(x)- 2xsin(x)- 2cos(x)] x- 1 + [½sin(x)] x2 = -sin(x)- 2x- 1 cos(x) =;,- y = -sin(x) - 2x-1 cos(x) + c1 + c2 x- 1 + c3 x2 .

38 5

C.4. CHAPTER 4 SOLUTIONS

15. x3y111 + 3x2 y " + 2xy'= 1. For x3 y"' + 3x2 y" + 2xy ' =0, x > 0, y = xr => P(r) = r(r2 + 1) = 0 => y 1 = 1, Y2 =cos[ln(x)] and y3 =sin[ln(x)]. With f(x) = x\, sin[ln(x)] 1 cos[ln(x)] - ...L' sin[ln(x)] cos[ln(x)] W(x) = 0 3 - x 0 }2 sin[ln(x)] - }2 cos[ln(x)] - x\ cos[ln(x)] - x\ sin[ln(x)] � 1 (x)= f(x) [¾] = ;4, � 2 (x)= -f(x) cos[ln(x)l} = -}4 cos[ln(x)], � 3 (x) =f(x) {-� sin[ln(x)l} = - }4 sin[ln(x)], u1= ¾ dx= ln(x), x u2 = - cos[;( )j dx= -sin[ln(x)], U 3 = - sin[l;(x)] dx= cos[ln(x)] => YP =ln(x) => y =ln(x) + c1 + c2 cos[ln(x)] + c3 sin[ln(x)].

¾





J

J

J

16. x3y111 + 3x2y" = x2 sin(x). For x3y111 + 3x2y"= 0, x > 0, y =xr => P(r) = r(r2 - 1) =0 => Y1 = 1, Y2 =x and y3 = x-1 . With J(x) =sin?), 1

X-l

X

0 1 -x- 2 = 2x-3 , � 1 (x) =f(x)[-2x-1] =-2x- 2 sin(x), 0 0 2x-3 1 2 � 2 (x) =-J(x)[-x- ] = x-3 sin(x), � 3 (x)= f(x)= x- sin(x), u1 = - J xsin(x) dx= xcos(x) -sin(x), u2 = ½ sin(x) dx= -½cos(x), u3 = ½ x2 sin(x) dx =-½x2 cos(x) + xcos(x) dx =-½x2 cos(x) + xsin(x) + cos(x) => YP =[xcos(x) -sin(x)] - ½x cos(x) + [-½x2 cos(x) + x sin(x) + cos(x)] x-1 =x- 1 cos(x) => y =x- 1 cos(x) + c 1 + c2 x + c3 x-1.

W(x) =

J

J

J

- 3y" - y' + 3y = 8e2 x . By Exercise 1, P(r) = (r - l)(r + l)(r - 3) ⇒ (D - l)(D + l)(D - 3)y = 8e2x ⇒ (D - 2)(D - l)(D + l)(D - 3)y = (D - 2)[8e2x ] = 0 ⇒ y = C 1 e2x + C2 ex + C 3 e-x + C4 e3x ⇒ YP = ae2 x , y� = 2ae2x , Y"P = 4ae2x ' yP"' = 8ae2 x ' and yP111 - 3yP" - yP' + 3y = 8e2x => -3a = 8 => a= _§.3 ⇒ YP = -ie2x _

17.

y '"

18.

y

= lOex . By Exercise 4, P(r) = (r + l)(r2 - 2r + 2) => (D+l)(D -2D+2)y = lOex => (D-l)(D+l)(D 2 -2D+2)y = (D-l)[l0ex ] = 0 => y = c1ex + c2 e-x + c3ex cos(x) + C4ex sin(x) => YP = aex , � =yi =yi' = aex , and yi' - yi + 2y =lOex => 2a = 10 => a= 5 => YP = 5ex . 111

y" + 2y 2

-

y

p

19.

= 20x3 . By Exercise 6, P(r) = r2 (r - 1) => D2 (D - l)y = 20x3 => D D (D - l)y = D 4 [20x3 ] = 0 => D 6 (D - l)y = 0 ⇒ Y= c1 + c2 x + C 3 X 2 + C 4 X 3 + c5x4 + c6x5 + C7ex => YP = ax2 + bx 3 + cx4 + dx5, 4 2 3 2 3 y� = 2ax + 3bx + 4cx + 5dx , yi = 2a + 6bx + 12cx + 20dx , 111 = 111 6b + 24cx + 60dx2 , and y - yp" = 20x3 ⇒ Y y

1 11

4

p

-

2

y"

p

[6b+24cx+60dx2 ]-[2a+6bx+12cx2 +20dx 3 ]= 20x 3 ⇒ d= -l, 60d-12c =0 => c = -5, 24c - 6b = 0 => b= -20, 6b - 2a = 0 => a= -60 => YP = -60x2 - 20x3 - 5x4 - x5 .

20. y"' - 2y" = 5sin(x). By Exercise 7, P(r) = r2 (r - 2) ⇒ D2 (D - 2)y ⇒ (D2 + l)D2 (D - 2)y = (D 2 + 1)[5sin(x)] =0 ⇒

= 5sin(x)

386

APPENDIX C. SOLUTIONS c1 + c2 x + C3 e2x + C4 cos(x) + c5 sin(x) ⇒ YP = acos(x) + bsin(x), y� = -asin(x)+bcos(x), yi = -acos(x)-bsin(x), yi' = asin(x)-bcos(x), and yi' - 2yi = 5sin(x) ⇒ [asin(x) - bcos(x)] - 2[-acos(x) - bsin(x)] = 5sin(x) ⇒ 2a - b = 0 and a + 2b = 5 ⇒ a= l and b = 2 ⇒ YP = cos(x) + 2sin(x). y=

21.

3y " + 2y' = 8xe 2x . By Exercise 2, P(r) = r(r - l)(r - 2) ⇒ D(D - l)(D- 2)y = 8xe2x ⇒ (D - 2)2 D(D- l)(D - 2)y = (D- 2)2 [8xe2x] = O ⇒ D(D - l)(D - 2)3 y = 0 ⇒ y = C1 + c2 ex + c3e2x + c4xe2x + c5 x2 e2x ⇒ YP = (ax+bx2)e2x , y� = (a+2bx)e2x+2(ax+bx2)e2x = [a+(2a+2b)x+2bx2]e2x, Yi= [(2a + 2b) + 4bx]e2x + 2[a + (2a + 2b)x + 2bx2 ]e2x = [(4a + 2b) + (4a + 8b)x + 4bx2 ]e2x, 2 2x 2x yi'= [(4a + Sb) + 8bx]e + 2[(4a + 2b) + (4a + 8b)x + 4bx ]e 2 2x = [(12a + 12b) + (Sa + 24b)x + 8bx ]e , and y;' - 3y; + 2y� = 8xe2x ⇒ [(12a + 12b) + (Sa + 24b)x + 8bx2 ]e2x - 3[(4a + 2b) + (4a + 8b)x + 4bx2 ]e2x+ 2[a+(2a+2b)x+2bx2 ]e2x = 8xe2x ⇒ 2a+6b+4bx = 8x ⇒ b = 2 and 2a+6b = 0 ⇒ a= -6 ⇒ YP = (2x2 - 6x)e2x .

22.

4sin(x) + 4cos(x). By Exercise 5, P(r) = (r - l)(r2 + 1) ⇒ (D - l)(D + l)y = 4sin(x) + 4cos(x) ⇒ (D2 + l)(D - l)(D2 + l)y = (D2 + 1)[4sin(x) + 4cos(x)] = 0 ⇒ (D - l)(D2 + 1)2 y = 0 ⇒ y = c1ex + c2 cos(x) + c3 sin(x) + C4Xcos(x) + c5 x sin(x) ⇒ YP = axcos(x) + bxsin(x) = x[acos(x) + bsin(x)], y� = [acos(x) + b sin(x)] + x[-asin(x) + bcos(x)], y; = [-2asin(x) + 2bcos(x)] + x[-acos(x) - bsin(x)], y;' = [-3acos(x) - 3bsin(x)] + x[asin(x) - bcos(x)], and y;' - y; + y� - YP = 4sin(x) + 4cos(x) ⇒ (-2a - 2b)cos(x) + (2a - 2b)sin(x) = 4sin(x) + 4cos(x) ⇒ 2a - 2b = 4 and -2a - 2b = 4 ⇒ a = 0 and b = - 2 ⇒ YP = - 2xsin(x).

y "' -

y "' - y " + y ' - y 2

=

23. y111 - 2y" + y' = -2ex cos(x). By Exercise 3, P(r) = r(r - 1)2 ⇒ D(D - l) 2 y= -2ex cos(x) ⇒ (D2 - 2D + 2)D(D - 1)2 y = (D2 - 2D + 2)[-2ex cos(x)] = 0 ⇒ x x x x x y = c1 + c2 e + C3 Xe + C4e cos(x) + C5e sin(x) ⇒ YP = e [acos(x) + bsin(x)], x y� = e [(a + b)cos(x) + (b - a) sin(x)], x x y;= e [(a + b)cos(x) + (b - a) sin(x)] + e [-(a + b)sin(x) + (b - a)cos(x)] x x = e [2bcos(x) - 2asin(x)], y i' = e [(2b - 2a)cos(x) - (2a + 2b)sin(x)], and x x x yi'- 2y i+y� = -2e cos(x) ⇒ e [(-a - b)cos(x)+(a - b)sin(x)] = -2e cos(x) x ⇒ a - b = 0 and -a - b = -2 ⇒ a= b = l ⇒ YP = e [cos(x) +sin(x)]. 24.

y" = 2. By Exercise 8, P(r) = r2 (r2 - 1) ⇒ D2 (D2 - l)y = 2 ⇒ D (D- l)(D+ l)y = D[2] = 0 ⇒ y = C1 + C2 X + C3X2 +c4ex +cse-x ⇒ YP = ax2 , 4 4 Y'P = 2ax ' yP" = 2a ' yP"' = yP( ) = 0 ' and yP( ) - yP" = 2 ⇒ -2a = 2 ⇒ a= -l ⇒ 2 YP = -x .

4 y( ) 3

-

387

C.4. CHAPTER 4 SOLUTIONS

25. y(4) + y" = 6x. By Exercise 9, P(r) = r2 (r2 + 1) ⇒ D2 (D2 + l)y = 6x ⇒ D4 (D2 + l)y = D2 [6x] = 0 ⇒ y = c1 + c2 x + c3x2 + c4x3 + c5cos(x) + c5sin(x) 4 ⇒ yP = ax2 + bx3' y'P = 2ax + 3bx2 ' y"P = 2a + 6bx ' y"'P = 6b,p y ( ) = 0' and Yt) + yi = 6x ⇒ 2a + 6bx = 6x ⇒ a= 0 and b = 1 => YP = x3. Chapter 4 Exercises - page 119 1. y"' - y" (a)

- 2y' = 12. y" - 2y' = 0, y = e For y

=> P(r) = r(r + l)(r - 2) = 0 => Y1 = 1, e-x e 2x 1 y2 = e- x and y3 = e2x, and W(x) = 0 -e-x 2e 2x = -6ex . With e-x 4e 2x 0 2 2x x x J(x) = 12, 6. 1 (x) = J(x)[3e ] = 36e , 6. 2 (x) = - f(x)[2e ] = -24e x, b. 3 (x) = J(x)[-e-x] = -12e-x , U 1 = - f6dx = -6x, U2 = 4 Jexdx = 4ex , U3 = 2 J e- 2xdx = -e-2x => YP = 3 -6x,or YP = -6x =? Y = -6x + C1 + C2 e-x + C3e 2x. 111

-

rx

(b) P(r) = r(r + l)(r - 2) ⇒ D(D + l)(D- 2)y = 12 ⇒ D2 (D + l)(D- 2)y = D[l2] = 0 => y = C1 + C2 X + c3e-x + C4e2x => YP = ax,y� = a,Yi= yi' = 0, and yi' - yi - 2y� = 12 ⇒ -2a = 12 ⇒ a = -6 ⇒ YP = -6x ⇒ Y = -6x + C1 + C3e-x + C4e 2x .

(c) y"' - y"

-

- -

- y' - 2y =

2y' = 12 => y" y' 2y = 12x + c 1 . For y" y = erx => P(r) = r2 - r - 2 = (r - 2)(r + 1) = 0 => Y2 = e-x . YP = ax + b => y� = a, yi = 0 and yi y� 2yp -a - 2[ax + b] = 12x + c1 ⇒ a = -6 and b = 3 - � = c2 ⇒ =? Y = -6x + C2 + C3e 2x + C4e-x .

- -

0,

Y1 = e2x and = 12x + C1 => YP = -6x + c2

2. x3 y111 + x2 y" = x. For x3 y"' + x 2y" = 0, x > 0, y = xr => P(r) = r(r - 1)2 = 0 1 x x ln(x) 1 => Y1 = 1, Y2 = x and y3 = x ln(x), and W(x) = 0 1 ln(x) + 1 X .!. 0 0 I 1 With J(x) = :2 , 6.1(.1:) = J(x)[x] = ¾, 6. 2 (x) = - f(x)[ln(x) + l] = - n(;l+ , I 6.3(x) = J(x)[l] = :2 , u1 = J1 dx = x,u2 = - J n(�+I dx = -½[ln(x)] 2 -ln(x), 2 U3 = J ¾ dx = ln(x) ⇒ YP = x - { ½[ln(x)] + ln(x)} x + ln(x)[x ln(x)], or YP = ½x[ln(x)] 2 ⇒ y = ½x[ln(x)]2 + c1 + c2 x + c3xln(x).

3. x3 y"'+x2 y" =�-By Exercise 2, W(x) = ¾- f(x) = x� => 6.1(x) = f(x)[x] = x�' 6.2(x) = - J(x)[ln(x) + l] = _4 In(;� +l, 6.3 (x) = J(x)[l] = �, x u1 = J ..i..2 dx = - 1x , u2 = -4f In(x3) +l dx = 2x- 2 ln(x)+x-2 +2x- 2 = 2 ln(2x ) +2x2 , x x x 1n x ( [x ln(x)] = _ _lx => U3 = J ..xi.. dx = _-1_ => yP = _ _'!x + [2 x ) + x x - 2 x x 3

y

2

= -¾ + c1 + c2 x + c3x ln(x).

2

2-] 2

2

388

APPENDIX C. SOLUTIONS

4. Y111 - y " = 20x3 => y " - y ' = 5x4 + C1 => y ' - y = x5 + c1 x + c2 . I(x) = e-x => (e-xy )' = (x5 + C 1 X + c2 )e-x => e-xy = J(x5 + C 1 X + c2 )e-x dx = -(x5 + C1X + c2)e-x +f(5x4 + c1 )e-xdx = -(x5 + C1 X + c2 )e-x - (5x4 + c1 )e-x +f20x3 e-x dx = -(x5 + C1 X + c2 )e-x - (5x4 + c1 )e-x - 20x3 e-x +J60x2 e-x dx = -(x5 + C1 X + c2 )e-x - (5x4 + c1 )e-x - 20x3 e-x - 60x2 e-x +J l20xe-x dx = -(x5 +c 1 x+c2 )e-x - (5x4 +c 1 )e-x- 20x3 e-x- 60x2 e-x - l20xe-x -12Qe-x +c3 => y = -(x5 + c1x + c2 ) - (5x4 + c1 ) - 20x3 - 60x2 - 120x - 120 + c3 ex = -x5 - 5x4 - 20x3 - 60x2 + C5 + c4 x + C3 ex . 4 5. x4 y ( 4) - 4x3 y"' + 12x2 y" - 24xy' + 24y = x . 111 4 (4) 3 2 For x y - 4x y + 12x y" - 24xy' + 24y = 0, x > 0, y = xr => P(r) = (r- l)(r- 2)(r- 3)(r- 4) = 0 =>Yi = x, Y2 = x2 , y3 = x3 and y4 = x4 , x x2 x 3 x4 x x2 x4 x x2 x3 1 2x 3x2 4x3 3 and W(x) = = -6 1 2x 4x + 24x 1 2x 3x2 0 2 6x 12x2 0 2 12x2 0 2 6x 24x 6 0 0 = -6[x(16x3 ) - 10x4 ] + 24x[x(6x ) - 4x3 ] = 12x4 . With J(x) = 1, x2 x3 x4 6 1 (x) = - 2x 3x2 4x3 = -[x2 (12x4 ) - x3 (l6x3 ) + x4 (6x2 )] = -2x6 , 2 6x 12x2 x x3 x4 3 4 6 2 (x) = 1 3x2 4x3 = x(l2x4 ) - 6x5 = 6x5, 6 3 (x) = -6x , 6 4 (x) = 2x , 0 6x 12x2 2 2 u1 = -lfx dx = _l.18..x3 ' u2 = l.Jxdx = lx = -lx 6 2 4 ' u3 = -lfldx 2 2 ' 4 4 U4 = ½ J � dx = ln(x) => YP = -;�x + ¼x ln(x), or YP = ¼x4 ln(x) => 4 2 4 3 y = ½x ln(x) + C 1 X + C2X + C3X + C4X .

i

6. x3y "' - x2 y" + lOxy' = 156x2 sin[3ln(x)]. For x3 y111 - x2 y " + lOxy' = 0, x > 0, 2 r y = x => P(r) = r(r - 4r + 13) = 0 => r 1 = 0 and r = 2 ± 3i => Y 1 = 1, sin 3 n Y2 = x2 cos[3ln(x)] and y3 = x2 sin[3ln(x)]. With J(x) = 155 [ � (xl], W(x) = x2 cos[3ln(x)] x2 sin[3ln(x)] 1 0 2xcos[3ln(x)] - 3xsin[3ln(x)] 2xsin[3ln(x)] + 3xcos[3ln(x)] = 39x, 9cos[3ln(x)] - 7sin[3ln(x)] 0 -7cos[3ln(x)] - 9sin 3ln(x)] 3 2 (x) = (x)[3x = 468x sin[3ln(x)], 61 ] f 6 2 (x) = - J(x){2xsin[3ln(x)] + 3xcos[3ln(x)]} = -312sin2 [3ln(x)] - 468sin[3ln(x)] cos[3ln(x)] = -156 + 156cos[6ln(x)] - 234sin[6ln(x)], 6 3 (x) = J(x){2xcos[3ln(x)] - 3xsin[3ln(x)]} = 312sin[3ln(x)]cos[3ln(x)] - 468sin2 [3ln(x)] = 156sin[6ln(x)] - 234 + 234cos[6ln(x)], u1 = 12Jxsin[3ln(x)] dx. x = et => u1 = 12 Je2t sin(3t)dt, and integration by partsg ivesu1 = i�e2t sin(3l)-i�e2t cos(3t) = i�x2 sin[3ln(x)]-��x2 cos[3ln(x)],

389

C.4. CHAPTER 4 SOLUTIONS

u2 = J -� + 4cos[6�n (x)] - 5 sin [6�n(x)] dx = -4ln(x) + �sin[6ln(x)] + cos[6ln(x)], u3 = J 4sin[6�n(x)] - � + 5 cos[6�n(x)] dx = -� cos[6ln(x)] - 6ln(x) + sin[6ln(x)] => YP = i!x2 sin[3ln(x)] - f�x2 cos[3ln(x)]+ x2 cos[3ln(x)] {-4ln(x) + �sin[6ln(x)] + cos[6ln(x)l} + x2 sin[3ln(x)] {-�cos[6ln(x)] -6ln(x) + sin[6ln(x)l}. Ignoring the first two terms because they are homogeneous solutions, YP = �x2 { sin[6ln(x)] cos[3ln(x)] - cos[6ln(x)] sin[3ln(x)]} +x2 { cos[6ln(x)] cos[3ln(x)] + sin[6ln(x)] sin[3ln(x)]} -4x2 ln(x) cos[3ln(x)] - 6x2 ln(x) sin[3ln(x)] = �x2 sin[3ln(x)] + x2 cos[3ln(x)] - 4x2 ln(x) cos[3ln(x)] -6x2 ln(x) sin[3ln(x)], oryp = -4x2 ln(x) cos[3ln(x)]-6x2 ln(x) sin[3ln(x)] =>y = c 1 +c2 x2 cos[3ln(x)]+ c3 x2 sin[3ln(x)] - 4x2 ln(x) cos[3 ln(x)] - 6x2 ln(x) sin[3ln(x)]. 7. Let P(r) = a0 +a1r+a2 r 2 +· · -+an rn , with ai, 0 :s; i :s; n, real. Then P(z) = 0 => ao + a1z + a2 z 2 + · · · + an zn = 0 => ao + a1z + a2 z 2 + · · · + an zn = 0 => a0 + a1z + a2z2 + · · · + an:zn = 0 since ai = ai, 0 :s; i :s; n . Thus, P(z) = 0. 1 X 8. No, because W(x) = 0 1 0 0 in contradiction of Theorem

x3

3x2 6x

= 6x = 0 at

x = 0 and W(x) =I- 0 for x =/- 0,

4.8.

9. y"' + 3y" + 3y' + y = 2xe- x cos(x). For y"' + 3y" + 3y' + y = 0, y = erx => P(r) = (r + 1) 3 = 0 =>Yi= e-x , Y2 = xe-x and y3 = x2 e-x , and e-x xe-x x2 e-x x x x W(x) = -ee- - xe2xe-x - x2 e-x e-x -2e-x + xe- x 2e-x - 4xe-x + x2 e-x 1 X x2 = e-3x -1 1 - x 2x - x2

1 x - 2 2 - 4x + x2 = e-3x[(l - x)(2 - 4x + x2 ) - (x - 2)(2x - x2 ) + x(2 - 4x + x2 ) - x 2 (x - 2) + x(2x - x2 ) - x2(1- x)] = 2e-3x_ With f(x) = 2xe-x cos(x), � 1 (x) = f(x)[x(2x - x2 ) - x2(1- x)Je-2x = x2 e-2x f(x) = 2x3 e-3x cos(x), � 2(x) = - f(x)[(2x - x2 ) + x2 Je-2x = -2xe- 2x f(x) = -4x2 e-3x cos(x), � 3(x) = J(x)[(l - x) + x]e-2x = e- 2x J(x) = 2xe-3x cos(x), u1 = J x3 cos(x) dx = x3 sin(x) -3 J x2 sin(x) dx = x3 sin(x) + 3x2 cos(x) -6 J xcos(x) dx = x3 sin(x) + 3x2 cos(x) - 6xsin(x) -6cos(x), u2 = -2 J x2 cos(x) dx = -2x2 sin(x) + 4 J x sin(x) dx = -2x2 sin(x) - 4xcos(x) + 4sin(x), u3 = J xcos(x) dx = xsin(x) +cos(x) => YP = [x3 sin(x) + 3x2 cos(x) - 6x sin(x) -6cos(x)Je-x+ [-2x2 sin(x) - 4xcos(x) + 4sin(x)] xe-x + [xsin(x) +cos(x)]x2 e-x

390

APPENDIX C. SOLUTIONS

= -2xe-xsin(x)- 6e-xcos(x) => 2 y = -2xe-x sin(x) - 6e-x cos(x) + c 1 e-x+ c2 xe-x+ c3x e-x. 10. By Exercise 9, P(r) = (r + 1) 3 . xe-x cos(x) and xe-x sin(x) are solutions of a fourth-order equation for which the indicial equation has the roots r = -1 ± i of multiplicity 2 => (r + 1) 2 = -1 => r2 + 2r + 2 = 0 and (r2 + 2r + 2) 2 = 0 is the indicial equation => (D 2 + 2D + 2)2 is the annihilator of xe-x cos(x) of minimal degree. Hence, (D + l)3y = xe-x cos(x) => (D2+2D+2) 2 (D+l) 3 y = (D 2+2D+2)2 [2xe-x cos(x)] = 0 =>y = c 1 e-x cos(x)+ c2 e-x sin(x) + C3Xe-x cos(x) + C4Xe-x sin(x) + cse-x+ c5xe-x+ C7x2 e-x => YP =(a+ bx)e-x cos(x) + (c+ dx)e-x sin(x) , v; = be-x cos(x) - (a+ bx)e-xcos(x)- (a+ bx)e-xsin(x) + de-x sin(x)­ (c+ dx)e-x sin(x)+ (c+ dx)e-x cos(x) = [(b- a+ c) + (d- b)xJe-xcos(x) + [(d- a - c) - (b+ d)xJe-x sin(x), v;= (d - b)e-xcos(x)- [(b- a+ c) + (d- b)x]e-xcos(x)[(b- a+ c)+ (d - b)x]e-xsin(x) - (b+ d)e-x sin(x)[(d- a- c) - (b+ d)xJe-x sin(x)+ [(d - a - c) - (b+ d)xJe-x cos(x) = [(2d - 2b - 2c) - 2dxJe-x cos(x) + [(2a- 2b- 2d)+ 2bxJe-x sin(x) . v;' = -2de-x cos(x) - [(2d - 2b- 2c) - 2dx]e-x cos(x)[(2d- 2b - 2c) - 2dx]e-x sin(x) + 2be-x sin(x)[(2a - 2b - 2d)+ 2bx]e-x sin(x) + [(2a - 2b - 2d)+ 2bxJe-xcos(x) = [(2a+ 2c- 6d)+ (2b+ 2d)xJe-x cos(x) + [(6b+ 2c- 2a)+ (2d-2b)xJe-x sin(x) , and v;'+ 3y;+ 3y; + YP = 2xe-x cos(x) => [(2a+ 2c- 6d)+ (2b+ 2d)x]e-xcos(x)+ [(6b+ 2c- 2a)+ (2d-2b)x]e-xsin(x) + 3[(2d - 2b - 2c) - 2dxJe-x cos(x) + 3[(2a - 2b- 2d)+ 2bx]e-xsin(x)+ 3[(b - a+ c) + (d - b)x]e-x cos(x) + 3[(d- a - c) - (b+ d)x]e-x sin(x)+ (a+ bx)e-x cos(x) + (c+ dx)e-x sin(x) = 2xe-x cos(x) => [-(3b+ c) - dxJe-x cos(x) + [(a- 3d)+ bxJe-xsin(x) = 2xe-x cos(x) => b = 0, d = -2, a - 3d = 0 =>a = -6, 3b + c = 0 =>c = 0 => YP = -6e-x cos(x) - 2xe-x sin(x) => 2 y = -6e-x cos(x) - 2xe-x sin(x) + cse-x+ C5Xe-x+ C7X e-x.

C.5

Chapter 5 Solutions

Exercises 5.1 - page 168 -5 6 x = -5x+ 6y 1. { y ; = -3x + 4y } => A= ( -3 4 ) ' =>,\1 = 1, A2 = -2. For ,\ 1

= 1,

For ,\2

= -2,

( � ( �

I ,\+ 5 3

=� ) ( � ) = ( � ) =� ) ( � ) = ( � )

-6 ,\ - 4

=>a= b =>

v1

=>a= 2b =>

I = (,\ _ l) (,\+ 2) = 0 = ( � ). v2

= ( � ).

391

C.5. CHAPTER 5 SOLUTIONS x1(t) = el ( � ), x2(t) = e-2l ( � ), x(l) = C1X1(t) + C2X2(t) ⇒ x(t) = ciel + 2c2e-2t, y(t) = c 1 e l + c2e-2t.

2. { ;: : ��; y } A 1 = 1, A2 = 2. For A1

=

1, (

-



⇒ A= -

( -� � ),

I

A; 3

1 �

I

= (A - l)(A - 2) = O

� ) ( �)= ( � ) ⇒ b = -2a

- � ) ( �) = ( � ) ⇒ b For A2 = 2, ( �

=

-a

⇒ v2 =

et ( , x2(l) = e2l ( ! ), x(t)= C1X1(t) 1 !2) t 2l l x(t) = c1e + c2e , y(i) = -2cie - c2e2L.

x1(l)

6.

{

=

-

⇒ vi =

( !2).

(

+ C2X2(l)

!l ). ⇒

=X y , x(O) = , x = 2, y(O) = 1 ⇒ A = ( 4 2 y' = 2x + 4y } (O) A 1 = 2, A2 = 3. I _::-2 A� 41 = (A - 2)(A - 3) = 0 ⇒ A1 X

1

1 -1 )

( 2 ). 1



392

APPENDIX C. SOLUTIONS

9. { �; : � 2x + 3y ' } z' = -x + y + 2z ,\-1 0 0 2 ,\ - 3 0 1 -1 ,\-2

u =� 0 0) n

ForA1 = 1,

v,

=

⇒ A = ( -� � � ) , -1

1 2

= (,\ - 1)(,\- 2)(,\ - 3) = 0 ⇒

J

⇒ x1(t) = e'

)

(

) =

0

)

>-1

=

1,

>-2

= 2,

,\3

= 3.

⇒ a = b and a - b - c = 0 ⇒

393

C.5. CHAPTER 5 SOLUTIONS

-1 0 2 11. A= ( -2 1 2 ) ,

0 0 1

).+1 0 ). - 1 2

0

0

-2 -2

.A. -1

= (). - 1) 2 (). + 1) = 0

⇒ ). 1 = 1,

394

APPENDIX C. SOLUTIONS

0 -1 ) 13 _6 ,

. A l = >.2 +6>.+ 13 = 0 ⇒ >., >.= -3 ± 2i. I _ 13 >.+6 For >. = -3+ 2 , ( � t/ ) ( � ) = ( � ) ⇒ b = (3 - )a 3

13. A = (

i

v= ( � ) 3 2

i

x 1 ( t ) = e _3 t

x(t)

I

(

i

⇒ z(t) = e-3t [cos(2t)+i sin(2t)]

2 -2 ), 14. A = ( l O For>.= 1+ , (

v = ( 1 �i)

x1 (t) =e

t

(

2i

i

=}

[( � )+ ( � )] ⇒ 2 cos(2t) sin(2t) t x (t) · =e _3 3cos(2t)+2sin(2t) ) ' 2 3sin(2l)- 2cos(2t) ) '

= C1X1(l) + C2X2(t).

i

1 2i

I

-

2

2 A -2 1 = >. - 2>.+2= 0 -l >.

�t

i

l

!i

(

⇒ >., ->.= 1 ± i..

) ( � )= ( � )

⇒ z(t) = et [cos(t) + isin(t)]

t cos(t)-sin(t) () () ) , x2 t =e cos t

(

⇒ a= (l + i)b ⇒

i

[ ( � ) + ( �) ] sin(l)+cos(l) . ), sm (·l)



C.5. CHAPTER 5 SOLUTIONS cos(l) - sin(l) sin(l) + cos(l) _ X(t) = et ( ). cos(t) sm(l)

17. A =

(

0 0 1 ) 0 1 0 ,

-1 2 0

,\ 0 -1 0 ,\-1 0 1 -2 ,\

395

APPENDIX C. SOLUTIONS

396

0 ,\ 0 0 0 1 -1 0 ,\ - 1 1 0 1 -1 0 l 1 0 ) ' _ 0 -1 ,\ - 1 0 0 1 0 ,\ 0 0 0 = l t i. l 1 i : = (� � � _=For ,\ 1 = i , ) 1 i �l 1 0 0 i d 0

0 0 lB. A = ( 0 -1 i ⇒ A , °Xi =

v,

(a) O) r( n

±(\ ,\2 ,12

( = �) ⇒

z,(t)

=

[cos(t)+ i sin{t)]

cos(t) X1(t)= (

- sin(t) l+ i

ForA2 = l+i,

⇒ v, = (

) ,x2(i)= (



n

(



1

sin(t) �

cos(t) -1

O O � � � l ) 0 0 l+ i

+i

⇒ b= c = 0, d=ia ⇒ � ( )]



(a) r ( D u) J

).

:

d

⇒ z,(t) = e•[cos(t)+ i sin(t)]

0

=

( �) 0

+i

⇒ a=d=0,b=ic ⇒

397

C.5. CHAPTER 5 SOLUTIONS 0 0 sin(l) l cos(l) X3(l) = e ( co (l) ) ' X4(t) = e ( sin(t) ) ' � Q 0 cos(t) sin(t) 0 0 0 -e l sin(t) e l cos(t) X(t) = ( ) 0 0 e l cos(t) e t sin(t) · - sin(t) cos(t) 0 0 l

19. {

t:;

m1 = 2.

-

+ 3Y } => A= ( � { ),

I

A� l

>,

-=_\ I = (>- - 1)

2

= 0 => A1 = 1,

For A1 = 1, ( � -�) ( �) = ( �) => b = 0 => v1 = ( � ) =>

x1(t) = e t ( � ) , x2(l) = e l (tuo + u1), uo = vi, (A - >-1I)u1 = uo => t ( � � ) ( �) = ( � ) => 3b = 1 => ll1 = ( 1�3 ) => Xz(t) = e ( 1;3 ) , x(l) = c1xt(l) + c2x2(l) => x(l) = e t (c1 + Czl), y(l) = ½cze l .

398 x

22. { ;: : ! +-2i - z } z' = 3y - z >.- 3 1 0 -1 >. - 2 1 0 -3 >. + 1 FoTA,

= 2,

=> x 1 (t) FoTA, v, =

= e"

= 1,

(�)

(A - A,I)u,

x3(t)

(

= e'

⇒A= (

3 -1

=i j D ( n 0)

n

=

=> 0

uo =>

( ;l �

= e'

\ ),

=

( � ) , x,(t)

=> b

= 1, C

=>

= 2a , c = 3 a

= e'(t11o + u1), u0 = v2, =

x(t)

>-2

=b=

=! =i O ) 0 ) U -i =D O ) U )

=> x,(t)

=

O

= (>.- 1)2 (>.- 2) = 0 ⇒ >.1 = 2,

( D

(

i -� -� ) ,

APPENDIX C. SOLUTIONS

=> u

= c,x1(t) + c,x2(t) + c3x3(t) =>

,

x(t) = c1e2t + et [c 2 + c3(t + 1)], y(t) = c1e2t + et [2c 2 + c3(2t + 1)], z(t) = c1e2t + et (3c 2 + 3c3t).

m2

= 2.

v, = (

D

=>

=

0)

=>

0 � n O) 0)

399

C.5. CHAPTER 5 SOLUTIONS

=>

x3 (t)

= e"

(

=

1�2) , x(t)

=> c

=

0, 2b

=

1 => u 1

= c1 x 1 (t) + c2x2 (t) + c3 x3 (t).

= (

1:2) =>

400

APPENDIX C. SOLUTIONS

>-+2 -1 0 1 >0 0 0 >-+1

0 1) -2 27. A = ( 1 -1 0 , -1 0 0 m1 = 3.

FofA,

=

=> x 1 (t)

=

-1, ( -: � : D

= e-' (

!),

(] � D O)

x2 (t)

u) u) n

-1 >- + 2 0 -1 >- + 1 0 1 >0

x,(t)

= (

=> a

= c = 0 => v, = (

n

= c'(tuo +u,), u0 = v,, (A - A,I)u, = Uo =>

a

= = C

I

=>

u,

= (

D

=>

= e_, ( : ) , x3 (t) = e-• (lt2 u0 + tu 1 + u2 ), (A - A 1 J)u2 = u 1

=>

=>

C.5. CHAPTER 5 SOLUTIONS

(

29· A=

1 -1 0 0

1 0 1) 1 -1 0 0 1 1 ' 0 -1 1

401

A - 1 -1 0 -1 1 A-1 1 0 0 0 A - 1 -1 0 0 1 .A.-1

402

APPENDIX C. SOLUTIONS

Exercises 5.2 - page 186

O 4 4 _ _ _ 4cos(2t) 2. Af ->-2 +4- 0⇒,\,>--±2i, ( -1 0 ) , (l)- (-2sin(2t)) ' I ,\1 -,\ , _ _ - _ · i v = � l , z(t) = [cos(2t) + i sin(2t)] [ �l + i � ) ( ( ) ( )] , X(t) = -2sin(2t) 2cos(2t) X(t)_ 1 = .! -sin(2t) -2cos(2t) 2 ( cos(2t) -2sin(2t) ' ( -cos(2t) -sin(2t)' ) ) O (2t) t) -sin(2t) 4cos(2 -2cos = X(t)-lf(t) = .! 2 ( cos(2t) -2sin(2t) ) (-2sin(2t) ( 2' ) )

C.5. CHAPTER 5 SOLUTIONS J X(l)- 1 f(t) dl = (

it ) ,

403

+(

x(l) = ( -2 sin(2t) 2c�s (2t) ) [( c 1 ) c2 -cos(2l) - sm(2t) = ( -2c 1 sin(2t) + 2c2 cos(2l) + 4tcos(2t) ) -c 1 cos (2t) - c2 sin(2l) - 2l s in(2i)

0 2i )]

404 (t + 3)et _ - ( 2(t + l)e 2t 6. A = ( �

-

)

APPENDIX C. SOLUTIONS .

�), f(t) = (

i ),

Xo = ( �),

I

:

1

;

I

= A'+ 1 = 0 c>

A,X = ±i, v = (; ), z(t) = [cos(t) +isin(t)] [ ( �) +i ( �)]. - sin(t) c?s(t) 1 - sin(t) c?s(t) X -l ( cos(t) sm(t) ) ' (t) f(l) = ( cos(t) sm(t) ) ( 1 ) = cos(t) - s� n(t) s� n(s) + cos(s) ( cos(t) + sm(t) ) '}0 X(s)-if(s) ds = ( sm(s) - cos(s) ) = 0 _1 _ 0 1 2 _ 1 sin(t) + cos(t) - 1 ( sin(t) - cos(t) + l ) ' X (O) x0 - ( 1 O ) ( 1 ) - ( 2 ) ' 1 sin(t) + cos(l) - 1 - sin(t) cos(t) x(t) = ( cos(t) sin(t) ) [( 2 ) + ( sin(l) cos(t) + 1 )] 3cos(t) - 1 sin(t) + co s (t) sin(t) co s (t) ( ( ( ) ) cos(t) sin(t) sin(t) - cos(t) + 3 3sin(t) + 1 ) · X(t) =

t I

f

-1 _ -2 _ , f(·t) _ - e -4 t 7. A ( 4 6)

( 12 ) ,

x0

_ -

o ),

( 1

I

). + 2 1 _4 ). + 6

I -(

- >. + 4)2

= 0 c> A1 = -4, v1 = ( ; ), (A - A1/)u1 = Uo = v, c> u, = ( �l ), t 1 1- 2t l 1 e-4t = X(t) = e-4 t 2 2t - ), X(t)-lf(t) = e4t ) 1 2 -1 2)

( � ), l

(

X(s J-' f(s) ds = ( ; )

1: i ) , = (

(

(

X (0)- 1 xo = ( ; _ � ) ( � )

+ = ( �l ) ' x(t) = e-« ( ; 21 � 1 ) [ ( �l ) (

t )]

= e- ( ) 4' � ·

C.5. CHAPTER 5 SOLUTIONS t

O

=

(

4 05 t

0

2e- l - 2

),

Chapter 5 Exercises - page 186 l. y" - 2y' + y

=

0.

(a) y = e rt ⇒ 2r - 2 r + 1 c te + 2c et t . y ( l) = 1

(b) X1 c ) A (

=

y and x 2

y'

1) 2

= 0⇒ r = 1 ⇒

U1

x1(t)

e t ( � ) , x 2(t)

=

e l(tuo

= ( �) ⇒ x 2(t) = e t ( = (c1 + 2c t) e t, x 2(t) = [ 1c +

(d) W [ 1Y (t) Y2(]l) W[x1(t) x 2(t)]

=

I

=I

=

Y1()l

e t, 2Y (t)

=

te l,

⇒ x� = y' = x 2 and x; = y" = 2y' - y = 2x 2 - x1.

= ( _ � � ), I ; ,\ -=_\ I = ( ,\ -

⇒ x1(t) = ⇒

=

= (r -

Y ( t) Y�1 ( ·t)

Y 2()l y�(t)

x1(t) x 2(t)

=

0 =?

=

u1 ), uo

l1 l) ⇒ x(t)

=

,\ 1

1, v1

t

el e t e (t+ l) e l l

I = I ::

e

(t: : ) e t

2. Let X(t) be a fundamental matrix for the system x'

= (�)

vi, (A - A1I ) u1

=

C1X1(t)

I -- e

2t

I=

=

+ C 2X 2(t)

+ l)]e t, and x1(i) = y ( t).

2c (t

1-1 -

+

1 )2

uo



and

e 2t

=

W[y1(t) Y2(t)].

= Ax.

(a) Since X(o)- 1 is nonsingular, Y()l = X(t)X(o)- 1 is a fundamental matrix by Theorem 5.4, and Y(0) = X(0)X(o)- 1 = I. The solution of x' = Ax, x(0) = xo, is x(t) = Y(t)x o.

(b) Since X(to )- 1 is nonsingular, Z(t) = X(t)X(to)- 1 is a fundamental matrix by Theorem 5.4, and Z(t0) = X(ol )X(io )- 1 = I. The solution of x' = Ax, x(to) = xo, is x(t) = Z(l)x o. Part (a) is the special case where ot = 0.

406

APPENDIX C. SOLUTIONS ), (a) A= ( � -�

I

,\

�/

,\

12

1

= (,\ - 2)(,\ + 2) = 0 ⇒ ,\ 1 = 2,

(�

),x 2 (t) = e- 2t ( � ), = -2, v 1 = ( � ), v 2 = ( � ),x 1 (t) = e2t x(t) = c 1x 1 (t) + c2x2 (t) ⇒ x(t) = 4c 1 e2t, y(t) = c 1 e2t + c2e- 2t. (b) x' = 2x ⇒ x(l) = c1 e2t ⇒ y' = c1 e2t -2y ⇒ y'+2y = c 1 e2t ⇒ (e2t y)' = c 1 e4t ⇒ e2ty = ¼c 1 e4t + c2 ⇒ y(t) = ¼c1e2t + c2 e- 2t and, replacing the arbitrary ,\2

constant c 1 by 4c 1 , the solution coincides with the one in part (a).

= x - 2 y ⇒ y" = x' - 2y' = 2x - 2(x - 2 y) = 4y ⇒ y" - 4y = 0. (d) y = e ⇒ r2 - 4 = (r - 2)(r + 2) = 0 ⇒ y(t) = c 1 e2t + c2e-2t ⇒ x(t) = 2y(t) + y'(t) = 2[c1e2t + c2e- 2t] + [ 2c1e2t - 2c2e- 2t] = 4c 1 e2t, as in

( c) y'

rt

part (a).

4. x'

= Ax, A =

(

-7 12 ) 7 .

_4

C.5. CHAPTER 5 SOLUTIONS

407

408

APPENDIX C. SOLUTIONS

lt2 lt3 2 6 t lt2 ) 1 t 1 0

=

7 If c1x1(t)+c2x2(t) 0, then t = 0 ⇒ c1a+c2b = 0, and l = 213 . ⇒ (er+ c�)b =0 ⇒ C1 = C2 = 0 since b-/- 0.

C.6

Chapter 6 Solutions

Exercises 6.1 - page 205 1. {5, 7, 9, 11, · · ·} = {2n + 3}�=l 2. {-6, -3, 0, 3, · · ·} = {3n}�=- 2 1 3. { 1, -21 , 3'

_ _! 4,

. } = { (- 1:-1} .. n=l OO

} 1 .!.9' 116' 4. { 1, 4, ... = { n\} :;=l 1 l 1 5. {1, 3' 9' 27' ... } = { 3�} :;=0 6.

{o'

00 1 - 12' 1 - .3' ! 1 - 14' ...} = {1 - 1n} =l n

2 9. {../2nn++1 l}

oo

n=O

2 v'55' v'Io 10 \/'17 17 ···} ={1, /2 3 5 7 , 9 '

⇒ c2a-c1b =0

409

C.6. CHAPTER 6 SOLUTIONS

+ n- l n+ 1 11 . bn = an+l + 2nn fOr a11 n > _ 1 =? bn+l _ 0 and bn-1 - an 2'11 - an+2 + 2n+1 f01· n > for n 2:: 2.

1 l 1. oo _ l _ 1 oo _ l oo 12 · { 3l ' 5' 7' g, · · · } - { 2n+l } n=l - { 2n+3 } n=O - { 2n-l } n=2

---t 0 =} { an}

13. an = n1,,

14. an = n-\ _ = -n3

16. an = 2-n 11. an = 18. an = 19. an = 20. an =

---t -oo =} { an}

---t 0 =} { an}

15. an = }n

=



converges. diverges.

converges.

---t 0 =} { an} converges.

e90 r ---to=⇒ {a } converges. < � = (-�r ---t O ⇒ {a } converges. 2

n

n

n

-;

; ( ! )n

lim a (-!r, and n-+oo

=

n

does not exist⇒ {an} diverges.

}n ---t O =} { an} converges.

1, 21. an = 10n0

000 , 000 , 000, ooo 000000000 1

---t O =}

{an} COnVerges.

---t 3 =} { an} converges.

22. an = 3 - ;2

+

23. an = n + 3n1

---t 00 =} { an} diverges.

}n

3n2 -2n-3 _ 3-¾-� 24 · an = 2n 2+3n+l , - 2+1 + J n �

---t 23

=? {an} Converges.

3 -2n-3 25. an = 3n 4n2 +n+l

-2n-3 26. an = 3n 2n3 +2n-l 2

---t 1 =} { an} converges.

3n2 -2n-3 27. an = J2 n4+3n+l 3n2 -2n-3 28. an = J2n 5-2n+l 3n2 -2n 29. an= J2 n3-n+l

3 - 2 - 3 ;r72 � � 2-3'-+;:!sn n



---t O =} { an}

converges.

410

APPENDIX C. SOLUTIONS

✓n2 + n - ✓n2

31. an=

+n =

n-

2 ✓n

3 ✓n

+ 1 - n = ( ✓n3 + 1

=

(n -

{an } converges.

33. an

=

2 n Jnz+n+JnT=n = = ( ✓n2 + n - ✓n2 - n) J n2+ n+Jn2 -n Jn + n+Jn2 -n

n -2 n 34. an - (3nn)2 -- ( - 32 ) 1n2 Hence, { an } converges. 35. an - -nconverges. _

2

-+ 1 => { an } converges.

Ji+¾! ✓i="¼

32. an

-n

(-lr I

-t

-t

✓n

2

v'n2+n + n ) nn++J n2 + n

Jn3"+l+n - n) M, n +l+ n

=

-n

A

n . 0 by Theorem 6.1 since ( - 32 )

2 In(x)

=

X

X-+00

38.

=0

lim �

n-+oo e

39. an

=

lim �

= 0=>

lim [In( �)]P n

n-+oo e

lim lX

X-+00

X x .

e

=

- n O and ( nl)

0=> lim

n-+oo

[In(nll 2 n

= 0 by Theorem 6.3.

By L'Hopital's rule, lim �

-t

0.

O by

-t

x:. e

By L'Hopital's rule, lim

x-+oo

x: e

=

Hence, 0=>

lim _!_ X

X---+00 e

x-+oo e

2 lim 3�

x-+oo

e

=

= 0 by Theorem 6.3. Hence, {an } converges.

lim 6;

x-+oo

e

·f l p n q/p) -t 0 . Let 0 1•f' and on1 y 1 {an/ } - lnn( A P · l/x · � · � ln(xIP) - 11m 11m · l's ru 1 e, 11m B y L'Hop1ta = X--->00 -= 0=> qxq/p X---+00 X X---+00 !l.x P

> 0 , q > 0 . an

-t

= 0 by Theorem 6.3. Hence, { an} converges.

I ( ) = x 1 I = e-x-.By L'Hopital's rule, lim � = lim 1- = 0 lim n 1/n = 1 by Theorem 6.3. Hence, {an } converges. = e0 = 1 => n---+oo

41. an = n 1 In_ Let f(x) => lim e¥ x-+oo

=

lim n!

p

40.

n-+oo

-t

by Theorem 6.3. Hence, { an} converges.

!. Let f(x)

x-+oo e

l

=

Let f (x)

n

e

O and 1n2

= [ln(xX )] 2 . By L'Hopital's rule, lim [ln(xX )]2 = lim [2 ln(x)J¾ X--->00 X--->00 1

lim �

X-+00

{an } converges.

-t

-l n 0 by Theorem 6.2 smce I ( )n 11 -- _ l � 0. Hence, {an } n

n)] 2 . Let f(x) 37. an = [ln( n lim

-t

i+Ji+¾

n3/2+_1_2 _ nl/2 3 n2 � i+.J,,.+ _1_ n +l+n n� nl/2

.!. - (-l) n l - -l n 36. an = (n ) -t O by Theorem 6.1 since .!. n n n Theorem 6.2. Hence, { an } converges.

=

-1

n+Jn2+n

x

ln(x)

X-+00

X

X-+00 X

42. an = [ln(n)] 1/ 1 u(n)_ Let f(x) = [ln(x)]1 /ln(x) = elnfxi ln [ln(x)]_ By L'Hopital's I I . . . . _._ ln(x) Tii(xj I l _ 1 ru 1e, 11m J.n[Inln(x · )] _ x _- 11m r - 0 _._ .....,- 11m e l n [ n(x)l = l .....,- 11m ---rX n() X x () X-+00 X---+00 X---+00 x-+oo lim [ln(n)]1 /ln(n) = 1 by Theorem 6.3. Hence, {an } converges. n---+oo

C.6. CHAPTER 6 SOLUTIONS

411

43. an = [ln(n)]1/n_ Let f(x) = [ln(x)]1/x = e ½I n (In(x)]_ By L'Hopital's rule, lim I n[lnX (x)] = lim -l 1(-) = 0 =} lim [ln(x)]l/x = 1 =} lim [ln(n )Pin 1 by X->(X) X ->(X) x n x X->(X) n->oo Theorem 6.3. Hence, { an } converges. 1 n) n) l n (n) = e for all n � 2 ⇒ hm an = lim e =e. ln(4 4. an= n1/1 n (n) =e ] n (n1/ ln( ) = e -

n---+oo

Hence, {an } converges.

n---+oo

< cos(n) < 1 =} _l < cosn(n) 4 =} - 1-2-3-. 1-2 3 4 n-1 n - 2 n 2n . (n-l)·n 3e 3e 0 :S an :S 2n and lim O = lim 2n = 0 ⇒ lim a n = 0 by Theorem 6.4. Hence, n---+oo n---+oo n---+oo { an } converges.

_ 47. an -

ne

n!

_ (2n)! _ 1 . 1·2·3--·(n-l)·n . (n+l)·(n+2)---2n < l . 1 . 2n _ 1 =} 0 < a < :;:;-1 and 48. a n - n- :;:;n· n-n--· n· 2n n2 n - 2n n·n·n--·n·n n - 2n :;:;lim O = lim .n!. =0 ⇒ lim an = 0 by Theorem 6.4. Hence, {a n} converges. n---+oo

n---+oo

n---+oo

49. an = 23n > 2n3+ 1 =an+1 ⇒ { an} is decreasing for n � 1. 50. an = 2 - ¾ < 2 - n!l = an+l ⇒ {an} is increasing for n � 1. 1 1 1 - > _.!_ ⇒ 1 - -- > 1 - l ⇒ .!. ⇒ -51. n + 1 > n ⇒ n+l < n n+l n n+l n -1 1_ n+l > 1 n ⇒ { an} = { 1 n is increasing for n � 2.

:::1

:::1}

52 . an _ - l nn(n)' Let f(X ) -_ l nx(x)' Then f'(X) _ increasing on (e, oo) ⇒ {an} is increasing for n

l (n (x)-1 l n (x)]2 > 0 c10r � 3 > e.

X

> e

=}

f lS

nl n) x l x) 53. a n = ;� . Let f(x) = ;! = xe-x ln(x). Then f'(x) =e-x[ln(x)-x ln(x)+l] = e-x[(l - x) ln(x) + 1], f'(l) > 0, f'(2) > 0, and f'(x) < 0 for x � e ⇒ {a n } is decreasing for n � 3 > e (but not for n � 2).

54. an = s,��n) =} { an} is not monotone. 55. a n = ✓nn+l � 0 ⇒ { a n} is bounded below by 0. lim � = lim � = n->oo v n+11 n->oo V l+,;;n ⇒ { a n} is not bounded above ⇒ {an} is unbounded.

oo

APPENDIX C. SOLUTIONS

412

L�}

56. an = 2� - ln(n) is decreasing since both and {- ln(n)} are decreasing. Since a1 = ½, an �½ for all n 2'. 1 ⇒ {an} is bounded above by ½. Since lim an = -oo, {an} is not bounded below ⇒ {an } is unbounded. n--+oo

= 1_-\ ⇒ {an} is increasing for n 2'. 1 by Exercise 51 with n replaced by n +1. a1 = - 2 and lim an -1 ⇒ -2 �an �-1 ⇒ {an} is bounded. n--+oo

57. an

n+l

=

58. an = n2 e-n. Let f(x) = x2 e-x. Then f'(x) = e-x[2x - x 2 ] = xe-x(2 - x) ⇒ f'(x) > 0 on (0, 2) and J'(x) < 0 on (2,oo) ⇒ f(x) � f(2) = 4e-2 on (0,oo) ⇒ 0 �an � 4e-2 for n 2'. 1 ⇒ {an } is bounded.

=

59. If n n+l

l, then 1

1

n

n(n+1)(2n+l) 2 Then Assume '\""' u i -6 6 i=l n(n+l (2n+l ) n(n+l)(2n+ )+6(n+l ) 2 _ 2 � ; + (n + l) =

= L i2

1(1+1)(2+1)

i= l

L i 2 = L i2 + (n + 1)2 i=l i=l (n+l )[n(2n+l)+6(n+l)] _ (n+l)[2n2 +7n+6] -_ (n+l )(n+2)(2 n+3) _ (n+l)[(n+l)+l)[2(n+l)+l] 6 6 6 6 n

60. a1 = 1 < 3, and if an < 3, then an+l = 1 + Fn < 1 + v3 < 3 ⇒ an < 3 for all n 2'. 1, by induction. Hence, {an}�=l is bounded above. a1 = 1 < 1 +JI= a2 , and if an < an+l, then Fn < � ⇒ 1 + Fn < l + � ⇒ an+l < an+2 ⇒ {an }� =l is increasing for n 2'. 1, by induction. Since {an }� =l is increasing and bounded- above, it converges by Theorem 6.6. Let L = lim an . Then

an+l

L

=

= 1 + Fn

3

L

=?

=

l+

v'I

> ±-/5_ Since an_ > l for all n 2

61. If lim an n�oo

=

n�oo

⇒ (L - 1) 2 = L ⇒ L 2 - 3L + 1 l' L > 1 and ' hence' L = 3+2Js_

1... ⇒ L =1- -£2 ⇒ L2 -L+ 2 L, then an+l = 1- .an

=

=0⇒ L=

0

l

=?

±p

is complex, which is impossible since every an is real. Hence, L does not exist, i. e. , {an} diverges.

=

2 ⇒ 1 � a1 � 2, and if1 �an � 2, then½ � a� � 1 ⇒ -1 �- a� � -½ =? 1 � 2 - � � ⇒ 1 � an+1 � 2 ⇒ 1 � an � 2 for all n 2'. 1, by a 1- ⇒ 1- ⇒ _.1... . < induction. a1 = 2 > �2 = a2 , and an > an+l ⇒ .1.. > -an an an+l an+l 1 2 - .1... > 2- - ⇒ an+l > an+2 ⇒ an > an+l for all n 2'. 1, by induction. an an+! lim an ⇒ L = 2 - z ⇒ Hence, {an } covergcs by Theorem 6.6. Then L n--+oo L2 - 2L + 1 = 0 ⇒ (L - 1) 2 = 0 ⇒ L =1.

62. a1

!

Exercises 6. 2 - page 240 1. 2.

L 00

n=O

1

2

n,

SQ

L= ¼, 81 00

n l

= 1, S1

=

1 +½, 82 = 1 +½+

¼, 83 = 1+½+ ¼+ ½, 84 = 1+½+¼+ ½ + /6 •

=1, 82 = 1+½, 83 = 1+½+½,

84

= 1+½+½+¼, 85 = 1+½+½+¼+½-

413

C.6. CHAPTER 6 SOLUTIONS . S 3 . S = 1lffi n-+oo n

7. � L; _!_ 3n n= 2 8. 9.

11.

1·lffi 3n-l n-+oo 2n+l

=

= = n� L; (.!3 ) n = � 1-3 =2

I: 5. 23n 3-2n = 5 I: 00

00

n=O

n=O

L=l5. 3 00

2n

n

L=O 4

2 -4 n

00

C96 n=l

= 5L 00

= nL ( 2n� l =O

n2}8n+3

l6.

(�r = 51�!!9 = 45.

00

n

3

= 2- .

l - 2 n�3 => L 4n 2 }8n+3 n=O

r=

9

5l�"._g__ 16

=

4;.

- 2 n�3 ), Sn

=

(l -

=

E.� S

n

=

1.

½)+(½- ¼)+· · ·+(2n�l - 2n�

2+ 2+ 12. I: 2 ✓nn 3 diverges by then th-term test because lim 2 ✓nn 3 n->oo n=l

13. 14.

16. 17. 18.

I:

= 2 -=/= 0.

00

· cos(n) • . 2( n ) -=/= 0. by thenth -term test because hm 2 +sm +cos(n) . 2c n ) diverges sm n > oo n=O

2

L=O [1 - (-l) 00

n

L=l 00

n

n2Jn

=

L 00

n=l

L=l :Ji = L=l 00

00

n

I:

n

00

n=l

] diverges by then l /1-term test because lim [1 - (-lt]-=/= 0. n->oo

n

nr�i ,

n

ni/ 2

converges because it is a p-series with p = �

> 1.

i/ 2 converges because it is a p-scries with p = � > 1.

r > 0, converges because it is a p-series with p = r + 1 > 1.

3)

=

414

APPENDIX C. SOLUTIONS 00

19. I: nr-l n=l

20.

=

L 00

n=l

r, r � 0, diverges because it is a p-series with p

n/

ft 2x1:_

00

diverges by the integral test because I: 2n1:_ 1 n=l 00

2 21. I: 2ne-n converges by the integral test because

n=l

2

- e-x ,� = e-1 < 00. 22.

1

dx

=

=

l - r :S 1.

= oo.

½ ln(2x-1)/�

ft 2xe-x

2

dx

=

L n[ln(n)]P, p < l, diverges by the integral test because ft x[ln(x)]P dx = n=2 00

1 P 1�P [ln(x)] - l;i = oo. 00

23. I:

n=2

1 �P [ln(x)]

24.

p>

n(In(n)]P,

(a) s

=

1 P -

l;i

l,

converges by the integral test because

=

P

(b) Rn :S

.;;4 �

f; ;4

00

_1_ 25. (a) s = � u n,/n n=l A :S R3 :S

(b) Rn :S 26.

� 1 [ln(2)] 1 -P < oo.

00

L n=l

ft x[ln(x)]P dx =

s2

= 3 + 136 = f!,

ft

;4

dx :S

R2

:S

ft

;4



S3 =

1

+

1 __

2v'2

+

_1_

3v13'

1, i.e., 1 :S R3 :S 1·

f; x-3/2 dx =

J4oo x-3/2 dx -< R

)n _1

:S

R2

:S ½-

3

}n :S 0.0001 if n � 400, 000, 000.

converges since it is a p-series with p

n=2

2\

J3oo x- /2 dx ⇒

< 3 -

00

00



dx = ;3 :S 0.001 if n � 10.

L nfn+3 converges by the comparison test because nfo+3 n=O

27. I:

dx

=�>

l.

diverges by the comparison test because

diverges since it is a p-series with p

=

:S n}/ 2 and

>

fo-1

n1\2

and

½ :S 1.

L n}/ n=l 00

L

n=l

2

n/12

!: 1 converges by the comparison test because 4!:1 4 n=O converges since it is a geometric series with r = ¾ and lrl < 1. 00

28. I:

00

29. I: n=O

2

J_

3

converges by the comparison test because 00

J_

2

3

:S

L 2 2n n

n � 2 and I: ;2 converges since it is a p-series with p = 2 > 1. n=l

.;;2 for

415

C.6. CHAPTER 6 SOLUTIONS 30.

E 2;t 1 converges by the comparison test because 2;t 1 00

n=O

rnr

00

and E n=O 31.

00 E

2n

n=O

n � 0 and lrl < 1. 32.

E 00

n=l

E 00

I: (�r

> n!n

00 .jn - - -jn-l y'n l 1 33. " L.,, n+l · an - n+l rv n vcn - bn n=l

diverges

2

< 1.

2 2n = --;n

converges since it is a geometric series with r

n=O



= �

00

-->... --r

L.,, bn 1·--->oo lm an bn - 1 ' 0 < 1 < 00, and " n=l

n

⇒ E an diverges by the limit comparison test.

OO 34. " n3 +2n-l an L.,, n4 +n3 +2· n=O

· n3 +2n-l rv !I:_3 n4 n4 +n3 +2

=

=

l n

and

for n � 1

00

n=l

for

= 1 � 1.

¼ diverges since it is a p-series with p

n=l

�'t for n � 0

2

2n n 2n+2n � � e!

-fo+n diverges by the comparison test because -fo+n

and

=

converges since it is a geometric series with r = � and lrl

converges by the comparison test because

n

e!

n n � 2 ;;;,2

=

bn



lim ab n n

n--->oo

=

1, 0

< 1 < oo, and

E bn diverges ⇒ E an diverges by the limit comparison test. 00

00

n=l

n=O

OO .3� 3� 1- - - � rv n2 / 3 35 " � · L.,, n2 +n -l · a n - n2 +n -l n2 n4/3 - bn n=l

=}

- 1, 0 < 1 < 00, and l'lm !!n. b -

n-+oo n

E bn converges ⇒ E an converges by the limit comparison test. 00

00

n=l

n=l

00

=

1 . a 36. " � n2/3 Jn(n) n n 2

Eb 00

n=l 00

diverges

n

n-+oo

Ea

n

00

n=l

n

=

ln(n) _ l'1m """""I/2 an T hen 1.1m b n n-+oo n

oo and

lim ln[(l

+ ln t] = ln(e) = 1, 0 < 1 < oo, and

n=2

n

converges by the limit comparison test.

0 and

n-+oo

converges by the limit comparison test.

00

n-+oo

n l /3 1· ) . :_, � ln(n n

n=2

Ea

converges⇒

=

1· a 1 n. Then n.:_,� b:

00

ln(n) Let b n n2 .

37. E ��)n=2

=

⇒ E an diverges by the limit comparison test.

l

lim �n

1 n2 /3 Jn(n). Lct bn

Eb ex,

n=l

n

Eb 00

n=l

n

converges ::::}

416

APPENDIX C. SOLUTIONS

39 � . L.,

n=2

3

00 L bn n=l

40.

3 Let bn - 4/3 1 . Then 1·1m an - [ln(n) ] -n n3/2 .

[ln(n)] n 'n . v ·,t,

converges

00 Lan n=2



!!n b -

n-+oo

n

_

f (;;,2t converges by the alternating series test since

n=l

decreasing for n � l and lim

= 0.

bn

44. 45.

46.

f (� converges by the alternating series test since

n=2 y ln(n)

decreasing for n � 2 and lim

f

n=2

n-+oo

decreasing for n � 2 and lim

f f

n=l

n=l

n-+oo

n

(-i� -i l +(

00

n -l

n=l

OO (a) L

n=O

n=l

(-�)

= 0.

< bn+l =

~

< bn+l

f ,;2 and f (-�)

n=l

=

en l� n)

(

>

n

n=l

(n+l\3+1 � l,�Ol

diverges. Hence, 49.

00 ( 1) n+1 L n..fii, n=l

50.

I: (-�: ../n

n=O

1

oo

( l)n

L ;;,_ 1 n=2

1 16 .

I R2 I 0, = � n(n)

f ¾ diverges.

converges and

n

(-l) n 1 l _ 11 n3+1 ~S2-l-2+9-18'

(b) IRnl

48.

f

bn

_j_ 0 1 .

bn

= 0.

converges since both

(-l) _ ~ -n2 - ~S3- 1

(b) IRnl

47.

diverges since

n ) n :;

(a) L

bn

converges by the alternating series test since

�;:-�l�n)

0 and

3

nI/6

bn

. (-l) n 1· (- ) n d"1verges by the n th- term test smce n-=-.� � 2+si l 41 . � 2+sin(n) n(n)

43.

n-+oo

[ln(n)]

converges by the limit comparison test.

n-+oo

42.

1·1m

converges.

f I (-�: ..fo I

n=O

1

00 ..fii, - L n+l n=O

417

C.6. CHAPTER 6 SOLUTIONS 51. 52.

I: n=l

(-�r diverges by the n n

2 ( l)n ( n) j �n I: n=2

{bn}

2)n � (n_

1

� 3

I: n=2

r

2 -term test since lim (-n n->oo

=J. 0.

converges by the alternating series test since

is decreasing for n::::: 8 and

diverges, 53 .

th

(-l)

n

!�n

(n)

2

)

1

���

bn

= 0. Since

converges conditionally.

. converges abSO1 ute 1y smce



;;:l

I

(-2) n n_

1

3

-

;;:l 3

2n

n_

1

56.

00

3n I n2 � 2 n +i(n+l)4 diverges by the n=O 3"(n+l) 2 2n +1(n+l)4 _ � rlffi n +2 3n 1 n 2 - 2 (n+2)4 n->oo 2

ratio test since lim

> 1.

00

E

n

n=O

3 2 2 l + � -

n.

,

converges by I: (n�l)! n=O lim 2(n+l) (n+l)! 0 2n n->oo (n+2)!

=

58.

59.

60.

61.

I: n=l

r

n

n

� ({:)!

� 3_6

n=l

n----+oo

0 < 1·

the ratio test since lim

n->oo

1. n V5 2 v'5 2 2n y15

C. 7. CHAPTER

C.7

7

419

SOLUTIONS

Chapter 7 Solutions

Exercises 7.1 - page 251

2.

4 3 2 n � (x- 2) = + x-2 + (x- 2) + (x-2) + (x- 2) • l J3 L, Jn+l y'2 2 y'5 n=O

3.

4 3 2 n � (x+l) 1 _ x+l + (x+l) _ (x+l) + (x+l) 3 9 27 81 . L, ( -3)" = n=O

4.

n (x+2) 4 (x+2) 3 (x+2) 2 (x+2) 5 � (x+2) _ - ( X + 2) + + + 16 + 25 . L.., 4 9 n2 n= l

5.

6-

7.

oo

L

=O

n

x2" x4 x6 x2 xs Jn+l = l + y'2 + J3 + 2 + v15·

3n (x-3) 15 (x- 3) 6 (x-3) 9 3 (x-3) 12 _ � (x-3) + 3) + v'2 + 3✓3 + . L, nfo = (X 8 5 v'5 2

n=l

00

L=O 2

n

n

( x - l)71 +

00

La

8.

a

9.

3 + 2x

10.

n=O

n

(x

-

� x n l (

n

- a)'1 + (3 2

5x +

00

Lb =O

00

Lb =O

00

n

=O

n

n

x

n

=

00

L= C

00

n

X

n

+ n�l ] (x

00

L= (aa

n

O

n

+ f3b

>i

(x-3)

2

n

-

)(x

n

l )71.

- a)'1.

=? Co = 3 + bo, C1 = 2 + b1, C 2 = -5 + b 2,

=

n

00

LC

3

⇒ Co=2+bo, C1 = l+b1,

4

2

X =Co+C1 X+C 2X +C3X +C4 X + C 5 X 5 + · ··= (Co+ C 2X + C 4 X 4 + · · ·)+ 5

.

O

n

n (x-3)1 1 O n=O C3 = b3 ' C4 =-1 + b4) Cn = bn) n 2". 5. n

3 (c1x+ c 3x + C 5 X + ···) =

12

- a)'1 =

n

L= b

00

L [2 n= O

l)71 =

(x

n

4

LC

n

-

n

2+(x-3)-(x-3) + C 2 = b 2)

11.

00

L=O

I� I -

00

I: c

k

=O

2k

2k

x

+

00

I: C k+1x k+l_

k=O

2

2

1 1 1 � 2"xn R - 1·lffi � n+2 - 1 1· - lffi n+l 2n+l - 2. j X j < 2 =? -2 < X < 2. At L, n+l , n__,oo n--,oo Cn+1 n=O 00 00 · _ 1 (-l) n 1 1 1 . x _ 12, ) I _ [ n+l converges, and at x n+l diverges 2, 2, 2 . = n O n=O

--

L

-

I:



- -

420 13.

APPENDIX C. SOLUTIONS

E

n=O

f+1,

;l 2 +1 = 1. Ix/< 1 ⇒ -1 < x < l. At lim /_fn._/ = lim (n+ R = n-+oo Cn +1 n-+oo n +l

n

x = ±l,

oo ± n ( l) converges⇒ I= [-1, 1]. n2 +i

E

n=O

� vn(x-6) n , R - I'lill 14 · L.J n3 +l n-+oo n=O

< X -6 < 1

-1 00

E

n�l

n=O

15.

)

n

)

n=O

n

⇒ 0 < x < 2.

1=(0,2]. 16.

E (x1,�

n=O

)

n

, R

' R = lim l_fn._l n-+oo en+ ! At x = 0,

=



_.!,

4

=

00

E 2n�

n=O

l

� = LJ

oo

E

-

yn(-l)n

n3 +l

n=O

lim 2n+3 n-+oo 2n+l

= 1.

2

00

;! 1 = 2.

Ix - 11

/ X + lI



converges, and at x = 7,

⇒ -1 < X - l < 1

oo ( 1. )2 (2n+2)I. -->oo (2n+2 )(2n+l) n

=

n

2 l m I (n!)2 (x-1) 2nt2 {2n)! � ((n-1)!] (x - 1 2n lim an+1 I I I i ) ' � (2n)! -->oo (2n+2)! >oo an [(n-1)!) 2 (x-1)2n -n 2 1. n2 lx-ll 2 _ lx-11 1 1 and only 1f X - l 2 2. R n-=-.m (2n+2)(2n+l) - -4 n

•f


oo Cn +I



22n (n+l)! _

lill I 2n+2 - lill n-+oo n. 2 n-+oo

(fo (b) [J; (a)

0

Cn =

a

n

X

n

) (�b ,

an(x n

E akb k=O

ar]

n

-k =

n

[t n

x

= fa Cn

)

• I

X

n

ar]

bn(x -

n

n

, Cn

I< ⇒ =

=

1

akbn-k

=

1

=

an-kbk.

fa en(x - ar,

E an-kbk. k=O

Exercises 7.2 - page 279 1 . f(x )

=

1)2x

J(x)

=

2�x

2.

3 · j'(x)

= =

00

E (-2xr n=O

=

00

l �;f 2

= 3 I: n

=O 00

00 n

E (-2r x , I - 2xl =O

n

(�r = I:

u

- 1 1 - 1 '\""' ( - x) - 1 - 3 3 1+;£ 3+x 3 3 n=O

00

n

n

=O

n

/n x ,

l�I

I -3 I 3n+1 n=O n

lxl


(X)

Hence, J(x) =

lxl < 4 = R.

, lx2 1 < 1 ⇒ lxl < 1 = R.

(- x ) 2

3

8. f(x) =

2n



, l¼I < 1

n

1 � �:1

3

1-

= ½-

1� =½I: (-�r +½I: (2;r for lxl < �+ 2 3 n =O n=O n n l 3 2 _fn_l) "hRn Hence, f() -1"-> X -� - u [(lffi l - . 2n H + 3n+l ] X , Wlt 2 n = Cn +1 n=O

9. f(x) = 2�x + 3l2x =

: �i = -(x +1) L=O x

10. f(x) = 2

2

2

-1-

L x2n - L x2n (X)

(X)

n=l

n =l

12. f(x) = e-3 x = 13. J(x)

=e

x

=

R= oo.

= - L x2n _ L x2n+2 = - L x2n _ L x2n = n n =O n=O n =O n=l = -1 - L 2x2n , jx 2 1 < 1 ⇒ R = 1. n=l (X)

(X)

2n

(X)

(X)

(X)

I: c-:r x , R = oo. I: ;h(-3xt = n=O n

e2ex -2 = e2

e - 3x

(X)

n =O

2 15. f(x ) = e-3x =

16. J(x ) =

(X)

L=O ;h(x - 2r = L=O �(x - 2r, (X)

2

n

n

I:=O ;h(-3x t = I:=O c :r - x 2

n

=

n

e - 3e - 3 ( x -l )

e-3

L (X)

n=O

2n

R = 00.

, R = oo.

�[-3(x - l)] n

k k 2k +1 (-1) 22k+1 2k+1 R = oo. ( 1) 17. f( x) - sin(2x) =� =� u (2-k+l)! (2x) u (2k+ l)! x ' k=O k=O

(X)

423

C. 7. CHAPTER 7 SOLUTIONS _ - cos(3x) 18. J(x) _

(-l) k ( x. 2k _ 3 ) -

oo

I: (2k)!

k=O

(-l}k3 L ( k}! k=O

2k

oo

2

_ oo. x2k , R·. -

19. J(x) = sin(x) = sin[1r + (x -1r)] = sin(1r) cos(x -1r) + cos(1r) sin(x -1r) � (- 1 ) k +1 · _ 2k - L.., (2k l}! (X - 7r ) + l , R - 00. -sm(X - 7r) _ + k=O

20. J(x) = cos(x) = cos[1r + (x -1r)] = cos(1r) cos(x -1r) -sin(1r) sin(x - 1r) = 00 ( ) k +l "\"' ....=]:__(x -1r)2k ' R - cos(x -1r) = L.., (2k}! · = oo. k=O

f ����'x

21. tan-1(x) =

2n+1

n=O

, jxj < 1,

⇒ tan -1(2x)=

� (-l} n22ntJ x 2n+l l ' 2xj < 1 ⇒ lxl < 21. = R· 2n+1 L..,

f ���;(2x)

2n+1

n=O

=

n=O

22.

00

L=O(-lt(x -2r, Ix - 21 < 1, ⇒ 00 ( l} 1 - (x - 2)n l + C X= 2 ⇒ C= 0 ⇒ ln(x -1) = J x- dx = "\"' L.., n+l ' = 1+(!-2) = x�l

n

n

f

ln(x - 1) = 23.

n=l

1

+

n=O

(-l�n -i (x

- 2) n, R = 1.

_1 � -x-2 n = � (-l)n 1 _ 1 1 -21 < 1, ( 2 ) 2) n ' I -x-2( L.., 2n +i X 2 L.., (x-2} - 2 l+x 22 n=O n=O 00 n (-l} n+ l "\"' 2n+l (n+l} (x -2) + C ' x = 2 ⇒ C = ln(2) ⇒ ln(x)= .!.x dx = L.., l

x

2+

ln(x) =

J

n=O

f=

n=O

2J;}l: 1 )

00

n

tan-1(x -2)

2

X

[1

+

00

2

tan-1 (x - 2) _

26 . f(x) =

(x -2r+ 1 + ln(2), jx -21 < 2 = R.

L=O [-(x - 2) t = L=O(-lt(x - 2)

24. 1+(L2)2 =



J l+(x-1 )

2 2

= L.., �

n=O

n

_ dx -

(-l} (x -

2n+ l

n

, jx - 2j < 1,

2n

00



_2 ⇒C_ n+l + C, x -0 ⇒ L=O (-l} 2 l (x -2)

n

n

n+

2)2n+ l ' R

2

= 1.

k ... (3k-2} k > l ⇒ x = x = x2(l+x)-l/3 ' a=_.!.3 ' (ak )= (-1) ,1.4.7 lfJ:+x 3 k! ' - ' 3/f+x � (-l) k ·l-4-7---(3k-2} k _ .2 � (-1 )k , 1 ,4.7,.. (3k-2) k +2 I _ X ] - X + X , X I < 1 - R. L.., 3kk! 3kk! L.., 2

2

k=l

k=l

424

APPENDIX C. SOLUTIONS

00

I:

k=O

( 2k +l)! x 3k+l , l x l < 1= R. 2k(k!)2

2

29. e x =l+ x+½x 2+ ¾x 3+·· · and sin(x)= x - ¾x 3+··· =>ex sin(x)= ( 1+ x+ ½ x 2+ ¾x 3+· · ·) ( x - ¾x 3+··· ) => T4 (x) = T3 (x)= x+ x 2+ ½ x 3.

00

00 ( l)n - l)n+l C n n � 32. 1.x = l+( x1-.1 )= � + ' LJ (-l) (x - l) => ln(x) = J .!x dx= LJ +l (x n n=O n=O OO ( )n I x = l => C= 0 => ln(x) = L -=¼-(x - 1r, x = l + (x - 1) => x ln(x) =

n=l

[1 + (x - l)][(x - 1) -½(x - 1) 2 + ½(x - 1) 3 - ¼(x - 1) 4 + . . •]= (x - 1) +½(x - 1) 2 - ¾(x - 1) 3+ ;2 (x - 1) 4 +·· ·, 00 x ( x) 1 . 3 2 n In x = � _1_ = = u (x - l) = 1+ (x - 1) + (x -1) + (x - 1) + . . . => 22-x 1-( x-l) =O n [(x-l)+½(x -1) 2-¾(x-1) 3+ 1\(x-1) 4 + • • l[l+(x -l)+(x -1) 2+(x -1) 3+· · ·] 2 4 => T4 (x) = (x - 1) + �(x - 1) + !(x - 1) 3 + g(x - 1) . _ �( (-l) k · k · oo (-lj k 2k+l s· 1) _ oo 33. s·m(x) _ => m( 2 - I: 2 2 k+1( 2k +l)! - 0 -1) bk 1s alternatmg, - I: ( 2k +l)! x k=O k=O k=O 1 IRk l < bk+l= 2 2k+3( 2k +3 )!, k= l ⇒ IR 1 I < 2i51 � 0.00026 < 0.001 ⇒ sin(½) � s 1 = ½ - }8= �� � 0.47916. The value by calculator is 0.47942, with an absolute difference of 0.00026 < 0.001.

34.

fl � (-l)k k � (-l)k 2k r;;.) = 0 ( 2k )! X => Jo COS( yX r;;.) dX = => COS( yX ( 2 k )! X k=O k=O . � (-l)k x k +11l - � (-l)k - � ( )k . alternatm g, IRkI < bk+l - 0 ( k +1)( 2k )! - 0 -1 bk lS u ( k +l)( 2k)! O k=O k=O k=O 1 ( k +2 )( 2k +2 )" k= 2 => IR2 I < 2 80 � 0.0003 < 0.001 => J�1 cos( vx) dx � S 2 = l - ¼+ i2= �� � 0.7638, with the required accuracy. COS

(X) =

0

l

C. 7. CHAPTER 7 SOLUTIONS

425

(2k+2)! k _ _ � - L..., (-1) bk i· s alternat·mg, I Rk I < bk+l - (sk+9) 210k+11[(k+l)!]2, k=O k = 0 ⇒ IRol < b1 = g.�10 = 92\6 � 0.000108 < 0.001 ⇒ 1 2 0 ✓i�xs dx � so = ½ = 0.5, with the required accuracy.

(-l) (2k)! L (Sk+l)210k+l(k !) k=O 00

k

2

J

2 3x - 1)2 COS2(X) _ 11· x ( 3x+-29 x 2+... )2(l --2J x 2 +.. ) . X(e sm . 3x 36. 11m - m ) 1 3+.. ( ) x-+0 x-+O ( x -6x . 2 3+2 x +.. ) (·l -21 x +·.. ) ( 9 lim = 9. 3 1 2 +.. ) x-+O ( J-6x 2

2

(x+2)3 (-x-½x 2 -... ) . )3 Jn(l-x) 1· 38. 11m (x+2 = -4. = . 1mO (2 x -6x s 3+... ) ( l -2X 9 2+ · ) x-+O Stn(2 x) cos(3x) x-+ 1 4· •- l n+l ' x = l ⇒ 0 < z < l ⇒ 0 < e z < e l /3 < e ⇒ < _e (x)I 40. I 1P Ln (n+l)! lx 3 - -3 - < 3 n+l(n+l)! ::; 0.0001 if 3n+1 (n + 1)! � lOOOOe � 27,182.8, and n = 4 ⇒ 3n+l (n + 1)! = 35 5! = 29,160 > 27,182.8 ⇒ ffe � 1 + ½ + }g + 6-17 + 24\ 1 � 1.395576. The value by calculator is 1.395612, with an absolute difference of 0.000036 < 0.0001.

IRn (½) I

L

-;/xn and x = 1� ⇒ ln(0.9) = L n�b n · f(n+l)(z) = n=l n=l _ f(n+l)(z)x n+l _ p _ -x n+l -1 1 -n! P ( ) 1 X (1 -z)n+l ⇒ Ln - (l -z)n➔ 1 (n+l) ⇒ 1 Ln ( 10 ) - ( l -z)n+ 1 (n+l)1Qn+l · (n+l)! n+l 0::; z::; /o ⇒ 0::; l�z ::; 190 ⇒ gn+l(n+l) < 0.001 if 9 (n + 1) > Uo) 1,000, and n = 2 ⇒ 9n+l (n + 1) = 93 · 3 = 2,187 > 1,000 ⇒ ln(0.9) � -/0 - 2�0 = - 22010 = -0.105. The value by calculator is -0.10536, with an absolute difference of 0.00036 < 0.001.

41. f(x) = ln(l - x) =

(X)

(X)

IRn

I ::;

42. J(x) = sin(x) = sin[a + (x - a)] = sin(a) cos(x - a) + cos(a) sin(x - a) oo (-l)k (x - ) 2k + cos(a) oo (-J)k (x oo e (x - )n , with _ L s·m(a) L L ( 2k+l)! - a) 2k+l a a n ( 2k)! k=O k=O n=O _ k k (-l) cos(a) _ (- l) sin(a) and C2k C2k+l - (2k+l)! , k � O , and R _ 00. (2k)! 43. J(x) = cos(x) = cos[a + (x - a)] = cos(a) cos(x - a) - sin(a) sin(x - a) 00 k oo c (x - ) n with 2k - s · (a) oo (-1)" (x - ) 2k+l _ L cos(a) L (-l) m a a , L ( 2k+l)! n ( 2k)! (x - a) k=O k=O n=O k k 1 1 _ (-l) cos(a) and _ (-l) sin(a) > 0, and R C2k C2k+1 - (2k+l)! , k _ _ 00. (2k)!

426

APPENDIX C. SOLUTIONS

'°"'

'°"'

'°"' 00

00

00

l (ix)2 k + _1 - i 2 k+1 = _ 44. eix = L.J .ln!.(ix)n = L.J (_ 2 k)! L.J ( 2k+l)! ( x) n=O k=O k=O � (-l) k 2 k � i(-l) k 2 k+l _ - cos(X) + ism · · (X ) . L.J ( 2 k)! X + L.J ( 2k+l)!x k=O k=O (a) If ab� 0, then ab= labl =lallbl. If ab< 0, then ab= -labl Hence, ab :s; lalJbl.

4 5.

< labl= lalJbl.

(b) (a+b)2= a 2 +2ab+b 2 :s; a 2 +21allbJ +b2= lal2 +2lallbl +Jbl2 =(Jal +lbl)2 2 2 ==? J(a + b) :s; J(lal + lbl) ==? la + bl :s; lal + lbl. (c) Since

E la x 00

n=O

lxl < R 2 ,

n

n

l converges for lxJ

E Jb x 00

< R 1 and

n =O

n

n

E (la x l + lb x l) converges for lxl < 00

n

n =O

n

n

n

l converges for min{R1, R 2 }- Since

l(an +bn)xn l= lan xn +bn xn l :s; lan xn l+lbn xn l by part (b),

L=O l(a 00

n

converges for Ix! < min{R1, Rd by the comparison test. Hence, R � min{R1, R2 }-

C.8

n n +bn )x l

Chapter 8 Solutions

Exercises 8.1 - page 297 1. Iff is periodic with period T, then J(x+T)= J(x) and f(x) =f((x- T)+T) = f(x - T), f(x ± 2T) = f((x ± T) ± T) = f(x ± T) = f(x), etc. Hence, f ( x + nT) = f ( x) for every integer n. 2. f(x) =x2 + 1 for -1 :s; x< 2 and f(x + 3) = f(x) for all x. Since f has period 3, f(5) = f(-1) = 2, J(7) = f(l) =2, J(-4) =J(-1) = 2, J(783) =f(0) =1 and J(-291) = J(0)= 1. 00

3. Multiplyf(x) = T+I°:: [an cos( nzx ) + bn sin( nlx )] by sin( mzx ),wherem� 1 n =l is any integer, integrate from - L to L and reverse the order of summation and m x m x integration to obtain L f(x) sin( z ) dx= T L sin( z ) dx + f1 [an

J:L cos(

J:

J:

m x l ) sin( z ) dx + bn

n x

J:L sin(

J:

l ) sin( mzx ) dx].

n x

-::7r

cos( mzx ) l�L = 0. The first integral on the right is L sin( mzx ) dx= Employing the trigonometric identity cos(a) sin(b) = ½ [sin(a + b) - sin(a - b)] with a= n1rLx and b = mL1rx ' we obtain

J!L cos(

l ) sin( mzx ) dx = ½

n x

J:L { sin [ ( +7)1r n

x

] - sin [ ( n -7)1rx]} dx = 0. If

n =/- m, then the trigonometric identity sin(a) sin(b) = ½ [cos(a - b) - cos(a + b)] with a= nt and b= mzx gives

427

C.8. CHAPTER 8 SOLUTIONS J_�L sin (

l )

n

x

sin ( mz

x

)

dx

{

½ f!:L cos [ (n-7) 1rx] - cos [ (n+7) 1rx] } dx

=

= 0.

If n = m 2'. 1, then, employing the trigonometric identity sin 2 (a) = ½ [1-cos(2a)] with a = mz , we obtain f!:L sin 2 ( z ) dx = ½ f!:L [1 - cos (2�7rX )] dx l [x - 2-__ sin ( 2 m 1rx)] I L = L. Thus b m = l.L J-LL f(x) sin (mL1r ) dx m > 1. L 2 2 m1r -L m

x

x

'

= { �: -� �:: � }

4. f(x)

t

' -

x

and J(x + 4)

=

f(x) for all x. T

=

4, L

2 = ½ 2 f(x) COS ( ; ) dx = ½ }� COS ( ; ) dx = nln sin ( ; ) I� = 0, n n 2 2 2 nn ) (-l) l. cos (nn2 ) 1 0 = [l-n dx = _....!... a = l r 1dx = 1 b n = l r sin ( 0 2 Jo ' 2 Jo 2 n7r 7r 11-�-:,1 ) n] sin ( ; ). Fourier series of J is ½ +

an

n

00+ 8. f(x) = { (a) T

0, -1 < -x,

X

+ nn 7""4"

fi { �;-�r +

= sin(x) on [0,3] is

14. The cosine series of J(x) ao

12[1��--:; 1)n]}

T

l.

cos(n1r x).

+La n n=l

cos (n;x ),

= i f03 sin(x)dx = -icos(x)I� = i [1- cos(3)], 3 3 i f0 sin(x) cos ( ; ) dx = ½ ]� { sin [(1+ 31r) x] + sin [(1 -

t) x]} dx 3 _ .! cos[(l+ n3")x] cos[(1- nt)x] {cos(3+n1r) -l cos(3-n1r)-l }{ l+}+-��� + n,r n,r - -��3 1-3+ n 1r 3- n 1r 3 3

an

n

=

1

_ {(-l)n cos(3)-l 3+ n 1r

x

n

3

n

1

0

=

0

n 1- + __) 1 + (-l) 3-cos(1r3)-l} = [l _ (-l)n cos(3)] (= 3+ 1r 3- 1r n

n

B[l-(-l)n cos(3)l n > l The cos ine series is , - . 9-n21r2 n cos(3)] COS (n1rx). � 6[1-(-l) .3! [1- cos(3)] + L.., 9-n2 1r2 3

n

n=l

15. J(x) = x2 on [-2,2].

J off to JR is displayed in Figure

(a) The graph of the 4-periodic extension C.6. y 4

-4

-6

-2

0

8

6

4

2

JO X

Figure C.6: The 4-periodic extension f of f to R

J

(b) The Fourier series of f converges to (x) for all x because everywhere and ]' is piecewise continuous. -

(c) Since f is even, the Fourier series off is the cosine series a0

=

r Jo

2

2 x dx

r x sin __i_ n 1r Jo z

=

x 1

3 2

3 O

1rx (n ) dx 2

=

=

§3'

an

s __ n27r2 x cos

= (

r x cos Jo 2

n 1rx

n 2 1. The cosine series is f + ;�

2

)

2

2

2

) dx

_ r2 cos 1 0 - _s n27r2 Jo

I: (

n=l

(

n 1rx

-,,T cos ( . n

; ) x

= (

J is continuous 00

T+La n=l

n

cos ( n ;x ), 2

..2. x2 sin ( n21rx ) 1 n 1r

n 1rx

2

)dx

=

O

16(-l)n ' n27r2

435

C.8. CHAPTER 8 SOLUTIONS 16· f(x)

=

{

0 :s; 1, 1 :s;

X,




n> 1. -

436

APPENDIX C. SOLUTIONS The cosine series is ¾ +

f(x) for all x in [0, 2].

L 00

n=l

� [cos n2 2

( n21r

) - 1] cos ( n;

x

),

and converges to

y

-6

-5

-4

-3

-2

0

-1

3

2

5

4

6

X

Figure C.8: The graph of the even, 4-periodic extension J off to IR. (f) At x = 0, the series converges to f(0) = f(O) = 0. At x= l, the series converges to ](1) = f(l) = 1. At x= 2, the seri�s converges to ](2) = f(2) = 1. At x = 78, since f is 4-perioclic, the series converges to ! (78) = !(2)= f (2)= 1. At x = 79, since J is 4-periodic and even, the series converges to ](79)= ](-1)= !(1)= f(l) = 1.

1 17.f(x)={ ' 0'

x �:S; - 0· (2k)! (2k)! ' _ _ -a1 _ � _ 3 _ a1. , n _ - -a1 n - 1 ⇒ a3 - - -, n - ⇒ a5 _ - � - 5 ⇒ a7 _

2

· genera1, a2k+l _ and, m -

6

24 4 k _ (-1) a1 k

2-4·6'

_ � n _ > _ 0 . y - 6 anx 2_4_6 ___(2k) - �, n=0 (-l) k 2 k+l (-1) k 2 k k!x2k 2k � � 2 k+l _ � = � x + al 6 � u a2kx + u a2k+lx - a0 6 (2k)! k=0 k=0 k=0 k=0 a0 y 1 + a 1 y2 is the general solution, and y 1 and Y2 are two independent solutions.

(c) y1

=

1 - x2

(-lla1

5 7 4 6 3 - ...!_x + · · ·. + lx + · · · ' y2 = x - .!x - 1...!_x + .!x 5 2 8 48 3

2 (-l) k 2k +l - � ..!_ - X 6 k! (-x2 ) ( d ) Y2 - � 6 �x

k

k=0

k=0

2

= xe-\ .

(e) Since the equation has no singular points, R

= oo for both series.

(f)

Thus, y

6. (1

y(O)

= 0 ⇒ a0 = 0 and y'(O) = 2 ⇒ a1 = 2.

= 2y2 = 2xe-2. x2

+ x2)y" - 4xy' + 6y = 0.

(a) y

=

Lax 00

n=0 2

n

n

, y'

=

L na x -i, y" = L n(n- l)a x 00

n=0

n

+ x )y" - 4xy' + 6y = 0 ⇒ 00 (1 + x 2 ) L n(n - l)anxn - 2 - 4x (1

n=0

00

n

n

n =0

n

-2,

L na x -l + 6 E a x 00

n=0

n

n

00

n=0

n

n

=0

=}

441

C.9. CHAPTER 9 SOLUTIONS

L n(n - l)a 00

n=O

L 00

n=O

n 2 nx -

+

L n(n - l)a 00

n =O

nx

00

n =O

nx

n

L 6a 00

+

n=O

[(n + 2)(n + l)an+2 + n(n - l)an - 4nan + 6an] xn = 0

L [(n + 2)(n + l)an+2 + (n2 - 5n + 6)a 00

n =O

( b)

L 4na

-

n

n ]x

nx

= 0 =}

n

=}

= 0 =}

n

= 0 ⇒ an+2 = -�:��)�:�3{t, n 2:: 0. n = 0 ⇒ a2 = -3a0 , n = 2 ⇒ a4 = 0, n = 4 ⇒ a6 = 0, etc. Hence, a2 k = 0 fork 2:: 2. n = 1 ⇒ a3 = -½a1, n = 3 ⇒ a5 = 0, n = 5 ⇒ a7 = 0, etc. Hence, a2k+1 = 0 for k 2:: 2. Thus, y = a0 + a1x + a2x2 + a3x3 = a0 +a1x-3a0x2 -½a1x3 = a0(1-3x2)+a1 (x - ½x3 ) is the general solution, with a0 and a1 arbitrary, and y1 = 1-3x and y2 = x-½x3 are two linearly (n + 2)(n

(n - 2)(n - 3)an

+ l)an+ 2 +

2

independent solutions.

(c) y(0) = -3 ⇒ ao = -3 and y'(0) = 1 ⇒ a 1 = 1. Hence, y = -3(1 - 3x2) + (x - ½x3) = -3 + x + 9x2 - ½x3 .

= 0.

7. y" + xy' - 2y y" + xy' - 2y

= 0::::}

L n(n - l)a n

n 2

00

=2

nx

-

=

y

Eax 00

n

n=O

n

L n(n - l)a

n=O

+

nx

L (n - 2)a 00

nx

n =O

=

, y'

n

n 2

-

E na 00

n=O

nx

n

=

-l, y"

E n(n - l)a 00

n=O

+ L nan xn - L 2an xn = 0 00

00

n=O

n =O

= 0 =} (n + 2)(n + l)an+

2

+

nx

n 2

- '

=}

(n - 2)an

=0

⇒ an+2 = - ( n� ��a.;2 ), n 2:: 0. n = 0 ⇒ a2 = ¥.'# = ao, n = 2 ⇒ a4 = 0, n = 4 � ⇒ a6 = 0, etc. ence, a2 k = 0 for k 2:: 2. n = 1 ⇒ a3 = ;.� = �t, n = 3 ⇒ 3a _ _ 7 ⇒ a _ -5a7 _ -3-5a1 _ g _ .=Qcl _ 5 ___,_ a5 _ 9 - 8-9 - 9! , n ----,,,- a7 _ - -6-7s - 3a1 -� 7! , n 4- 5 - 5! , n -

___,_

_

_

3-5-7a1 - 7ag · genera1 , a2k+l _ ----,,,- au - 10_11 - (1 l)! , etc., and , m (-l)k+l.1. 3.5.7...(2 k-3) a1 _ (-l)k+L.1-3 -5-7---(2k- 3) a 1 2-4-6---(2k-2) _ (-l)k+1(2k-2)! a 1 (2k+l)! (2k+l)! 2-4-6···(2k-2) - (2k+l)!2 k -1(k-l)!'

k 2:: 1. The general solution is y (-l)k+1(2k-2)! _ oo 2k+l Y2 - x + (2k+l)!2k-1(k-l)!x k=l

L

8. (2 + x2 )y" + 2xy' - 2y = 0. y = y"

=

L n(n - l)a 00

L 2n(n - l)a 00

nx

n=O

-

nx

n 2

L 2n(n - l)an x

n 2

n=O 00

n =O

.

=

+ a 1 y2,

a0 y 1 ·

·

La 00

n=O

n n x , y'

=

L=O na 00

n

- , (2 + x2 )y" + 2xy' - 2y

- +

L n(n - l)a 00

n=O

L (n 00

n=O

2

=

1

+ x2

are two lmearly mdependent solutions.

n 2

+

where y1

nX

n

nx

n

-l,

= 0::::}

+ L 2nan Xn - L 2anXn = 0 ⇒

+ n - 2)an x

00

00

n=O

n =O

n

= 0 =}

2(n + 2)(n + l)an+2 + (n + 2)(n - l)an = 0 ::::} an+2 = -;(:)i)n, n 2:: 0. n = 1 ⇒ a3 = 0 ⇒ a5 = 0, etc., and a2k+l = 0 for k 2:: 1. 3a1 Qll - -=!!c2. - 4 ⇒ a6 - .=Q11 n ----,,,- a4 ----,,,- a2 - -2.5 - 0 ___,_ - 2. , n - 2 ___,_ 2_3 - 22. 3 , n -

and

442

APPENDIX C. SOLUTIONS

9. y" + exy = 0.

(a) Y = ao + a1x + a2 x 2 + a3 x3 + a4 X4 + a5 x5 + · · ·, y' = a1 + 2a2 x + 3a3 x 2 + 4a4 x3 + 5a5x4 + · · ·, y" = 2a2 + 6a3 x + 12a4x 2 + 2Oa5x 3 + • • •, 1 5 ... ex = l + x + l2 x 2 + l6 x3 +. _!._ x4 + 24 12 0 x + x 2 e y = ao + (ao + a1)x + (½ao + a1 + a2 ) x + ( ¾ao + ½a 1 + a2 + a3 ) x 3 + • • • , y" + e xy = 0 =} (2a2 + ao) + (6a3 + ao + a1)x + (12a4 + ½ao + a1 + a2 ) x 2 + ( 2Oa5 + ¾ao + ½a1 + a2 + a3 ) x3 + · · · = 0 =} a2 = =?, a3 = -ao5-ai, a4 = -;_;1 , a5 = � - �b :::::} y = ao + a1x - �x 2 - ¾(a0 + a 1 )x3 - n-x4 + a1 5 ... 4 0 60) X + 5 (b) ao = 1 and a1 = 0 ⇒ Y1 = 1 - ½x 2 - 16 x3 + _!_x 40 + · · · , a0 = 0 and a 1 = 1 1 3 1 4 1 5 + ... ⇒ Y2 = x - -x - -x - -x 12 6 60

(.'!Q _

10. Legendre's equation of order v, (1 - x2 )y" - 2xy'

(a)

Y

=

La x 00

n=0

n

n

, y'

=

L n(n- l)a x 00

n =0 =}

n

n 2

+ v(v + l)y = 0.

L na x -l, y" = L n(n - l)a x -2:::::} =0 00

n=0

-

n

00

n

L n(n- l)a x 00

n =0

n

n

n

-

n

L=0 2na x 00

n

n

n

n

+

L v(v + l)a x = 0 00

n=0

n

n

I: n(n - l)anx - + L [v(v + 1) - n(n + 1)] anx = 0 00

n 2

n=0

00

n

n=0

:::::} L {(n + 2)(n + l)an+2 + [v(v + 1) - n(n + 1)] an } xn = 0 00

n =0

_ n(n+l)-v(v+l) an , n _ ---'> O. ----r an+2 (n+l)(n+2 )

(b) If v = m � 0 is an integer, then the recursion relation becomes n(n+l)-v(v+l) _ n(n+l)-m(m+l) an+2 _ (n+l)(n+2) an - (n+l)(n+2 ) an , n > - 0' and n = m ⇒ am+2 = 0, n = m + 2 ⇒ am+4 = 0, n = m + 4 ⇒ am+6 = 0, etc., which shows that one of the series solutions truncates at amxm . If m is even, then y 1 = ao + a2x2 + · · · + amxm is a polynomial, and if m is odd, then Y2 = a1x + a3 x3 + · · · + amxm is a polynomial.

(c) If m = 0, then P0(x) = a0, and P0 (1) = 1 :::::} a0 = l. Hence, P0(x) = l. If m = 1, then Pi(x) = a1x, and P1(1) = 1 ⇒ a1 = l. Hence, P1(x) = x. If m = 2, then P2 (x) = a0 + a 2x 2, with a 2 given by the recursion relation with n = 0 : a2 = -;_�;}ao = -3ao :::::} A(x) = ao - 3aox2 = ao (1 - 3x 2). Then A (1) = 1 ⇒ -2a0 = 1 :::::} a0 = -½ :::::} P2 (x) = + !x2 = ½ (3x2 - 1). If m = 3, then P3(x) = a1x + a3 x3 , with a3 given by the recursion relation with n = l: a3 = 1· 22 4 a1 - -ia1 :::::} P3 (x) = a1 (x - ix3 ). Then A(x) = ½ (5x3 - 3x). A(l) = 1 :::::} a1 =



�t

-! :::::}

443

C.9. CHAPTER 9 SOLUTIONS

Exercises 9.2 - page 332 1. f(x) = x 2 - x + 2 on [O,2] and f(x + 2) = f(x) for all x. f(x) =

2

°+ I::[an cos(mrx)+bn sin(mrx)],ao = f0 (x 2 -x+2)dx = (; - ; + 2x) / n=l O 14,a = f 2(x 2 - x + 2) cos(mrx) dx = ,;'If (x 2 - x + 2) s i n(mrx)I� n 3 0 1 f 2 (2x-l) sin(mrx) dx = 2�2 (2x-l) cos(mrx)l�- l lf2 f 2 cos(mrx)dx = 2�2 , n 0 n ' n n7f 0 n 2'. 1, bn = f0\x 2 - x + 2) sin(mrx) dx = - n� (x 2 - x + 2) cos (mrx) I� + 2 2 2 = n� J0 (2x-1) cos(mrx) dx = - n1f + n2� 2 (2x-1) sin(mrx)I�-)7f2 J0 sin(mrx) dx -;'If + n}'lf 3 cos(mrx)I� = -;'If , n 2'. 1. y" + 3y = J(x) =} 3

2

a2

�+ L [cn cos(mrx)+dnsin(mrx)],co = � = \4,Cn = _�11 7f2 = n=l dn = ___Qn__ 3- n27f2 = n7r(-2- n2 7r2). 3 2 2. f(x) = x - 2x on [-1,3] and f(x + 4) = f(x) for all x. J(x) = y

00

=

3

2

n2 7r2 (

� n2 7r2 )'

3

2 3 _x ) 3 �+ � [an cos(n;x)+ bn sin(n;x)],ao= ½f_3 1 (x 2 - 2x)dx= (x6 2 , 1 1 -l 3 3 2 n x �3' an = .!.2 f-1 (x 2 - 2x) cos(n'lf2 x)dx = ...L(x 2x) sin( " ) 2 -1 n,r 3 3 4 4 n x n = dx ...L f (2x - 2) sin( "x) dx = 2 ,r2 f-1 cos (�) 2 2 n2-(x ,r2 - l) cos( "2 ) -1 - nn,r -1 3 2 _ (n ") > (n'lf2 x)dx = _!_§ n2 2 cos 2 ' n - 1 , b n = .!.2 f-1(x - 2x) sin 3 2 n x 4 ( n "2 x) 1 3-1 1 -...L(x +...L -2x) cos (n"x) n,r J-1 (2x-2) cos( 2" )dx = n2,r2 (x-1) sin n,r 2 -1 4 f3 sin(n "x)dx = -lB sin(n") , n > 1. y" + 4y = f(x) =} - n-2 2 -1 2 2 n2 2 00

1

1

,r

y

3

,r

=

3

�+

a Cn = _4 n2n rr2 4

,r

n x ; ) + dn s in ( ; )] , Co = 1 = ¼, -6si 64cos(!!f) 4 n(!!f) bn = n2 ,r2(6 1 n- 2,r 2),d n = 4_ n2rr2 = n2 ,r2 (16- n2 2 ) ·

L [Cn n=l 00

COS (

n x

,r

4

3. f(x) = x - x 2 on [O,1], odd, 2-per i odic. f(x) = L bn sin(mrx), n=l 1 bn = 2 JO (x - x 2) s in(mrx) dx = -; (x - x 2) cos(mrx)I� + ;" JO1 (1 - 2x) cos (mrx) dx = 1r2 (1" - 2x) sin(mrx) I� + n2�2 J; sin(mrx) dx 00 1 4(1-(-l)nj , n _ /I dn Sln > - n3,r4 3 COS( n1rx) 1 O 3 L.., =} y = ,;;:-" X . = y y 1 . . ( n1rX), 3 3 J ( ) n ,r n=l 00

i

f,

n x 4. f(x) = { �' � 2 x, � � �; }, odd, 4-per i odic. f(x) = 1 bn s in ( ; ), bn = n2�2 sin ( nr) by Sect ion 8.2, Exercise 9. y" + 3y = f(x) ⇒ 2 in(¥) . n,rx s � b Y = L.., d n sin ( -2- ), d n = - ,� rr2 = n2,r2(1 2 - n2 ,r2 ) · 4 n=l 3

3

444

APPENDIX C. SOLUTIONS 00

5. J(x) = 1 - x on [0, 1], even, 2-periodic. f(x) = � + Lan cos(mrx), a0 = 1

=

and an

Y=

2 [1

:I�J l l , n 2 1, by Example 8.20 in Section 8.2. n

n= l

3 + L Cn cos(mrx), Co = ao = 1, Cn =

a17r2 1 -n

n=l

y" + y

=

f(x) =>-

(- 1 . n;l;( 1r 1r 1 -n(t )

=

X, 0 � X < 1 � } , even, 4_pen.od"1c. J( x) _ 6. f( x) = { l +� an cos (n1r2 x ) , - QcO. 2 , 1�x�2 ao = �' an = n2�2 [cos ( nr) - 1], n 2 1, by Exercise 16(e) in Section 8.2.

Y

11

+ 4y

f(x) =>- Y =

=

n2 1r2(4��)

[cos

00

T + L CnCOS ( ; n=

1] =

1r

(� ) -

n x)

l

n2 1r2(l��n21r2)

7. f(x) = 2x - x2 on [O, 2] and f(x

J(x)

�+

=

f

, Co

[cos

= �'

= �o

1r

(� ) -

= 4_:'2"2 4

Cn

1].

+ 2) = f(x) for all x. Since f is even,

an cos(mrx), ao = 2 f0 (2x - x2 ) dx 1 2

=

2 ( x2

I� -

-

x;)

i:

i,

an = 2 f0\2x - x2 ) cos(mrx) dx = n1r2 (2x - x2 ) sin(mrx) 2 2 2 n1r 0 (2- 2x) sin(mrx) dx = n2�2 (1- x) cos(mrx) I�+ n2�2 0 cos(mrx) dx = 00 n 2 1. y" + 2y = J(x) =>- y = T + L Cn cos(mrx), Co = �=!,

f

f

n;-!2,

n=l

8. f(x) = sin(2nx) on [0, 1], even, 2-periodic. J(x) = �

ao

=

f1

2 0 sin(2nx)

=

0. If n =/= 2, an

=

f1

__ { cos[(2 +n)1rx] • [( 2 _ n ) nx] }dx Jfo { sm • [( 2 + n ) nx] + sm ( 2 +n)1r +

n [ 1 -(-l) ] ( 2 -n)7r

=

4 -(-l) n

[l ]. 1r(4-n2 )

If n

n=l

2 0 sin(2nx) cos(nnx) dx =

l

n [ 1 -(-l) ] (2 +n) 1r

00

+ Lan cos(nnx),

=

2'

a2

+

cos[( 2-n)1rx] ( 2 -n)1r } 1

1

0

1 = 2 Jro sin(2nx) cos(2nx) dx

f01 sin(4nx) dx = 0. y" = f(x) =>- y = T + n=I: Cn cos(nnx), 0 ·Co = ao = 0 =>- co

· 1s arb"t 1 rary,

_ a2 _ C2 - _4 1r2 -

_ 0 , Cn -

00

l n _ -4 (1-(-l) ] 2 2 2 3( 4-n2) , n -n 1r an

1r

n _j_ r2

Index Absolute convergence, 234 Complex eigenvalues, 135 Adjoint Conditional convergence, 234 classical, 340 Connected region, 30 Alternating series, 229 Continuous approximations, 231 function of two variables, 26 remainder, 231 Contrapositive, 214, 233 test, 230 Convergence Alternating series test interval of, 247 proof, 343 of a sequence, 193 Analytic function, 278 of a series, 208 Annihilator, 112 ·radius of, 247 Approximations Converse, 215 of alternating series, 231 Cramer's rule, 68 of series, 219 Associated homogeneous equation, 57, 103 Decreasing sequence, 200 Dependent variable, 3 Associated homogeneous system, 171 Diagonalizable matrix, 187 Average value of a function, 288 Difference equation, 244 Direction field, 5 Bernoulli equation, 21 Divergence Binomial of a sequence, 193 series, 265 of a series, 208 theorem, 263, 265 Domain, 24 Binomial theorem proof, 346 Eigenvalue, 126 Bounded sequence, 201 complex, 135 Eigenvector, 126 Cauchy-Euler equation, 52, 98 Chain rule, 27 complex generalized, 164 special case of, 27 generalized, 147, 158 Change of index, 256, 319 Elementary function, 321 Characteristic polynomial, 146 Envelope, 40 Equation Classical adjoint, 340 Cofactor, 340 associated homogeneous, 57, 103 Bernoulli, 21 Comparison test, 223 Cauchy-Euler, 52, 98 proof, 341 difference, 244 Complex conjugate, 50 445

446

INDEX

Euler, 52, 98 exact, 29 homogeneous, 15, 44, 87 indicial, 48, 52, 88 Legendre's, 328 linear, 18, 44, 87 nonhomogeneous, 44, 87 nonlinear, 18, 44, 87 of order n, 3 ordinary point of, 317 separable, 9 with constant coefficients, 48 Euler equation, 52, 98 Euler's identity, 50 Euler-Fourier formulas, 286, 315 Even extension, 307 function, 292 Exact equation, 29 Existence and uniqueness theorem Picard's, 41 Existence theorem Peano's, 41 Extension, 299 even, 307 even, periodic, 307 odd, 304 odd, periodic, 304 periodic, 300

average value, 288 elementary, 321 even, 292 extension, 299 odd, 292 of two variables, 24 periodic, 283 periodic extension, 300 piecewise continuous, 288 potential, 29 singular point of, 278 singularity of, 278 vector, 122 vector-valued, 122 Fundamental matrix, 124 period, 283 set, 124

F ibonacci sequence, 193 F inite series, 213 Fourier convergence theorem, 289 cosine coefficients, 286 cosine series, 295, 307 series, 286 series on [a, b], 300 sine coefficients, 286 sine series, 294, 305 Frobenius method, 326 Function analytic, 278

Identities trigonometric, 285 Imaginary part, 51 Implicit solution, 17 Increasing sequence, 200 Independent variable, 3 Independent variables, 24 Indicial equation, 48, 52, 88 Indicial polynomial, 114 Induction principle of, 203 Inductive hypothesis, 203 Inductively-defined sequence, 193

General solution, 4, 43, 87, 124 General term of a sequence, 191 Generalized eigenvector, 147, 158 complex, 164 Geometric series, 210 Homogeneous equation, 15, 44, 87 solution, 57 system, 121

INDEX

'-

Infinite series, 208 Initial condition, 5, 43, 87, 125, 177 Initial-value problem, 5, 43, 87, 125, 177 Integral test, 217 Integrating factor, 19, 33 Integration by partial fractions, 339 by parts, 335 by substitution, 336 techniques, 335 Interval of convergence, 247 Inverse of a matrix, 340 Irred�1cible quadratic, 339 Left-hand limit, 288 Legendre polynomials, 328 Legendre's equation, 328 Limit comparison test, 227 proof, 342 Limit of a sequence, 193 Linear equation, 18, 44, 87 operator, 91, 112 system, 121 Linearly dependent functions, 44, 45 vector-valued functions, 123 Linearly independent functions, 44, 45 vector-valued functions, 123 Lower bound, 201 Maclaurin series, 259 Matrix inversion, 340 Method of undetermined coefficients, 57 Minor, 340 Monotone sequence, 200 Multiplicity, 91 n choose k, 263 nth -term test, 213 n-parameter family of solutions, 87 Nonhomogeneous equation, 44, 87

447 system, 121 term, 57 Nonlinear equation, 18, 44, 87 Odd extension, 304 function, 292 One-parameter family of circles, 11 of curves, 11 of ellipses, 14 of lines, 11 of parabolas, 13 of solutions, 4 Open region, 26 Operator linear, 91, 112 Ordered set, 191 Ordinary differential equation, 3 Ordinary point, 317 Orthogonal, 12 family, 314 functions, 314 trajectory, 12 p-series, 218 Parameter, 4, 11 Partial derivative second-order, 25 with respect to x, 25 with respect to y, 25 Partial differential equation, 3 Partial fraction decomposition, 339 Partial fractions, 339 Partial sum of a series, 208 Particular solution, 4, 43, 87, 124 Period, 283 fundamental, 283 Periodic extension, 300 function, 283 solution, 328 Piecewise continuous function, 288

448 Potential function, 29 Power series, 245 coefficients, 245 interval of convergence, 247 radius of convergence, 247 uniqueness, 259 Principle of induction, 203 Radius of convergence, 247 Ratio test, 236 proof, 343 Real part, 51 Recurrence relation coefficient, 320 Recursion relation coefficient, 320 Reduction of order, 47 Region connected, 30 open, 26 simply connected, 30 Relation, 24 Remainder formula, 276 of a series, 219 of a Taylor series, 269 of an alternating series, 231 Resonance, 330 Right-hand limit, 288 Root test, 239 proof, 345 Separable equation, 9 Sequence, 191 bounded, 201 bounded above, 201 bounded below, 201 convergence, 193 decreasing, 200 divergence, 193 Fibonacci, 193 general term, 191 increasing, 200

INDEX inductively defined, 193 initial value of n, 191 limit of, 193 lower bound, 201 monotone, 200 unbounded, 201 upper bound, 201 Series, 208 absolute convergence, 234 alternating, 229 approximations, 219, 231 binomial, 265 conditional convergence, 234 convergence, 208 cosine, 295, 307 divergence, 208 finite, 213 Fourier, 286 geometric, 210 infinite, 208 Maclaurin, 259 p-, 218 partial sum, 208 power, 245 remainder, 219 sine, 294, 305 sum, 208 Taylor, 259 telescoping, 212 trigonometric, 283, 286 Sigma notation, 203 Simply connected region, 30 Singular point of a function, 278 of an equation, 52, 317 Singular solution, 40 Singularity of a function, 278 of an equation, 52, 317 Solution, 3 general, 4, 43, 87, 124 homogeneous, 57 implicit, 17

--

INDEX particular, 4, 43, 87, 124 periodic, 328 singular, 40 Solution curve, 5 Squeeze theorem, 199 Standard form, 18, 44, 103 Sum of a series, 208 System of equations, 121 decoupled, 130 homogeneous, 121 in matrix form, 122 linear, 121 nonhomogencous, 121 with constant coefficients, 121 Taylor polynomial, 269 remainder formula, 276 series, 259 Telescoping series, 212 Transpose, 340 Triangle inequality, 282 Trigonometric identities, 335 identity, 285, 426, 427 integrals, 336 series, 283, 286 substitution, 338 Two-parameter family of solutions, 43 Unbounded sequence, 201 Undetermined coefficients method of, 57 Upper bound, 201 Variation of constants, 68 Variation of parameters, 68, 104 for systems, 172 Wronskian, 103, 124

449

'