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English Pages 544 [546] Year 1997
Table of contents :
Title Page
Dedication
Table of Contents
Preface
1 Definitions; Families of Curves
2 Equations of Order One
3 Numerical Methods
4 Elementary Applications
5 Additional Topics on Equations of Order One
6 Linear Differential Equations
7 Linear Equations with Constant Coefficients
8 Nonhomogeneous Equations: Undetermined Coefficients
9 Variation of Parameters
10 Applications
11 Linear Systems of Equations
12 Nonhomogeneous Systems of Equations
13 The Existence and Uniqueness of Solutions
14 The Laplace Transform
15 Inverse Transforms
16 Nonlinear Equations
17 Power Series Solutions
18 Solutions Near Regular Singular Points
19 Equations of Hypergeometric Type
20 Partial Differential Equations
21 Orthogonal Sets of Functions
22 Fourier Series
23 Boundary Value Problems
24 Additional Properties of the Laplace Transform
25 Partial Differentia1 Equations Transform Methods
Answers to Oddnumbered Exercises
Index
Elementary Differential Equations EIGHTH EDITION
Earl D. Rainville Late Professor of Mathematics University of Michigan
Phillip E. Bedient Professor Emeritus of Mathematics Franklin and Marshall College
Richard E. Bedient Professor of Mathematics Hamilton College
PRENT ICE HALL, UPPER SADDLE RIVER, NJ 07458
Library of Congress CataloginginPublication Data Rainville, Earl David Elementary differelll ial equa t i on ~ .  8th cd. / Earl D. Rainvil le, Phillip E. Bedient_R ichard E. Bedie nt. p. c. Includes index. ISBN 0 1350801 18 I. Di ffc relllial equations. I. Bedient, Philli p Edward. II. Bedient. Richard E. lI l. Title. QA37 J.R 29 1997 515.3'5dc20 963 1777
CIP
Acquisitions Edi to r: George Lobell Editorial Ass istant: Gale Epps Ed itorial Di rector: Tim Bozik Edi torinC hie f: Jerome Gra nt Ass i tant Vice Pre ident of ProducLion and Manufacturing: David R. Riccardi Edi torial/Producti on Supervisor: Robert C. Wa lters Managi ng Edi tor: Linda Mihatov Behrens Executive Managing Edi tor: Kathleen Schiaparelli Manu fac tu ring Buyer: Alan Fischer Manufactu ring Buyer: Trudy Pisciotti Marketing Manager: John Tweedda lc Marketing Assistant: Di ana Penha Creative Director: Pau la May lahn Art Manager: Gus Vibal Art Director: Maureen Eide Cover and Interior De igner: Ji ll Little Cover Photo: Spinning Schaji, by Alej andro and Moira Siiia S upplements Editor: Auclra Walsh © 1997 by PrenticeHa ll , Lnc. Simon & Schuster/A Vi acom Company Upper Sadd le Ri ver, NJ 07458 Previou edi tions copyright © 1989, 1981, 1974, 1969, 1964, 1958, 1952, and 1949 by Macmillan Publishing Company, a div ision of Macmillan, lnc. All rights reserved . No part of thi s book may be reproduced. in any fo rm or by any means, wi thout perm is ion in writing from the publisher. Prin ted in the Uni ted States of America 10 9 8 7 6 5 4 3 2
ISBN 0135080118 PRENTICEH AL L I NTERNATIONAL
(UK)
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Toron/o
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&
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For Esther, Marie, Betsy Kate, and Adam
Contents Preface 1
1
/
1
18
Separation of Variables I 18 Homogeneous Functions I 24 Equations with Homogeneous Coefficients Exact Equations I 29 The Linear Equation of Order One / 35 The General Solution of a Linear Equation Computer Supplement / 43
Numerical Methods 3.1 3.2 3.3
/
Examples of Differential Equations I Definitions I 2 Families of Solutions I 5 Geometric Interpretation I 10 The Isoclines of an Equation I 12 An Existence Theorem I 14 Computer Supplement I 15
Equations of Order One 2.1 2.2 2.3 2.4 2.5 2.6 2.7
3
xiii
Definitionsi Families of Curves 1.1 1.2 1.3 1.4 1.5 1.6 1.7
2
/
/
25
/
38
45
General Remarks / 45 Euler's Method / 45 A Modification of Euler's Method
/
48
v
vi
Contents
3.4 3.5 3.6 3.7 3.8 3.9 4
Elementary Applications 4.1 4.2 4.3 4.4 4.5
5
/
/
49
62
Velocity of Escape from the Earth / 62 Newton's Law of Cooling / 64 Simple Chemical Conversion / 65 Logistic Growth and the Price of Commodities Computer Supplement / 73
Additional Topics on Equations of Order One 5.1 5.2 5.3 5.4 5.5 5.6 5.7
6
A Method of Successive Approximation An Improvement on the Method of Successi ve Approximation / 51 The Use of Taylor's Theorem / 52 The RungeKutta Method / 54 A Continuing Method / 58 Computer Supplement / 60
69
/
75
Integrating Factors Found by Inspection / 75 The Determination of Integrating Factors / 79 Substitution Suggested by the Equation / 83 Bernoulli's Equation / 86 Coefficients Linear in the Two Variables / 89 Solutions Involving Nonelementary Integrals / 94 Computer Supplement / 97
Linear Differential Equations 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 6.10
/
/
99
The General Linear Equation / 99 An Existence and Uniqueness Theorem / 100 Linear Independence / 102 The Wronskian / 103 General Solution of a Homogeneous Equation / 106 General Solution of a Nonhomogeneous Equation / 107 Differential Operators / 109 The Fundamental Laws of Operation / 111 Some Properties of Differential Operators / 113 Computer Supplement / 115
vii
Con.tents
7
Linear Equations with Constant Coefficients 7.1 7.2 7.3 7.4 7.5 7.6 7.7
8
8.2 8.3 8.4 8.5 9
/
152
Introduction / 152 Reduction of Order / 152 Variation of Parameters / 156 Solution of y"+ y =ICx) / 161 Computer Supplement / 164
Applications 10.1 10.2 10.3 10.4 10.5 10.6 10.7 10.8
134
Construction of a Homogeneous Equation from a Specific Solution / 134 Solution of a Nonhomogeneous Equation / 137 The Method of Undetermined Coefficients / 139 Solution by Inspection / 144 Computer Supplement / 150
Variation of Parameters 9.1 9.2 9.3 9.4 9.5
10
Introduction / 117 The Auxiliary Equation: Distinct Roots / 117 The Auxiliary Equation: Repeated Roots / 120 A Definition of exp z for Imaginary z / 123 The Auxiliary Equation: Imaginary Roots / 125 A Note on Hyperbolic Functions / 127 Computer Supplement / 132
Nonhomogeneous Equations : Undetermined Coefficients / 8.1
/
/
165
Vibration of a Spring / 165 Undamped Vibrations / L67 Resonance / 169 Damped Vibrations / 172 The Simple Pendulum / 177 Newton's Laws and Planetary Motion / 178 Central Force and Kepler's Second Law / 179 Kepler's First Law I 180
117
viii
Contents
10.9 10.10 11
linear Systems of Equations 11.1 11.2 11.3 11.4 11.5 11.6 1l.7 11.8 11.9
r2
186
Introduction / 186 FirstOrder Systems with Constant Coefficients Solution of a FirstOrder System / 187 Some Matrix Algebra / 189 FirstOrder Systems Revisited / 195 Complex Eigenvalues / 204 Repeated Eigenvalues / 208 The Phase Plane / 216 Computer Supplement / 222
Nonhomogeneous Systems Arms Races / 228 Electric Circuits / 232 Simple Networks / 235
/
/
/
186
224
/
Preliminary Remarks / 243 An Existence and Uniqueness Theorem / 243 A Lipschitz Condition / 246 A Proof of the Existence Theorem / 246 A Proof of the Uniqueness Theorem / 250 Other Existence Theorems / 251
The Laplace Transform 14.1 14.2 14.3 14.4 14.5 14.6
/
224
The Existence and Uniqueness of Solutions 13.1 13.2 13.3 13.4 13.5 13.6
14
/
Nonhomogeneous Systems of Equations 12.1 12.2 12.3 12.4
13
Kepler's Third Law / 182 Computer Supplement / 184
252
The Transform Concept / 252 Definition of the Laplace Transform / 253 Transforms of Elementary Functions / 253 Sectionally Continuous Functions / 257 Functions of Exponential Order / 258 Functions of Class A / 261
243
ix
Conten.ts
14.7 14.8 14.9 14.10 15
Inverse Transforms 15.1 15.2 15.3 15.4 15.5 15 .6 15.7 15.8 15.9 15 .10
16
Transforms of Derivatives / 263 Derivatives of Transforms / 266 The Galmna Function / 267 Periodic Functions / 269
274
/
Definition of an Inverse Transform / 274 Partial Fractions / 277 Initial Value Problems / 280 A Step Function / 286 A Convolution Theorem / 294 Special Integral Equations / 298 Transform Methods and the Vibration of Springs The Deflection of Beams / 307 Systems of Equations / 310 Computer Supplement / 316
Nonlinear Equation
/
303
/
347
320
/
16.1 Preliminary Remarks / 320 16.2 Factoring the Left Member / 320 16.3 Singular Solutions / 323 16.4 The cDiscliminant Equation / 325 16.5 The pDiscriminant Equation / 326 16.6 Eliminating the Dependent Variable / 328 16.7 Clairaut's Equation / 330 16.8 Dependent Variable Missing / 334 16.9 Independent Variable Missing / 335 16.10 The Catenary / 338 17
Power Series Solutions 17.1 17.2 17.3 17.4 17.5 17.6
/
342
Linear Equations and Power Series / 342 Convergence of Power Series / 343 Ordinary Points and Singular Points / 345 Validity of the Solutions Near an Ordinary Point Solutions Near an Ordinary Point / 347 Computer Supplement / 356
x
Contents
18
Solutions Near Regular Singular Points
/
358
18.1 18.2 18.3
Regular Singular Points / 358 The Indicia} Equation / 360 Form and Validity of the Solutions Near a Regular Singular Point / 362 18.4 Indicial Equation with Difference of Roots N onintegral / 363 18.5 Differentiation of a Product of Functions / 367 18.6 Indicial Equation with Equal Roots / 368 18.7 Indicial Equation with Equal Roots: An Alternative / 374 18.8 Indicial Equation with Difference of Roots a Positive Integer: Nonlogarithmic Case / 377 18.9 Indicial Equation with Difference of Roots a Positive Integer: Logarithmic Case / 381 18.10 Solution for Large x / 385 18.11 ManyTerm Recurrence Relations / 388 18.12 Summary / 392
1Y
Equations of Hypergeometric Type 19.1 19.2 19.3 19.4 19.5 19.6 19.7 19.8
20
20.3
396
Equations to Be Treated in This Chapter / 396 The Factorial Function / 396 The Hypergeometric Function / 397 Laguerre Polynomials / 399 Bessel's Equation with Index Not an Integer / 400 Bessel's Equation with Index an Integer / 401 Hermite Polynomials / 402 Legendre Polynomials / 403
Partial Differential Equations 20.1 20.2
/
/
404
Remarks on Partial Differential Equations / Some Partial Differential Equations of Applied Mathematics / 404 Method of Separation of Variables / 406
404
xi
Con.tents
20.4 20.5 21
Orthogonal Sets of Functions 2l.1 21.2 21.3 21.4 21.5 21.6
22
23.3 23.4 23.5 23.6 23.7 24
/
411
418
422
Orthogonality of a Set of Sines and Cosines / 425 Fourier Series: An Expansion Theorem / 427 Numerical Examples of Fourier Series / 431 Fourier Sine Series / 438 Fourier Cosine Series / 441 Numerical Fourier Analysis / 443 Improvement in Rapidity of Con vergence / 444 Computer Supplement / 445
/
447
The OneDimensional Heat Equation / 447 Experimental Verification of the Validity of the Heat Equation / 453 Surface Temperature Varying with Time / 455 Heat Conduction in a Sphere / 457 The Simple Wave Equation / 458 Laplace's Equation in Two Dimensions / 461 Computer Supplement / 464
Additional Properties of the Laplace Transform 24.1 24.2
/
425
Boundary Value Problems 23.1 23 .2
/
Orthogonality / 418 Simple Sets of Polynomials / 419 Orthogonal Polynomials / 419 Zeros of Orthogonal Polynomials / 421 Orthogonality of Legendre Polynomials / Other Olthogonal Sets / 424
Fourier Series 22.1 22.2 22.3 22.4 22.5 22.6 22.7 22.8
23
A Problem on the Conduction of Heat in a Slab Computer Supplement / 416
Power Series and Inverse Transforms The En·or Function / 471
/
467
/
467
24.3 24.4 25
Bessel Functions / 478 Differential Equations with Variable Coefficients
/
Partial Differentia1 Equations Transform Methods 25 .1 25 .2 25.3 25.4 25.5 25.6
Boundary Value Problems / 481 The Wave Equation / 485 Diffusion in a SemiInfinite Solid / 488 Canonical Variables / 491 Diffusion in a Slab of Finite Width / 493 Diffusion in a QuarterInfinite Solid / 496
Answers to Oddnumbered Exercises Index
/
527
/
500
480
/
481
Preface In publishing a new edition of Elementary Differential Equations, we have two main goals. First, we hope to maintain the direct style that users of earlier editions have come to expect. Secondly, in response to changes in the nature of many differential equations courses, we have added some new geometric material, reorganized some sections, and added a computer component to the text. The new geometric material appears mainly in Sections 1.4 and 11.8. In the former we introduce students to the idea of a family of curves as solutions to a differential equation, and in the latter, the concept of the phase plane of a system of equations is presented. We have moved the treatment of systems of equations to an earlier place in the text. Of all areas of mathematics covered in a typical undergraduate cuniculum, the field of differential equations is arguably the most affected by the computer. Numerous packages have been produced which are either designed specifically for differential equations, or have subpackages designed for that material. We have made the rather arbitrary decision to present our computer examples using the Maple package. We could have equally well chosen any of the other Computer Algebra Systems such as Mathematica, MatIab , or Derive. Further, there are a number of numerical graphing packages that are more efficient for producing geometric results. Among the more commonly available are MacMath and Phaser. Each computer supplement contains an example from the conesponding chapter, worked out with the aid of Maple. Subsequently a set of computer exercises is presented which can be solved using whichever package is available to the snldent. It is our hope that these brief introductions will encourage users to look beyond the text for further computer explorations. The authors wish to thank the following reviewers of the Eighth Edition manuscript for their comments : Ebrahim Saichi, University of Nevada Las Vegas; J. P. Mokanski, University of Guelph; Thomas G. Berry, University of Manitoba; Giles Wilson Maloof, Boise State University; John H. Ellison, Grove City College; James L. Handley, Montana Tech; Baigiao Deng, Columbus College, and Jay Delkin, University of Western Ontario. Philli P E. Bedient Richard E. Bedient
xiii
Definitions; Families of Curves
1. 1
Examples of Differential Equations The construction of mathematical models to approximate real world problems has been one of the most important aspects of the theoretical development of each of the branches of science. It is often the case that these mathematical models involve an equat ion in which a function and its derivative play important roles. Such equations are called differential equations. As in equation (3), a derivative may be involved imp licitly through the presence of differentials. OUf a im is to find methods for olving differential equations; that is, to find the unknown function or fun ctions that satisfy the differential equation . The following are examples of differential equations:
dy
 = d
2
dx
y 2
(1)
cosx ,
dx
+k
2
(2)
y = 0,
(3)
(4) d 2i L.2
dl
di
I
dt
C
+ R  + i a2 v ax2
d
2W) 3 
( dx2
= Ewcoswt ,
aV
(5)
2
+
(6)
ay 2 = 0, dw
xydx
+w =
0,
(7)
1
1.2
Definitions
3 (I)
is an equation of "order two." It is also referred to as a "secondorder equation." More generally, the equation F(x , Y,Y,I · ··,Y (II » = 0
(2)
is called an "nthorder" ordinary differential eq uation. Under sui table restrictions on the function F , equation (2) can be solved explicitly for y( lI ) in terms of the other n + 1 variables x , y, y' , . . . , y (II  I), to obtain
y 0 , (2 / ..J3)x 3 / 2 is a solution of the equation of Exercise 6.
Families of Solutions Every student of calculus has spent a significant amo unt of time in finding the solutions of firstorder differential equations of the form
d y = f (x).
( I)
dx This antiderivative problem is often written
y=
ff
(x) dx
+c
(2)
and the student is asked to find a single function of x whose derivative is identical to f (x) on some interval. Having determined such a functio n it is proved that any other function that satisfies the differential equation (1) differs from that function by a constant for all x in the interval. This important theorem establishes the
6
Chapter 1
Definitions; Families of Curves
fact that solutions of equation (1) do not occur in isolation, but as oneparameter families of solutions, the parameter being the socalled arbitrary constant c of equation (2). If one considers the general firstorder differential equation
dy dx .  r(x ,
(3)
)1) ,
the problem of finding so lution , that is, functions ¢ (x) that satisfy the equation when substituted for the dependent variable y, is in general more difficult if not impossible. However, as we shall see, these solutions, when they exist, occur as oneparameter families of solutions. In Chapter 2 we shall study a number of methods for finding families of solutions for some particular types of firstorder equations, but in general there is no method of attack that will solve eve ry such equation. We content ourselves for the moment by illustrating what happens in a few simple examples.
EXAMPLE 1.1 The differential equation
dy = 8sin4x dx

(4)
has the family of solutions
y =  2 cos 4x
+ c,
(5)
a simple antiderivative having produced this result. If we wish to find one member of the family (5) that satisfies the additional condition that y = 6 when x = 0, we are forced to choose c = 8. We then say that
y = 2cos4x
+8
is the sol ution of the initial value problem
dy .  = 8 s1ll4x,
y = 6, when x =
dx
o.
•
EXAMPLE 1.2 From calculus we Jearn that the derivative of the function f ex) = ce 2x is f' (x) = 2ce 2x . Phrased in the language of differential equations, we say that the differential equation
dy

dx
=2y
(6)
/.3
Families ofSolulio17s
7
has the family of sol ution s
y = ce 2x .
(7)
If we seek a solution of equation (6) that sati sfies dy dx = 2y ,
y = 4, when x = 0,
(8)
then from eq uation (7) we see that c = 4 and the so lution of (8) is
y
= 4e 2x .
• EXAMPLE 1.3 Consider the second order equation
y" = 12x 2 .
(9)
Integration of both sides of this equ ation w ith respect to x yields
y' = 4x 3 + C I·
(10)
A eco nd integration produces ( I I)
In this example there are two arbitrary con stants, so we have a twoparameter family of solutions. T hi means that to si ngle out one member of thi s family we need to provide two pieces of information. These are usuall y given by pecifying the values of both y and y' for the same va lue of x . Fo r exam ple, suppose we want the solution to (9) that also satisfies y eO) = I , and y' (0) = 2. Substitu ting x = 0, and y' = 2 into (10) we see that C I = 2, so tbat 4 y=x +2x + C2.
Finally, substituting x is
= 0, y =
1, we see that C2
y=
X4
=
I so thatthe required soluti on
+ 2x + 1.
• EXAMPLE 1.4 Consider the o neparameter fa mily of curves
x3

3x 2 y =
C.
(1 2)
8
Chapter 1 Definilions ; Families of Curves
A d iffere ntiation of both sides of this eq uati on with respect to x yields 2
3x 
dy  6xy = 0 dx
?
3 x~
or
dy dx
x  2y x
when x
i= O.
(13)
If we had started this exampl e with equatio n (13) and tried to find the fa mily of curves given by eq uation ( 12), we would face a more chall engin g problem th an in the prev ious examples. We will learn how to solve eq uation (13) in Chapter 2. Here we simply point o ut that the value x = 0 creates a difficulty for both the differential equation (13) and for its fami ly of sol ution s
x3  c 3x
y =2
Jbtained from equation (12).
•
EXAMPLE 1.5 Co nsider the family of circles (x  2)2
+ (y + 1)2 =
c2 .
( 14)
A simple differentiation wilh respect to x yields
2(x  2)
dy dx
+ 2(y + 1) ~ = 0
or
dy dx
 (x  2)
y
+1
when y =f.  1.
'
(15)
We are forced here to think of the family of circles as consisting of two fa mili es of semicircle: the one family
y =  1 + Jc 2

(x  2)2
(16)
(x  2) 2.
(17)
and the other
y =  1
J c2 
In equat ion (16) we have a family of solutio ns of the d ifferential eq uation for y >  1, whereas in equatio n (17) we have a family of soluti ons of (15) for y <  1. To solve the initial value problem
dy dx
(x  2) y+1 '
y = 2 , when x =  1,
(18)
1.3
Families of Solutions
9
we must choose the parameter c from equation (16), since 2 >  1. We have 2 = 1 + .Jc2  9 or c = /I8. The fu nction y =  1+
j
18  (x  2)2
is therefore the solution we seek. Its grap h is that of a semicircle of radius
• Exercises
.JT8.
•
Solve the differenti al equation s in Exerc ises I thro ugh 6.
l.
dy
dy dx dy 5. dx dy 6. dx
dx = x 3 + 2x .
4
4.
3 dy 2. dx x dy 3.  = 4cos6x. dx
x2 
1
2
= x2 + 4 3
x2
+x
Solve the initi al valu e problems in Exerc ises 7 through 12.
dy = 3e X y = 6, when x = o. dx ' dy 8.  = 4e 3 X, y = 2, when x = o.
7.

elx
9.
dy
d x = 4 y,
y = 3, when x =
o.
lO.
dy dx = 5y , y = 7, when x =
11.

= 4 sJI12x ,
dy
= x 2 + 3 + e 2x ,
12. 13.
ely
dx dx
.
y
= 2,
o.
when x
= n / 2.
y =  1, when x
Show that the family of circles (x + 1)2 + (y  3)2 = c2 can be interpreted as two families of solutions of the differentia l eq uations
 ex + 1)
dy dx 14.
= o.
y  3
Show that the fa mi ly of parabolas y = ax 2 can be interpreted as two families of solutions of the differenlial equation
dy
2y
dx
x
10
Chapler I
Deiilliliolls; Families of Curves
then find the solution of the initial value problem dy
2y
dx
x
y=2 , whenx=l.
For what values of x is this solution valid? Notice also that there is no solution of this differential equation that satisfies the initial condition y = 2, when x = O.
1 .4
Geometric Interpretation In Section 1.3 we saw that a firstorder equation u ually has a fami ly of solutions. A useful technique in understa nding the nature of these solutions is to graph representative solutions from this family.
EXAMPLE 1.6 Graph several members of the family of solutions of the equation dy  = 8sin4x.
dx
(1)
Recall from Section 1.3 that the family of sol utions i y = 2cos4x
+ c.
(2)
Graphing the sol utions corresponding to c = 2, 1, 0, J we obtain F igure 1.1. II is not difficult to imagine what the rest of the fami ly looks like.
y
Figure 1.1
•
1.4
Ceomelric i nlerpreutliol1
11
The oneparameter family of solution curves of Example 1.6 satisfies an important property: Through each point in the plane there passes one and only one member of the family of solutions. We wi ll formalize this fact in Section 1.6, but for now, we claim that it is true fo r the sol ution of any suitably restricted firs torder differential equati on. If we specify a point in the pla ne, by the property above there will be exactly one solution passing through that point. The unique curve that resu lts is the solution curve of the initial value problem. This is the geometric ver io n of the process desc ribed in Section 1.3 . If we extend these ideas to higher order equations, we find that only part of the geometri c interpretation carries over.
EXAMPLE 1.7 Graph severa l members of the family of solutions of the equation (3) As we saw in Section 1.3, the family of solutions is
(4) Graphing the soluti ons cO ITesponding to the pairs of constants Cl = 2, C2 = 1; C I = 0, C2 = 0; and C I = 0, C2 = 1 we obtain Figure 1.2. Clearly these solutions do not satisfy the uniqueness propel1y of the firsto rder case. There are at least two sol utions passing through the points (0, 1) and a point near (112, 0). We note however that these pairs of solutions do not have the same slope at their point of intersection. That is, to specify a particular solution to a secondorder equation, we could specify both a point through which the so luti on y
"c:;;;;""+=~
Figure 1.2
x
12
Chapter I
Definitions: Families of Curves
passes, and the slope at that point. Given this infonnation there is, in this case, a unique solution.
•
The eco ndorder version of the geometric property stated above then becomes: Through each point in the plane there passes one and only one member of the family of solutions that has a given slope. This property wi ll be discussed in greater detail in Chapter 6 .
• Exercises
1.5
L.
For Exercises I through 6 of Section 1.3, draw a representative sample of sol ution curves.
2.
For Exercises 7 through 10 of Section 1.3, draw the grap h of the soluti on of the initial value problem.
The Isoclines of an Equation In Section 1.4 we noted some of the geometric properties of families of sol utions wh ich we had found by the analytic methods of Section 1.3. In this section we see that we can use geometllc methods to actually find solution curves. Con ider the equation of order one
dy dx
 =
f(x, y) .
(1)
We can think of equation (1) as a machine that assigns to each point (a, b) in the domain of f some direction with slope f(a , b). We can thus speak of the direction field of the differential equation. In a real sense any solution of equation (I) must have a graph, which at each point has the direction equation (1) requires. One way to visualize this basic idea is to draw a short mark at a number of point to indicate the direction associated with each of those points. This can be done rather systematically by first drawing curves called isoc lines, that is, curves along which the direction indicated by equation (1) is fixed.
EXAMPLE 1.8 Consider the equation dy
dx y .
(2)
The isoclines are straight line f(x, y) = y = c. For each value of c we obtain a line in which, at each point, the direction dictated by the differential equation is that number c. For example, at each point along the line y = I, equation (2) determines a direction of slope 1. In Figure 1.3 we have drawn severa l of these
1. 5
The Isoclines of an Equation
13
y
"........,t"' DEplotl ( diff(y(x) , x)=x A2+y A 2 , y(x), x=2 .. 2, y=2 .. 2 );
will produce Figure 1.5. To add the solution curves, we can add several selected poinls at which to start. The command > DEplotl
(dif f(y (x) ,x ) =xA2 +yA2 , y(x), x=2 .. 2, {[O,2].(O,O],[O,l)L y= 2 .. 2 );
produces Figure 1.6.
1.7
Computer Supplelllent
17
/"
....
 ,  ..
/"
/"
/
/"
/"
/ I I
/
/
f I I
I I
I f
I
/
I / I I
I
r I
....
/"
,
/
/"
/"
/
I I I / I
I I I I I
./
/"
/
/
/
/
I I
I
I I
I f I
/ f I
I
I f I
I I
I I
I I
I I I I I
f I I I I
Figure 1.6
• Exercises 1.
For Exercises 1 through j 0 of Section [ .5 use the computer software package of yo ur cho ice to produce slope fie ld s and representative soluti on cu rves.
2.
Consider the problem of drawing isoclines for the equation dy / d x = y si n (y + x) . This is a very hard problem. Now use the compu ter to draw the slope field and representative solution curves .
3.
What ca n you say about the curves drawn in Exercise 2 as x + x ')0  oo? Do your an. wers depend on the initial condi tions?
00 ,
as
Equations
2
of Order One
2.1
Separation of Variables In this chapter we study several elementary methods for solving firstorder differential equations. We begin by studying an equation of the form Mdx+Ndy = 0, where M and N may be functions of both x and y . Some equ ations of this type are so simple that they can be put in the form A(x)dx
+ B (y)dy = 0;
(1)
that is, the variables can be separated. Then a solution can be written at once. For it is only a matter of finding a fu nction F whose total differential is the left member of (1) . Then F = c, where c is an arbitrary constant, is the desired result.
EXAMPLE 2.1 Solve the equation dy
2y
dx
x
for x > 0 and y > O.
(2)
We note that for the function in equation (2), the theorem of Section 1.6 applies and assures the existence of a unique conti nuous solution through any point in the first quad rant. By separating the variables we can write dy
y
=
2dx x
Hence we obtain a family of solutions Inlyl =2 Inl xl+c
(3)
or, because we are in the first quadrant,
Y = ec x? .
18
(4)
2.1
Separation of Variables
19
If we now put Cl = e C , we can write ?
Y=
CIX ,
CI
> 0.
(5)
• EXAMPLE 2.2 Solve equation (2) of ExampJe 2.1 for x =1= 0. The argument now must be taken in two parts. First, if y =1= 0, we can proceed as before to equation (3 ). However, equation (5) must be written
Iyl =
2 CIX ,
CI
> 0.
°
(6)
Second, if y = 0, we see immediately that since x =1= 0, y = is a solution of the differential equation (2). As a matter of convenience the solutions given by equation (6) are usually written (7)
where C2 is taken to be an arbitrary real number. Indeed, this form for the solutions incorporates the special case y = 0. Several representative sol ution curves are show n in Figure 2.3. We must be cautious, however. The function defined by
g(x)=x 2 ,
x> o
x.::: 0, hown in bold in Figure 2.1, obtained by piecing together two different parabolic arcs could also be considered a solution of the differential equation, even though y
~~~~~x
Figure 2.1
20
Chapler 2
Equations of Order One
this function is not included in the family of equation (7). The uniqueness statement in the theorem of Section 1.6 indicates that as long as we restrict our attention to a point (xo, YO) with Xo =f. 0 and consider a rectangle with center at (xo, Yo) containing no points at which x = 0, then in that rectangle there is a unique solution that passes through (xo , Yo) and is continuous in the rectangle.
•
EXAMPLE 2.3 Solve the equation (l
+ l)dx + ( 1 +x2)dy =
0,
(8)
with the "initial condition" that when x = 0, y = 1. If we write this equation in the form
dy
(1
+ y2)
1+ x 2 '
dx
we observe that the right member and its partial derivative with respect to yare continuous near (0, 1). It follows that a unique solution exists for equation (8) that passe through the point (0, 1) . From the differential equation we get
~+~o 1+ x2 1 + y2  , from which it follows at once that mctan x
+ arctan y
= c.
(9)
In the set of solutions (9), each "arctan" stands for the principal va lue of the inverse tangent and is subj ect to the restriction
!n < arctan x < !n. The initial condition that y = \ when x = 0 permits us to determine the va lue of c that must be used to obtain the particular solution desired here. Since arctan 0 = 0 and arctan (1) = the oLution of the initial value problem is
in, arctan x+ arctan y =  in.
(10)
Suppose next that we wish to sketch the graph of (10). Resorti ng to a device of trigonometry, we take the tangent of each side of (10). Because tan (arctan x) = x and tan (A
+ B) =
tan A + tan B    1  tan A tan B '
2.1
Separation of Variables
21
we are led to the equation
x+y Ixy
   = 1 ,
or
xy  x  y  1 = O.
(11)
Now (11) is the equation of an equilateral hyperbola with asymptotes x = 1 and y = 1. But if we turn to (10), we see from arctan
x=  in  arctan y
that since ( arctan y) < ~n, arctanx
1. In this example we were forced onto the left branch, equation (10), by the initial condition that y =  1 when x = O. One distinction between equations (10) and (1]) can be seen by noting that a computer, given the differential equation (8) and seeking a solution that passes through the point (0, 1), would be constrained to follow the left branch of the curve in Figure 2.2. The barrier (asymptote) at x = I would prevent the computer from learning of the existence of the other branch of the hyperbola (II) .
•
y I I I I I I I I
\ \ \ \
"" ........
....
_
 T        
a
Figure 2.2
I
22
Chapter 2
Equations of Order One
EXAMPLE 2.4 Solve the initial value problem
°
2x(y
+ 1) dx 
(12)
y dy = 0,
where x = and y =  2. Separating the variables in equation (12), we obtain 2xdx =
(1 _1 _) + y
dy,
y
1
'1=1.
Integrating, we get a family of sol utions given implic itl y by
x 2 = y In Iy
+ 11+ c.
(13)
Since we seek a member of this family that passes through the point (0,  2), we must have
0= _ 2  In
I  11+ c,
or
c
= 2.
Thu s the solution to the problem is given implicitly by
x 2 = y In Iy + 11+ 2. The reader should note how the theorem of Section 1.6 app lies to th is problem to indicate that we have found implicitly the unigue so lution to the in itial value problem which is continuous for y < I. A representative sample of sol ution curves is shown in Figure 2.3 with the patticular soultion shown in bold. Note that some of these curves are not the graphs of functions, and mu st be split into separate arcs where they cross the line y = 0, much as we did in Example 2.2. y
 + + t++ 
Figure 2.3
x
•
2. J
Separation of Variables
23
Exercises In Exercises Lthrough 6, obtain the particular solution sati sfyi ng the ini tial condition indicated. In each exercise interpret your answer in the light of the existence theorem of Section 1.6 and draw a graph of the solution.
1.
dr/dt =  4rt; when t = 0, r = ro .
2.
2xyy' = 1 + y2; when x = 2, y = 3.
3. xyy' = 1 + y2; when x = 2, y = 3. 4.
2ydx = 3xdy; when x = 2, y = 1.
5. 2ydx = 3xdy; when x =  2, y = 1. 6.
2y dx = 3xdy; when x = 2, y =  1.
In Exerc ises 7 through 10, obtai n the particular solution satisfy ing the initial condition indicated.
7.
y' = x exp (y  x 2); when x = 0 , y = O.
8.
xy 2 dx+e X dy = 0; whenx ~ oo,y ~! . (2a 2 r2)dr=r3 sinede; whene = O, r = a.
9. 10.
v(dv/dx) = g; when x = xo, v =
Va·
In Exercises 1 1 through 37, obtain the general solution.
11.
(l  x)y' = y2.
24.
(l  y)y' = x 2 .
12.
sin x sin y dx+cos x cos y dy = O.
25.
x 2yy'
13. 14.
xy 3 dx
+ e dy 2ydx = 3xdy.
26.
tan y dy = sin 3 x dx.
27 .
y' = cos 2 X cos y .
15.
mydx=nxdy.
28.
y' = y secx.
]6.
y' = xy2.
29.
17.
dV/dP = VIP. ye 2X dx = (4 + e 2x ) dy.
30.
dx = t(1 + t 2) sec2 x dt. (e 2x + 4)y' = y .
18.
x2
= O.
= b(cosedr +rsined8).
19.
dr
20.
xy dx  (x + 2) dy = O. x 2 dx + y(x  l)d y = O.
21. 22. 23. 37.
3l. 32. 33. 34.
= eY •
2
a dfJ+fJ da+afJ(3 da+dfJ) = O. (l+ ln x)dx+(l+lny)dy = O. x dx  .Ja 2  x 2 dy = O. x dx + .Ja 2  x 2 dy = O. a 2 dx = x.Jx2  a 2 dy.
35. + tan y dy = O. 36. y In x In y dx + dy = O. xy 3 dx + (y + l )e dy = O. 2 2 (xy +x)dx = (x y 2 + x + y2 + l)d y . X
cos 2 y dx
X
2.3
Equations wilh Homogeneous Coefficients
25
Theorem 2 . 1 IfM(x, y ) and N(x, y) are both homogeneous and of the same deg ree, the function M(x, y)IN(x, y ) is homogeneous of degree zero.
Proof of Theorem 2. 1 is left to the studen t.
Theorem 2.2 Iff(x, y) is homogeneous ofdegree zero inx and y, j(x, y) is afunction ofylx alone. Proof. Let us put y = vx. Then Theorem 2.2 states that if f (x, y) is homogeneous of degree zero, f (x, y) is a function of valone. Now f(x, y )
=
f(x, vx)
= xOf (1,
v)
=
f(l, v),
(4)
in which the x is now pl aying the role taken by A in the definition (3). By (4), f(x, y) depends on v alone, as stated in Theorem 2 .2.
•
Exercises
D etermine in each exercise whether or not the function is homogeneous. Jf it i homogeneous, state the degree of th e function.
2.
4x 2  3x y + y2 . x 3  xy + y3.
3.
2y+jx 2
4.
Jx 
5.
eX .
6.
tanx.
7.
exp
8.
3y tan  . x
I.
2 .3
9.
(x
. . 10
X
2
+ y2.
y.
x 2 + 3xy x 2y
12.
x5 x 2 + 2y 2 '
13.
(u 2
14.
(u 2 _ 4v 2 )1 /2 .
+ v 2 )3/2 .
x IS . itan  . y (x2 + y2 ) 1/2 16. (x2 _ y2) 1/2 .
(~). + y2) exp
11.
l7 .
(2;) +
4xy.
. Y . x sm   y sm  . x y
a + 4b a  4b
x
18. In  . y 19. x In x  y In y. 20.
x In x  x In y .
Equations with Homogeneous Coefficients Suppose that the coefficien ts M and N in an equation of order one, M(x, y) d x
+ N(x , y) dy
= 0,
(1)
26
Chapter 2
Equations of Order One
are both homogeneous functions and are of the same degree in x and y. By Theorems 2.1 and 2.2 of Section 2.2, the ratio M / N is a function of y I x alone. Hence equation (1) may be put in the form dy dx
+ g (~) x
= 0.
(2)
This suggests the introduction of a new vru.iable v by putting y becomes dv
x
dx
+ v + g(v)
=
vx. Then (2)
(3)
= 0,
in which the variables are separable. We can obtain the solution of (3) by the method of Section 2.1 , insert y/x for v, and thus arrive at the solution of (1). We have shown that the substitution y = vx will transform equation (1) into an equation in v and x in which the variables ru.·e separable. The method above would have been equally successful had we used x = vy to obtain from (1) an equation in y and v. See Example 2.6.
EXAMPLE 2.5 Solve the eq uation (x 2

xy
+ i) dx
(4)
 x y dy = O.
Since the coefficients in (4) are both homogeneous and of degree two in x and y, let us put y = vx. Then (4) becomes (x 2

x 2v
+ x 2 v 2 ) dx
 x 2 v(v dx
+ x dv)
= 0,
from which the factor x 2 shou ld be removed at once. That done, we have to solve (1  v
+ v 2 ) dx 
v(v dx
+ x dv)
= 0,
or (I  v) dx  xv dv = O.
Hence we separate variables to get dx vdv  +   = 0. x v I Then from
1]
dx + [ 1 +  . . x v 1
dv =
°
2.3
Equations with Hom.ogeneous Coefficients
27
a family of solutions is seen to be In
Ix l + v + In Iv 
11 = In
lei,
or
xCv  l)e V = c. In terms of the original variables, these sol utions are given by
(~l)exp(~)
x
= c,
or (y  x) exp
(~)
= c.
•
EXAMPLE 2.6 Solve the equation
(5) Again the coefficients in the eq uation are homogeneou s and of degree two. We could use y = vx, but the relative simplicity of the dx term in (5) suggests that we put x = vy. Then d x = v dy + y dv , and equatio n (5) is replaced by
ul(vdy
+ ydu) + (u 2 l + i)dy =
0,
or
v(udy
+ ydu) + (u 2 + l )dy =
0.
Hence we need to olve uy du
+ (2u 2 + 1) dy
= 0,
+ 1) + 41n Iyl
= In c,
which leads at once to In (2v 2 or
l(2v 2
+ 1) =
c.
Thus the desired sol ution s are given by 2X 2 l ( yz + 1)
= c;
(6)
28
Chapter 2
Equations of Order Olle
that is,
(7) Since the left member of equation (7) can not be negative, we may, fo r sym metry's sake, change the arbitrary co nstant to writing
ci,
/(2x
2
+ /) = ci·
It is worthwhil e fo r the tuden t to attack equation (5) usi ng y = vx. That method leads directly to the equation (v
3
+ 2v) dx + x(v 2 + 1) dv = O.
Frequ ently in equations with homogeneous coeffi cients, it is q uite immaterial whether one uses y = vx or x = vy. However, it is so metimes easier to substitute for the variab le whose differential bas the si mpler coefficient.
•
• Exercises In Exerci es
j
through 21, obtai n it fami ly of sol utions.
1.
3(3x2+y2)elx2xydy=0.
2.
(x  2y) dx + (2x + y) dy = O. 2(2x 2 + y2) dx  xy dy = O.
3. 4. 5.
xydx_(x2+3y2)dy= O. x 2 y' = 4x 2 + txy + 2yZ.
14.
+ (x 2 + y2) dy = O. (x  y)(4x + y)elx +x(5x  y)dy = O. (5v  u) du + (3v  7u) dv = O. (x 2 + 2xy  4y2) dx  (x 2  8xy  4y2) dy x(x 2 + y2)2(y dx  x ely) + y6 ely = O. (x 2 + yZ) dx + xy dy = O. xy dx  ex + 2y)2 dy = O. v 2 elx + x(x + v)dv = O. [x csc (y/x)  y ] dx + x ely = O.
IS.
xdx+s in2(y/x)[yelxxely ] =0.
16.
(x  y l ny
O.
17 .
[x 
O.
6. 3xy dx 7. 8.
9.
10. 11. 12 . 13.
+ y ln x)dx +x(ln y  In x)dy = y arctan (y/x) ] dx + x arctan (y/x) ely =
18. yZ dy = x(x ely  y dx)e x / y . 19 . t (S2 + (2) ds  s(s2  [2) elt = O.
= O.
2.4
20. 21.
22.
Exact Equations
29
ydx = (x + jy2  x 2) dy. (3x 2  2xy + 3i) dx = 4xy ely .
Prove that with the aid of the substitution y = vx, you can solve any equation of the form y"l(x) dx
+ H( x,
y)(y dx  x dy) = 0,
where H (x, y) is homogeneolls in x and y. In Exercises 23 lhrougb 35, find tbe particu lar sol ution indicated.
23. 24 . 25. 26. 27. 28. 29. 30. 31. 32.
+ (3x + y) ely = 0 ; when x = 3, y = 2. (y  jx2 + y2) elx  x ely = 0; when x = 0, y = 1. (y + jx2 + y2) dx  x ely = 0; when x = .../3, y = 1. [x cos 2 (y/x)  y] dx + x ely = 0; when x = 1, Y = n/4 . (y2 + tx y + 16x 2) elx + x 2 ely = 0; when x = 1, y = 1. y2 dx + (x 2 + 3xy + 4y2) dy = 0; when x = 2, y = 1. xy dx + 2(x 2 + 2y2) dy = 0; when x = 0, y = 1. y(2x2  xy + y2) dx  x2(2x  y) dy = 0; when x = I , Y = y(9x  2y) dx  x(6x  y) dy = 0; when x = 1, y = 1. y(x2 + y2) dx + x(3x 2  5 y2) dy = 0 ; when x = 2, Y = 1. (x  y) dx
+ (3x + y) ely
4.
33.
(l6x + 5y) dx (1, 3) .
34.
v(3x
35.
(3x 2 
36 .
From Theorems 2.1 and 2.2, Section 2.2, it follows that if F is homogeneous of degree k in x and y, F can be written in the form
+ 2v) dx 2y2)y'
x 2 dv
= 2xy;
= 0;
= 0; the curve to pass thro ugh the point
when x
= 0,
when x
= 1, v = 2. y = 1 .
(A) Use (A) to prove E ul er' theorem that if F is a homogeneous function of degree k in x and y,
aF
aF
ax
ay
x+y=kF.
2 .4
Exact Equations In Sectio n 2. 1 it was noted that when an equation can be put in the form A(x) dx
+ B(y) ely =
0,
30
Chapter 2
Equations of Order One
a set of soluti ons can be determined by integration, that is , by finding a function whose differential is A(x) elx + B(y) ely. That idea can be extended to some equations of the form
M(x , y) elx
+ N(x, y) ely = 0,
(1)
in which se paration of varia bles may not be possible. Suppose that a function F(x, y) can be found th at has for its d ifferential the expression M elx + N ely; that i ,
el F
= M elx + N ely.
(2)
Then certainl y
F(x , y) = c
(3)
defines .impli citly a set of so]ution of (1). For, from (3) it follows that
elF = O, or, in view of (2),
Melx +Nely =0, as desired. Two th ings, the~ , are needed: (1) to find out under what co ndi tio ns o n M and N a function F ex ists suc h that its total differe ntial is exactly M elx + N d y; and (2), if those conditions are satisfied, actuall y to determine the fu nction F. If a f unction F exists such that
Melx
+ Nely
is exactly the total differentia l of F , we call equation (1) an exact equation. If the eq uation
M elx
+ N ely =
0
is exact, then by definition F exists such that
elF = Melx+Nel y . But, f rom calcu]us,
aF
aF
ax
ay
elF = elx + ely, so
aF
M= 
ax'
aF
N
 ay'
(1)
2.4
Exact Equations
31
T hese two equations lead to
aM ay
a2 p ayax
aN ax
and
Aga in from calcu lu s
a2 p
a2 p
ayax
axay'
prov ided that these partial derivatives are continu ous. T herefore, if (1) is an exact equation, then
aM ay
aN ax
(4)
T hu s, for (1) to be exact it i necessary that (4) be satisfied. Let us now show that if condition (4) is satisfied, (1) is an exact equation. Let ¢ (x, y) be a functio n for which a¢ _ M
ax 
.
The function ¢ is the result of integrat ing M dx with respect to x whil e holdi ng y constant. Now
a2¢ ayax
aM ay ,
he nce, if (4) is satisfied, then also (5)
Let us in tegrate bo th sides of equation (5) with respect to x, holding y fixed. In the integration with respect to x, the "arbitrary constant" may be any func tion of y . Let us call it B I(y), for ease in indicating its integra l. Then integratio n of (5) with respect to x yields
a¢ ay = N + B (y). I
(6)
Now a fu nction P can be ex hibited, namely, F
= ¢(x , y) 
B(y),
for which
a¢ dx + a¢ dy 
dP = 
ax
ay
,
B (y) dy
= M dx + [N + B'(y)] dy = M elx
+ N dy.
B I(y ) ely
32
Chapter 2
Equations of Order One
Hence equation (1) is exact. We have completed a proof of the theorem stated below.
Theorem 2.3 If M, N, aM/ay, and aN lax are continuous functions ofx and y, then a necessa ry and sufficient condition that (1)
Mdx+Ndy=O be an exact equation is that
aM ay
aN ax
(4)
Furthermore, the proof contains the germ of a method for obtaining a set of sol utions, a method used in Examples 2.7 and 2.8.
EXAMPLE 2.7 Solve the equation
3x(xy  2) dx
+ (x 3 + 2y) dy =
O.
(7)
First, from the fact that
aM 2 =3x ay
aN ? =3xax '
and
we conclude that equation (7) is exac t. Therefore, its solution is F = c, where
aF

ax
= M = 3x 2 y  6x
(8)
and
aF
3
=N=x +2y. ay
(9)
Let us attempt to determ ine F from eq uation (8) . Integration of both sides of (8) with respect to x, holding y constant, yields (10)
where the usual arbitrary constant in indefinite integration is now necessarily a function T(y), as yet unknown. To determ ine T (y), we use the fact that the function F of equation (10) must satisfy equation (9). Hence
x 3 + T'(y) = x 3 + 2y, T'(y) = 2y.
2.4
Exact Equations
33
No arbitrary constant is needed in obtainin g T(y) , since one is being introduced on the right in the solution F = c. Then
i,
T(y) =
and from (10)
Finally, a set of soluti ons of equation (7) is defi ned by
x 3 y  3x 2
+l
= c.
•
EXAMPLE 2.8 Solve the eq uation
(2x 3

xi  2y
+ 3) dx 
(x 2 y
+ 2x) dy
= O.
(1L)
Here
aM =2xy 

ay
aN
2= 
ax '
so equati on (1 1) is exact. A set of solutions of ( I I) is F = c, where
aF

ax
3
2
= 2x  xy  2y
+3
(12)
and
aF

ay
2
= x y 2x.
(J 3)
Because ( 13) is simpl er than (12) and, for variety's sake, let min ation of F from eq uation (13). At once, from ( 13),
+ Q(x) ,
F = !x 2 i  2x y
where Q(x) will be determined from (12). The latter yields
xi  2y
+ Q'(x)
= 2x 3
Q'(x) = 2x 3 Therefore,

xi  2y
+ 3.
+ 3,
LI S
start the deter
34
Chapler 2
Equations of Order One
and the desired set of solutions of (11) is defined implicitly by
~x2l 2xy
+ ~X4 + 3x =
~c,
or
•
•
Exercises
Test each of the fo llowing eq uations for exactness and solve the equati o n. The equations that are not exact may be solved by methods di sc ussed in the preced ing sections.
+ y) d x + (x
= O.
1.
(x
2.
(6x+i) dx+y(2x3y) dy = O. (2xy3x 2 ) d x+(x 2+y) dy = O.
3.
 y) dy
4.
(2xy+y)dx+(x 2 x)dy=0.
5.
(x 2y)dx +2(y  x)dy = O.
6.
(2x3y)dx+(2y  3x)dy
= O.
7 . Do Exercise 5 by another method. 8.
9.
Do Exercise 6 by another me thod. (i  2xy + 6x) dx  (x 2  2xy + 2) dy = O.
3) dL! + (3u 2 v 2  3u + 4v) d v = O. 11. (cos 2y  3x 2 i) dx + (cos 2y  2x sin 2y  2x 3 y) dy = O. 12. ( 1 +y2)dx+(x 2y + y)ely =O. 13. (1 + y2 + xi) dx + (x 2y + y + 2xy) ely = O. 14. (w 3 + wz 2  z)dw + (Z3 + w 2 z  w)dz = O. 15. (2xy  tan y) dx + (x 2  x sec 2 y) d y = O. 16. (cosxcosy  cotx)dx  sinxsin ydy = O. 17. (r + sine  cose) dr + r(sine + cose) de = O. 18 . x(3xy  4y 3 + 6) dx + (x 3  6x 2y2  1) d y = O. 19. (sin e  2r cos 2 e) dr + r cose(2r sin e + 1) de = O. 20. [2x + y cos (xy)] dx + x cos (xy) dy = O. 21. 2xyelx + (i +x2)dy = O. 22. 2xy dx + (y2  x 2 ) dy = O. 23. (xi + y  x)dx +x(xy + l)dy = O. 24. 3y(x 2 l)dx+(x 3 +8y3x)dy = 0; whenx=O,y=l. 25. (1  xy)2 elx + [y2 + x 2 (l  xy)2] dy = 0; when x = 2, y = l. 10. v(2uv 2
26. 27. 28.

(3 + y + 2y2 si n2 x) dx + (x + 2xy  y sin 2x) dy = O. 2x[3x + y  y exp (_x 2 )] dx + [x 2 + 3y2 + exp (_x 2)] d y = O. (xi+x  2y+3)dx+x 2ydy=2(x+y)dy; when x = l,y= 1.
2.5
2 .5
The Linear Equalioll of Order One
35
The Linear Equation of Order One In Section 2.4 we studied firstorder differential equations that were exact. If an equation is not exact, it is natural to attempt to make it exact by the introduction of an appropriate factor, which is then called an integrating factor. Indeed, in Section 2. 1 we multiplied by an integrating factor to separate the variables and thereby obtain an exact equation . In general, very little can be . aid about the theory of integrating factors for firstorder equations. In Chapter 5 we shall prove some theorems that wi ll give some as istance in a few isolated situations. There is one important class of equatio n , however, where the existence of an integrating factor can be demonstrated . This is the class of linear equations of order one. An equation that is linear and of order one in the depende nt var iable y l11 U t by definition (Section 1.2) be of the form dy A(x) 
dx
+ B(x)y
= C(x).
(1)
By dividing each member of equation (1) by A (x), we obtain
dy
d~
+ P(x)y =
Q(x),
(2)
w hich we choose as the standard form for the linear equation of order one. For the moment, suppose that there exists for equation (2) a positive integrating factor vex) > 0, a function of x alone. Then vex)
[~~ + P(X)y ] =
vex) Q(x)
mu st be an exact equation . But (3) is easily put into the form
Mdx
+ Ndy =
0
with
M = vPy  vQ and
N = v, in which v, P , and Q are functions of x alone. T herefore, if equation (3) is to be exact, it follows from the requirement
aM ay
aN ax
(3)
36
Chapter 2
Equatiolls oj Order Olle
th at v must satisfy the eq uation dv vP=.
(4)
dx
From (4), v may be obtained readil y, for dv v
Pdx =  ,
so
or v = exp ( / P
dX) .
(5)
That is, if eq uation (2) has a positive integrating factor independent of y, then that factor must be as given by equatio n (5). It remains to be shown that the v give n by equation (5) is ac tually an integra Lin g factor of dy
dx + P(x) y
(2)
= Q(x).
Let us mu ltiply (2) by Lhe integrating factor, obtaining exp ( / P dx )
~~ + P exp ( / P dx )
y = Q exp ( / P
dX) .
(6)
The left member of (6) is the derivative of the prod uct y exp ( / P d x ) ; the right member of (6) is a function of x only. Hence equation (6) is exact, which is what we wan ted LOshow. Of co urse, one integrating factor is sufficient. Hence we may use in the expone nt (j P dx) any functi on whose derivative is P. Because of the great importance of the ideas just discLlssed and the frequent OCCLlrrence of li near eq uations of first order, we summarize the steps involved in solving such eq uat ions: (a)
Put the equation into standard fOlm: dy

dx
+ Py= Q.
2.5
The Linear Equation of Order One
37
(b)
Obtain the integrating factor exp(j P dx).
(c)
Multiply both sides of the equation (in standard form) by the integrating factor.
(d)
Solve the resultant exact equation.
Note in integrating the exact equation that the in.tegral of the left member is always the product of the dependent variable and the integrating factor used.
EXAMPLE 2.9 Solve the equation
2(y  4x 2 ) dx
+ x dy =
O.
The equation i linear in y . When put in standard form it becomes
dy
2
dx
x
 +
when
Y = 8x
x::j: O.
(7)
Then an integrating factor is exp ( /
2~X) = exp (21n Ix l) = exp (In x 2 ) = x 2 .
Next apply the integrating factor to (7), thus obtaining the exact equation
dy
x2_
dx
+ 2xy =
8x 3 ,
(8)
which may immediately be written as
(9) By integrating (9) we find that ( 10)
This can be checked. From ( 10) we get (8) by diffe rentiation . Then the original differential equation follows from (8) by a simple adjustment. Hence (10) defines a set of solutions of the original equation.
•
EXAMPLE 2.10 Solve the equation y dx
+ (3x
 xy
+ 2) dy =
O.
38
Chapler 2
EquCllio17S o/Order One
Since the product y d y occurs here, the equation is not linear in y . It is however, linear in x. Therefore, we arrange the terms as in
ydx
+ (3 
y)xdy = 2dy
and pass to the standard form ,
dx dy
(3 ) y2
+ Y1
for y
x =
1= o.
( 1 I)
Now
f (~ 
1) d y = 3 In Iyl
 y + CI ,
so that an integrating fac tor for equation (11 ) is exp (31n Iyl
 y)
= exp (3 1n Iy l)e  Y = exp ( In ly I3 )e Y
= ly I3e y . It follows that for y > 0, y3 e  Y is an integrating factor fo r equation (11) and for y < 0,  y3 e Y serves as an integrating fac tor. In either case we are led to the exact equation
from which we get
xy3e  Y =  2
f
ie  Y d y
= 2ie Y
+ 4ye Y + 4e  Y + c.
Thu s a family of so luti o ns is defined implicitly by
xy3 =2i+4y +4+ ce Y .
•
I 2.6 I The Genera] So]ution of a Linear Equation In Section 1.6 we stated an existence and uniqueness theorem for firstorder differential equations. If the differential eq uation in that theorem happens to be a linear equation, we can prove a so mewhat stronger statement. Con ider the linear differential eq uation
dy dx
+ P(x)y =
Q(x) .
(1)
2.6
The General Sululion of a Linear Equation
39
Suppose th at P and Q are con tinuous funclions on the interval a < x < b, and that x = Xo is any num ber in that in terval. If Yo is an arbitrary real number, there exists a unique oJution y = y(x) of differential equation (1) that also satisfies the initial co ndi tion y(xo) = Yo·
Moreover, this solution satisfies equati on (1) th rougholl t the entire interval a < x < b.
The proof of this theo rem has essentially been obtained in Section 2. 5. Multiplication of equ ation (1) by the integrating fac tor v = exp (j P dx) and in tegration gives
Si nce v
yV=
f
vQdx+c .
y = vI
f
vQdx
i= 0, we ca n write
+ cv  I.
(2)
It is a si mple matter to show that since v i= 0 and v is continuolls on a < x < b, (2) is a famil y of solutions of equation (1) . It is also easy to see that given any Xo on th e interval a < x < b together with any number Yo, we can choose the constant c so thal y = Yo whe n x = Xo. The effect of our argument is that every equation of the form of equati on (I), fo r which P and Q have some co mmo n in terval of continuity, will have a unique set of solutions containing one co nstant of integration th at ca n be obta ined by introducing the appropriate integrati ng fac tor. Because we are ass ured of the uniqueness of these sol utions, we k now that any other solu tion obtained by any other method must be one of the fu ncli ons in our oneparameter family of solu tions. It is for thi · rea on lh at th is set of solutions is called th e general sol ution of equation (1). The word "genera]" is intended to mean th at we have fo und aU po sible solutions that satisfy the diffe rential equation on the in terval a < x < b.
• Exercises Tn Exercises J th rough 24, find the general so lution.
+ 3y)dx 
1.
(x 5
2.
y' =
3.
(y
X 
xdy = O.
2y .
+ I) dx + (4x 
y) d y = O.
4. udx + (I31£) x du =3u 2 e 3l1 du. 5.
udx
+ (1 
3u)xdu = 3udu.
6.
y' = X  4xy. 7. y' = cscx + y cotx . 8. y' = cscx  Y cotx. 9. (y  cos 2 x ) dx + cosx dy = O. 10. y' = x  2Y cot 2x .
40
Chapter 2
Equatiolls of Order One
13 .
+ xdy = O. 2(2xy + 4y  3) dx + (x + 2)2 dy = O. (2xy + x 2 + x 4 ) dx  (1 + x 2 ) dy = O.
14.
y'  my =
15.
y'  m2y=cleIllIX, wherecl,m l,ln2areconstantsandml i=m2.
16.
17.
+ (2x + 1  vx) dv = O. 2 x(x + 1)y' + 2y = (x 2 + 1)3.
19. 20.
dx  (1
18.
2y(i  x) dy = dx .
21.
y' =
22.
(l
11. 12.
23. 24.
(y  x +xycotx)dx
C I e",·r,
where
CI
and m are constants .
v dx
2y dx = (x 2

1)(dx  dy).
+ 2x tan y) dy = O. ] + 3y tanx.
+ cos x) y' = s in x (sin x + si n x cos x  y ). (x + a 2 ) dy = 2x[(x 2 + a 2 )2 + 3y] dx; a is a constant. (x + a)y' = bx  ny; a , b, n are constants with n i= 0, n i= 2
 1.
25 .
Solve the equation of Exercise 24 for the exceptional cases n = 0 and It =  l.
26.
In the standard form dy
+ P y dx
= Q dx, put y = vw, thus obtai ning
w(dv+Pvdx)+vdw= Qdx . Then , by first choosing v so that dv
+ Pvdx
= 0
and later determining w , show how to complete the sol ution of dy
+ Pydx =
Qdx.
In Exercises 27 th rough 33, find the particu lar solution indicated .
27. 28.
(2x y'
+ 3»)1' = y + (2x + 3)1 /2;
=x 3
di
29 . Ldt di
when x = ], y = O.
 2xy; when x = 1, y = l.
+ Ri
= E; where L , R , and E are constants, when t = 0, i = O.
31.
+ Ri = E sin wt; when t = 0, i = O. dt Find that solution ofy' = 2(2x y) w hichpa ses through the point (0,  1).
32.
Find that sol ution of y' = 2(2x  y) which passes through the point (0, 1) .
33 .
(1
30.
L
+ t 2 ) ds + 2t[st 2 
3(1
+ t 2 ) 2] dt = 0;
when t
= 0, s =
2.
• Miscellaneous Exercises In each exercise, find a set of solutions, unless the statement of the exercise stipulates otherwise.
1.
y' = exp (2x  y).
2.
(x 4 +2y)dxxdy=0.
2.6
3. 4. 5.
+ 3y  4) dx + (x + 1)2 dy (x + y) dx + x dy = O. y2 dx  x(2x + 3y) dy = O. (x 2 + l )dx +x 2y2dy = O. (3xy
6. 7. y' = x 3 8. 9.
The Gell.eral Solution ofa Linear Equatioll.

13. dx / eI t = cos X cos 2 t .
2
y(x+3y)dx+x dy=0.
= sec
2
ely / dx
11.
(L
+
12.
(2X2_2xy_y2)elx +xy dy = O.
18.
(2xy  3x ) elx + (x (x 3 + y3) dx + y2(3x
20.
2 l. 22.
23 . 24.
25. 26. 27. 28.
x
sec 3
10.
19.
= O.
2xy; when x = 1,y = 2.
sin e d r/de =  1  2rcose .
x 2 )y'
41
y.
= x4y4 .
14.
3x 3y' = 2y(y  3) .
15 .
xy(dx  ely) = x 2 ely
16.
+ si n x dy = o. (x + 2y) dx + (2x + y)dy = O.
dx .
(y  sin x) dx
17 .
+ 2y) dy = O. + ky) dy = 0; k a co nstant. 3 2 y(2x  x y + y3) dx  x(2x 3 + y3) ely = O. y(3 + 2xyZ) dx + 3(X 2y 2 + x I ) dy = O. y(x2 + y2) dx + x(3x 2  5y2) ely = 0; when x = 2, y = l. y' + ay = b; a and b co nstants. Solve by two methods. (x  y) elx  (x + y) ely = O. So lve by two methods. (siny  ysi nx)dx + (cos x +x cos y) dy = O. (I + 4xy  4x 2y ) dx + (x 2  x 3 ) ely = 0; w hen x = 2 , Y = (2y cosx + sin4 x) dx = inx ely; w hen x = ~n, y = 1. a 2(dy  elx) = x 2 ely + y2 dx; a constant. 2
+i
2
2
±.
In solv ing Exerci. es 29 through 33 , recall that the pri ncipal val ue arcsin x of the inver ·e s ine fu nctio n is restricted as fol lows: arcsin x ~ 1T. Exercises 30 through 32 refer La different arc segments in Figure 2.4 which shows the graph of the ell ipse
!1T :s
+ ~ dy =
:s
29.
)1 y 2 elx
30.
Solve the equation of Exerc ise 29 w ith the added condi tion that w hen x = 0, y = ~v'3. Solve the eq uation of Exe rcise 29 with the added condition th at when x = 0, y = ~ v'3. Show that after the answers to Exercises 30 and 3 1 have been deleted, the remain ing arcs of the ellipse
3 1. 32.
O.
42
Chapter 2
Equations of Order On e y
~~~ x
Figure 2.4
are not solutions of the differential equation Jl=Y2 dx
+ ~ dy =
O.
For this purpose consider the sign of the slope of the curve. 33.
For the equation
~h

y2 dx  ~ dy = 0
state and solve four problems analogous to Exercises 29 through 32.
+ 2uv  2u) dv. 36.  (xy + 2) d y = O. 37. 2 3xy2) dx + (y3  3x y) dy = O.
34.
v du = (e V
y(yZ  3x 2 ) dx
35. 38.
yZ dx
y' = y tan x
39.
(l x 2)y' =I x y3x 2 +2x 4 . (y3 x 3) dx = xy(x dx + y dy).
40.
(x 3 
41.
y' = secx  y tanx.
45.
ydx = (3x
46. 47 . 48. 49. 50.
(x 2

y2 dx
+ y3 
+ cosx.
X2 yl=y (lX).
43 . xy' 44.
= O.
=x
y2 dx
 y
+x
2
+ xy tanx.
dy = 2xy dy.
y2) dy; when x = 1, Y =l.
2xy  y2) dx  (x 2 + 2xy 
+ (xy + yZ 
42.
+ x 3 dy
yZ) dy
= O.
1) dy = 0; when x = 1, Y = l.
y' = co x  ysecx; when x = 0, y = 1. Find that solution of yl = 3x + y which passes through the point (1, 0).
Find that sol ution of yl = 3x
+ y which passes through the point (1,
1).
2.7
52 . 53.
54. 55 .
2 .7
I
43
(x 2  I
5l.
I
Computer Supplement
+ 2y) d x + (1  x 2) dy = 0; when x = 2, y = 1. (y 2 + y) dx  (y2 + 2xy + x) dy = 0; w hen x = 3, y = 1. (3x 4 y  1) dx + x 5 dy = 0; when x = 1, y = 1. (sinx sin y + tan x) dx  cosx cos y dy = O. (3xy  4y  1) dx + x(x  2) dy = 0; when x = 1, y = 2.
Computer Supplement In Chapter 2 we have begun the process of solving differential equat ions analytically. Most of the methods described involve integration in some way, and are therefore amenable to solution using Computer Algebra Systems (CAS) which can integrate symbolically. As an easy exam ple consider the separable differential equation in Exampl e 2.1 of Section 2.1: dy
2y
dx
x
The sol ution given in the text involves separating the variables and then integrating both sides ofthe resulti ng equatio n. The integrations can be acco mpli shed in Maple by the followin g command: >int( 1/y,y)=int(2/x,x ) +C i
In(y) = 2 In (x)
+C
This implicit solution ca n be solved for y and simplified by > sol ve ( " , y) ;
e 2I n (x J+C >s implify( " ) i
As a second example consider equation (7) in Example 2. 7 of Section 2.4, 3x(xy  2) dx
+ (x 3 + 2y) dy
= O.
Here the first step is to check whether th e equation is exact. Maple can do this as follows: >M : = 3*x * (x*y2) i
M := 3 x (xy  2) A
>N: = (x 3+2*y) ;
>diff (M, y ) ;
>di f f (N, x) ;
44
Chapter 2
Equatiol/s of Order Oll e
Thu the eq uation is exact. We cou ld then use the computer to assist in the remaining steps in the process. Fortu nately, most symboli c packages are designed to take care of all of these steps at once. First, return to the first example above. We can enter the differential equation as >diff(y( x ) , x ) =2 *y/ x; d 2y  y (x) = 
dx This can be solved in one command: >d solve ( " ,y (x) ) ;
x
The second example is just as easy: >(3 *x* (x *y  2) ) + (x A3 +2*y ) *diff(y (x ) , x )=O; 3 x (x y  2)
d dx
+ (x 3 + 2 y)  y (x)
= 0
>ds olve ( " ,y (x ) ) ;
Finally, the computer can also handle ini tial va lue problems. For examp le, con ider equation (8) in Example 2.3 of Section 2.1: (1
+i
)d x + (1 + x 2 ) dy =
0,
with the " initial condition" that when x = 0, y =  1. The equation is en tered as >d iff(y (x) ,x) = (1 +y( x ) A2)/ ( 1 +x A2 ) ; d
I ex y (x )
=
I
+ (y(x» 1+ x2
2
and then solved by entering >ds o l v e({", y( O) =l},y( x )) ; y(x) = tan (  arctan (x) 
~) .
• Exercises 1.
Use a Computer Algebra System to solve a repre entative sample of problems from the chapter. Be sure to try some with and some without initial conditions. You may find some problems that the CAS will not solve using basic techniques. See if your system has more advanced techniques to solve them.
2.
A CAS is capable of sol ving even equations as general as dy / d x Q(x). Try it on your system.
+ P (x ) y =
Numerical
3
Methods
3.1
General Remarks There is no general method for obtaining an expli cit formu la for the solution of a differential equation. Specific equations do occur for wh ich no known attack yields a sol ution or for wh ich the exp li cit forms of solution are not well adapted to computation. For these reasons, systematic, efficient methods for the numerical approximation to solutions are important. Unfortunately, a c lear grasp of good numerical methods require timeconsuming practice and the availability of an adequate computer. Thi s chapter is restricted to a fragmentary discussion of so me simpl e and moderately useful methods. The purpo e here is to give the student a concept of the fundamental principles of numerical approximation to solu tion. We shall take one problem, which does not yield to the methods developed earlier, and apply to it several numerical processes.
3.2
Euler's Method We seek to obta in that solu tion of the differential equation
y' = / _ x 2
( I)
for which y = 1 when x = O. We wish to approximate the so lution y = y (x) in the interval 0 S x S ~. Equation (1) may be written in differential fo rm as ely =
(i  x 2 ) clx.
(2)
F igure 3.1 shows the geometrical 'ignificance of the differential ely and of I:!. y, the actual change in y, as induced by an increment elx (or I:!.x) applied to x. In calculu it is shown that near a point where the derivative exists, ely can be made to approx im atel:!.y as closely as desired by taking I:!.x sufficiently mall. We know the value of y at x = 0; we wis h to comp ute y for 0 S x < Suppose that we choose I:!.x = 0.1; then ely can be computed from
1.
ely = (i x 2 )l:!.x.
45
46
Chapter 3
Numerica! Methods
Ind eed, d y = (I  0)(0. 1) = 0. 1. T hus for x = 0 + 0 .1, the approximate value of y is J + 0. 1. Now we have x = 0.1, y = 1.1. Let us choose L}.x = 0.1 aga in . The n
dy = [(1 .1)2  (0. 1)2] L}. x, so dy = 0. 12. Hence at x = 0.2 , the approx imate value of y is 1.22. The co mplete computation us ing L}.x = 0. 1 is shown in Table 3.1. The computations are carried o ut to six decimal places and then rounded off to three decimal places. The increment 6.x need not be constant throughout the interval. Where the slope is larger, it pays to take a s maller increment. For implic ity in computations, eq ual increments ,u'e used here. It is helpfu l to repeat the computation with a small er increment and to note the changes th at resu lt in the approxim ate values of y. Table 3.2 shows a computation with 6.x. = 0.05 throu gho ut. In Table 3.3 the value of y obtained from the computations in Tables 3. J and 3.2 and also th e values of y obtained by us ing 6.x = 0.01 (co mputation not shown ) are ex hi bited beside th e va lues of y correct to three decimal places. y
t __ dx =t:o.x_ r~~ x
0
Figure 3.1
x 0.0 0.1 0.2 0 .3 0.4 0.5
TABLE 3 . 1 ~x y2 y x2 1.000 1.100 1.220 1.365 1.542 1.764
1.000 1.210 1.488 1. 863 2.378
0.000 0.010 0.040 0.090 0.160
= 0.1 ci 
x 2) 1.000 1.200 1.448 1.773 2.2 18
dy
O. JOO 0.120 0.145 0.177 0.222
3.2
TABLE 3.'2 X
0.00 0.05 0.1 0 0.1 5 0.20 0.25 0.30 0.35
OAO OA5 0.50
When x
Y 1.000 L.050 1.1 05 1. 166 1.232 1.306 1.388 1.480 1.584 1.70 1 1.836
f:y,x
0.0 0 .1 0.2 0.3
?
x?
ci x 2 )
dy
1.000 L.102 l.22 1 1.359 1.5 19 1.706 1.928 2. 192 2.508 2.894
0.000 0.002 0.010 0.022 0.040 0.062 0.090 0. 122 0.160 0 .202
1.000 1.1 00 1.2 11 1.336 I A79 1.644 1. 838 2.069 2.348 2.692
0.050 0.055 0.06 1 0.067 0.074 0.082 0.092 0 .1 03 0.117 0.135
= 0.1 y
0.5
TABLE 3.3 f:y,x = 0.05 y 1.000 1.1 05 1.232 1.388 1.584 1. 836
47
= 0.05
y
1.000 1. 100 1.220 1.365 1.542 1.764
OA
~x
Euler's Method
f:y,x
= 0.0 1 y
1.000 1. 110 1.244 JA il 1.625 1.9 1 1
Correct y
1.000 1. 111 1.247 1.417 1.637 1.934
T he correct va lue are obtained by the method of Section 3.7. T he ir ava ilab ility is in a sense acc idental. F requently, we know of no way to ob tain the y value correct to a specified degree of accu racy. In uch in tance it is customary to resort to decreas ing the ize of the increment until the y values show c hange no larger than the errors we are wi lli ng to permit. Then it is hoped that the steady in g down of the y values is due to our bei ng close to the co rrect solution ra ther than (as is q uite possi ble) to the slow ness of convergence of the process used. For the more general initial val ue prob lem
dy  = fCx, y);
dx
when x = Xo,
Y = Yo,
(3)
the sequence of ap proximations desc ri bed above can be ex pressed in te rm s of the rec urrence relations
48
Chapter 3
NUlllerical Methods Xk + 1
=
Xk
Yk+l = Yk
+h + hf(Xb yd,
(4)
for k = 0, 1, 2, .... Here we have used h for the value .6.x. The technique de cribed above has come to be known as Euler's method, although it involves nothing more than the linear approximations of elementary calcul us .
• Exercises eac h of the fo llowing exerci es, use Eul er's method with the prescribed !::" x to approximate the solu tion of the initial value problem in the g iven interval. In Exerc ises I through 6, solve the problem by elementary methods and compare the approximate val ues of y with the correct values. [n
y' = x
2.
Use
3.
7.
y' = x + y; when x = 0, y = 2; .6.x = 0 .1 and 0 ::s x ::s 1. y' = x + y; when x = 1, Y = 1; ~X = 0.1 and I ::s x ::s 2. y'=x+y; whenx=2,y=  I; ~x = 0 .l and2::Sx ::s 3. y' = 2x  3y; when x = 0, Y = 2; ~X = 0.1 and 0 .::s x .::s 1. y' = e  xy ; when x = 0, y = 0; ~ x = 0.2 and 0 .::s x .::s 2.
8.
Use
4.
5. 6.
3.3
+ y;
1.
~x
~x
when x = 0 , Y = I;
= 0.05 in Exercise 1.
= 0.1 in Exercise 7.
y' =C1+x 2 +y2 ) I;
9. 10.
U e
11.
y' = (cosx
12.
Use.6.x = 0 .1 in Exercise 11.
13.
y'
~x
.6.x = 0.1 and 0 ::s x ::s 1.
whenx=O,y=O;
~x=0.2andO .::s x::s2.
= 0.1 in Exercise 9.
+ sin y)1/2;
x2 + y2 x2  y2 + 2'
when x = 0 , y = 1; .6.x = 0.2 and 0 ::s x .::s 2.
when x = 0, y = 0;
~X = 0.2 and 0 .::s x ::s 2.
A Modification of Euler's Method At each step in Euler's method as described by equations (4) of Section 3.2, the new approx imation Yk+l uses the slope f(Xb Yk). This slope is computed at the point (Xb Yk), a point that li es at the lefthand endpoint of the interval Xk ::s x .::s Xk + h. It is reasonable to expect that a better app roximation for the value of Yk+ I would be obtained if the slope was computed at the midpoint of the interval rather than at the lefthand endpoi nt. A modification of E ul er's method makes use of this observation. We p roceed in the following manner. Starting at the initial point (xo , YO) and using Eul er's method to determine the point (XI , YI), we start over aga in at the
3.4
A Method o.fSuccessive Approximation
49
initial point (xo , Yo). Th is time, however, we use Eu ler's method with inc rement s ize 2h and use the value of the slope at the point (x I , Y I), a point that lies at the midpoint of the new in terval
+ 2h .
Xo ~ x ~ Xo
The formulas for the modified E uler 's method are, therefore, XI
+ h, Yo + hf(xo, Yo)
= Xo
YI = a nd
+ 2h, k :::: 0, Yk +2 =)/k + 2hf(xk+ l, Yk+I),
Xk +2
= Xk
k ::::
o.
Applying the mod ified E ul er's method to the probl em
= 0,
Xo
Yo
= I,
prod uces the resu lts in Tab le 3.4. We see by com parison with Table 3 .3 that there is considerable improvement in the acc uracy of th e computed values fo r y .
TABLE 3.4
= 0.1
X
y
h = 0.05 Y
0.0 0. 1 0.2 0.3 0.4 0.5
1.000 1.100 1.240 1.400 1.6 14 1.888
1.000 1.1 00 1.245 1.414 1.63 1 1.922
When
h
h
= 0.0 1 Y 1.000 1.111 1.247 1.4 17 1.637 1.933
Correci Y 1.000 1. 1 I I 1.247 1.417 1.637 1.934
• Exercises In each of the exercises in Section 3.2, use the modified Euler's method to approx imat.e the solution of the given i nitial va lue prob lem in the given interva l. Co mpare the results w ith the results obtai ned by Euler's method.
I 3.4 I A Method of Successive Approximation Next let us attack the arne problem as befo re,
x = 0, Y = I ,
( I)
50
Chapter 3
Num erical Methods
with y desired in the interval 0 :'S x :'S ~, by the method suggested in the discussion of the existence theorem in Chapter 13 . Applyi ng the statements to be made in that discuss ion, we conclude that the desired solu tion is y = y(x ), where y (x) = ljm y,,(x ) 11 +00
and the sequence of f unctions y" (x) is given by yo(x) = I , and for n. ~ I, (2)
For the problem at hand ,
Next we obtain a second approximation , finding Y2 (X) from YI (x) by mean of (2). Thus we find that
1 x
Y2(X) = 1 +
[(1
+t

~t3)2 
t 2] dt ,
2 1 4 2 5 1. 7 = 1 +x+ x  6X TI x +6:i X'
Then Y3(X ), Y4(X) , .. . , ca n be obtained in a similar manner, each from the preceding element of the sequence y" (x). In Table 3.5 the values taken on by YI (x) , Y2 (X), and Y3 (X) at intervals ofO.l in x are shown be ide the corresponding va lues of y (x ) , correct to two dec im al places, as obtai ned in Section 3.7 . It must be realized that the usefulness of thi method is not dependent upon our being ab le to carry out the integratio n ' in a formal sense. It m ay be best to perform the integrations by some numeri cal process, suc h as Simpson's rule.
TABLE 3 .5 x
YI (x)
Y2 (X)
Y3 (X )
y(x)
0.0 0.1 0.2 0.3 0.4 0.5
1.00 1.l0 1.20 1.29 J.38 1.46
1.00 1.11 1.24 1.39 1.56 1.74
LOO 1.11
1.00 l. 11 1.25 1.42 1. 64 1.93
1.25 1.41 1.62 1.87
3.5
A ll
Improvement 011 the Melilod of Successive Approximation
51
• Exercises 1.
App ly the method of this section to the problem (Exercise 1, Section 3.2)
y' = x
+ y;
when x
= 0, Y =
1.
Obtain YI (x), Y2(X), and )l3(X).
°
2.
Compute a table of values to two decimal places of )I I , Y2, Y3 of Exercise I for x = to x = 1 at intervals of 0.1. Tabul ate also the correct values of )' obtained fro m the elementary solution to the prob lem.
3.
Obtain Yl (x), Section 3.2)
)'2 (x),
Y'
and Y3(X) for the initial val ue prob lem (Exercise 4,
= x + y;
when x
=
I, Y
=
1.
Hint: Express the integrand of the integra l in equation (2) in powers of t  1 before integrating. 4.
Compute a table of val ues to two decimal places of th e YI , Y2, Y3 in Exercise 3 for x = I to x = 2 at interval of 0.1. AI 0 tabulate the correct values of Y obtai ned from the elementary o lution to the problem.
I 3 .5 I An Improvement on the Method of Successive Approximation In the method used in Sectio n 3.4, each of the )l1/(x), where n = 0, 1,2, ... , yields an approximation to the solution Y = .v(x) . It is plausible that, usually, the more nearly correct a particular approximation Yk (x), the better will be its successor Yk+1 (x). The initial value problem we are treating is
x
= 0, Y =
1
and it tell s us at once that at x = 0, the slope is y' = 1. But in Section 3.4, by blindly following the suggestion from C hapter 13, we started out with Yo(x) = I, a line that does not have the correct slope at x = O. It is therefore reasonable to alter our initial approx imation by choosing Yo(x) to have the correct slope at x = 0 , Y = I . Hence we choose Yo(x) = I
+x
and proceed to compute YI (x), Y2(X), . . . , as before. The successive stages of approximation to y(x) now become )' 1(x)
= J+ = I
1,1;[(1 +
+x + x2 ;
t)2 
t 2 Jdt
52
Chapter 3
Nume rical Methods
TABLE 3 . 6 x
YI (x)
Y2(X)
0.0 0.1 0 .2 0.3 0.4 0.5
1.00 1. 11 1.24 1.39 1.56 1.75
1.00 1.1 1 1.25 1.41 1.62 1.87
Y2(X)
Y3(x)
y(X)
1.00
1.00 1.11 1.25 1.42 1.64 1.93
1.11
1.25 1.42 1. 64 1.92
= 1 + L'[(l+t+l 2)2_ r2]dt =
1
+ X + X 2 + ~X3 + ~X4 + kx5 ;
and so on. In Table 3.6 the values of YI , Y2, Y3 obta i.ned by thi s me thod are shown bes ide th e conect values of y .
• Exercises 1.
Ap ply the method of th i section to o btain approxi matio ns problem of Exercise 1 of Section 3.2.
2.
Tabu late to two decimal places)l l, Y2,)'3 of Exercise 1 beside the corre ponding valu es of the exact solutio n )/(x) = 2e x  I x .
3.
A ppl y the method of this section to obta in the approx im ati ons the problem of Exercise 3 of Sectio n 3.4.
4.
Compare the)ll, Y2, )'3 of Exercise 3 with the Tay lor series in powers of x  I fo r the exact solution
)I I ,
Y2, )13, fo r the
)'1,
Y2,)/3 for
y(x)=3exp(x 1)(x I)  2.
3.6
The Use of Taylor's Theorem For students fam ilia r wi th elementary calculus, the most natu ral approach to the approx imation of solu tion is to make use of Taylor's theore m. If we consider the initia l va lue proble m
/ = F(x, y):
x
= Xo, )I = Yo,
( 1)
we may be able to compute successive derivatives of the sol utio n y = y(x) at x = xo by usin g equation (1) . We adopt the notation
Yb = /(xo),
y~ = y" (xo) ,
3.6
The Use ojTay/or's Theo rem
53
and recall that Taylor's theorem suggests the approximation fo rmul a Y~ ~ Yo
+Y
1 (
o
. ~)
n
, ) x  .xo
+ Yo (x 2!
 Xo )2
+ . . . + Yo
n!
(x  Xo , )" .
(2)
One adva ntage in usi ng the approximat ion in (2) is th at we may be able to estimate the erro r in ou r calcul atio n by exa mi ning the valu e of the remainder term in Taylor's theorem. For relation (2) th is re mainder takes the form y(II+ I)(C)     (x  X )"+ 1 (n I)! ' . 0 ,
(3)
+
w here c i so me number between x and Xo . It shou ld be clear that the p ractica li ty of this techniq ue fo r approx imating solu tions will depend greatly on the difficulty of obtain i ng val ue of the derivatives involved. If the f unction F (x , y) is at all com plicated, there may be a great dea l of co mputation necessary to prod uce a reaso nable approxim ation usi ng Taylor's theorem . For the examp le
(4)
Xo = 0, Yo = I ,
it is relatively easy to show that
yn = 2yy'  2x, y'" = 2yy"
+ 2(y')2 
y(4) = 2yy'n
+ 6y' yn,
/5)
= 2yy(4)
2, (5)
+ 8y'y"1 + 6(yn)2
Thus we ca n obtain the val ues Yo = 1,
Yb =
1,
y~ = 2,
y~' = 4,
Y64 ) =
20,
and
Y65) =
96 .
Equation (2) thus becomes y ~ 1
+ x + x 2 + ~ x 3 + ~X4.
(6)
Some indicatio n of the accuracy of equatio n (6) is given in Ta bl e 3.7 . The valu es of y obtai ned fro m equation (6) for several values of x are ex hibited beside the val ues of y correct to two dec im al places. Another ind ication of the elTor in using equat ion (6) for estim ating the value of y for x = 0.5 can be obtained by exa mi ning the next term in th e Taylor's series expansion. T hat is, we may compute the val ue of (96j5 !)x s at x = 0. 5 and fi nd that the error is at least as big as 0 .02. A more careful study of the re mainder term given in equat ion (3) coul d be u ed to get a better estimate of the error in our res ul ts. In practice, however, it
54
Chapter 3
NUII/erical Me/hods
TABLE 3.7 y Correct y x
0.0 0.1 0.2 0.3 0.4 0.5
1.00
1.00 1.11
1.11
1.25 1.41
1.25 1.42 1.64 1.93
1.62 1.89
is extremely difficult to make these error estim ates because of the complex ity of the derivative formulas involved and the fact that the value of c is unknown to us .
• Exercises For each o f the following ini tial val ue prob lems, use Taylor's theorem, retaining powers of x  Xo suffici ently large to approximate the val ues of y acc urately to two decimal places o n
the given interval using the p re cribed increments in x . In Exercises I th rough 6, compare the estimated values with the co rreCl val ues obtai ned by solvi ng the problem exactly lI sing elementary methods.
1.
Exercise I, Section 3.2.
4.
Exerci e 6, Section 3.2.
2.
Exercise 3, Sectio n 3.2.
S.
Exercise 5, Section 3 .2.
3.
Exercise 4, Section 3.2.
6.
y' =
i + x 2;
when x = 0, y = 1;
/:;.x = 0.1 and 0 .:::; x .:::; 0.5.
7.
/= y2_ x 2;
whenx=O,y = l ;
6.x = O. l andO .:::;x.:::; O.S.
8.
Use Taylor's series to determine to three places the value of the solutio n of the problem
y' =
_ xy2;
w hen x = 0, y = 1,
for x = 0.1,0.2, and 0 .3 . Compare your results with the values obtained by solving the problem by elementary means.
3.7
The RungeKutta Method From a computational point of view, the major drawback in usi ng Taylor's series to estimate the values of so lutions of differential eq uation s is that each coefficient in the series involves a different derivative function. Thus each approximation requires computations of the values of several different functions . We now cons ider a widely used tec hnique that req uires the comp utation of a single fu nction at several points rather than the co mputation of several differe nt f unctions a t a si ngle point.
3.7
The RungeKurta Method
55
We consider the initial value problem
y' = F (x, y);
when x = x"' Y = YII'
(1)
and for convenience adopt the notation
F"
(2)
= F(xlI ' YII).
Let us begin by considering the tangent line to the solution curve at the point (XII ' YII). The equation of this line is given by
Y = Yn +
FII (x
 XII).
(3)
The value of Y for thi s tangent line at x = XII + h is thus Y YII + h F". If we define Kl = FIl and compute F at the point (x" + h, y" + hK 1 ) , we obtain K2 = F (XII + h, y" + h K 1). Thus K I and K2 represent the values of Y' at two points, the two endpoi.nts of a segment of a tangent line. If we consider tbe arithmetic mean of these values of y' . namely! (Kl + K2), and replace the tangent line with a new I ine through (XII ' Yll) having this slope, we obtain
Y For x =
XII
= y" + !C K ] + K 2 )(x
 x,,).
+ h, this line has a point whose y coordinate is (4)
where Kl =
F"
(5)
and
(6) The generalizations of th e idea above are the basis for the method of RungeKutta. Instead of choosing the tangent line as a means for approximating the value of y atx = XII +h, we choose a line whose slope is an average of the values of y' at several carefully chosen points . When only two points are used, as just shown, the idea can be pictured as in Figure 3.2. We now describe the intuitive idea behind the more elaborate scheme. Again we define K l = F" to be the slope at the point P. Th.i time, we define K2 to be the slope at the midpoint M of the line segment of the tangent line PQ. From equation (3) we find that M is (XII + ! h, y" + !hK 1) and thus K 2 = F(x"
+ ! h , YII + ! hK 1).
The line through P with slope K2 has equation
y = Yll
+ K 2(X
 XII),
56
Chapter 3
Numerica l Methods Y YII + hF" f        =...".r
~~~x
o
xlI + h
x"
Figure 3.2
and by etting x
XII
+
~h, we obtain a point on this seco nd line, nam ely
(XII + ~h, YII + ~hK2). Now, de finin g K 3 = F(xll + ~ h , YII + ~ h K 2), we consider a third line through P, thi one having slope K 3 . Its equation is
Y = Yll
+ K 3(X 
XII)'
The value of Y for thi s third line at X = X" + h is YII + hK3. We now define K4 = F(xlI + h, YII + hK3) a the val ue of y' at the fo urth point. Thus the numbers K 1 , K 2, K 3, and K4 representthe va.lue ofy' at four points, one wi th X = XII' two with x = XII + ~h, and one with X = X II + h. We now determine the weighted mean of these fo ur numbers, K = ~(KI
+ 2 K2 + 2K3 + K 4 ),
and consi der a line through P with lope K . Its eq uation is
+ K (x  XII) ' The value of Y for this fo urth lin e at X = X II + h is YII+ I = YII + hK, Y = y"
(7)
where K = ~( KI
+ 2K2 + 2K3 + K 4 ),
KI = FII ,
+ ~ h , y" + ~ hKI) ' F(xli + ~ h , Y II + ~hK2)' F(xlI + h , YI1 + hK3) ·
(8) (9)
K 2 = F(xli
(10)
K3 =
(1 1)
K4 =
(12)
3.7
The RungeKutta Met/lOci
57
TABLE 3.8
x
0
0.1
0.2
Y KI
1.00 l.00
1.11 1.22
1.25
+ ~h Y + ~hK I
0.05
0.15
1.05
1.17
K2
1.10
1.35
Y + ~hK2
1.06
1.1 8
K3 x +17 Y + hK3 K4 K
1.12 0.10 1.11 1.22 1.1 1
0.20 1.25 1.52
x
1.37
1.36
Formulas (7) through ( 12) are due to Runge (1856 1927) and Kutta (18671944). The particular weighting factors assigned to the KI, K 2 , K J , and K4 in equation (8) are chosen so that the value of )1,,+ 1 computed by the RungeKutta method and the value computed by a fiveterm Taylor formula, y,, + 1  )III
h 2 y"
h3y
h 4 y(4)
' + /1 )"/ + " " + 41' "21 + 31l!
differ by an amount that is proportional to h 5 . The proof of this fact will not be given here but can be found in various texts on numerical analysis. We notice in passing, however, that if F (x, y) does not explicitly involve the variable y, then the RungeKutta formulas reduce to the familiar Simpson's ru le of elementary calcu lus. (See Exercise 15 below.)
EXAMPLE 3.1 Solve the example of Section 3.2 by the RungeKutta method . We present in Table 3.8 the results of the computation for x = 0.1 and x = 0.2 and leave the remaining computations for the exercises .
•
• Exercises In each of the following exercises, use the RungeKutta method to approximate the sol ution of the initial va lue problem in the given in terval. In Exercises 2 through 6 compare the approximate va lues with the correct val ues obtained by elementary methods .
1.
Continue the computation of Example 3.1 to obtain approximate values of y for x = 0.3,0.4, and 0.5.
58
Chapter 3
Numerical Methods
2.
Exerc ise 1, Section 3.2.
8.
Exercise 7, Section 3.2.
3.
Exercise 2, Section 3.2.
9.
Exercise 8, Section 3.2.
4.
Exercise 3, Section 3.2.
10.
Exerc ise 10, Section 3.2.
5.
Exercise 4, Section 3.2.
II.
Exercise 11 , Section 3.2.
6.
Exerci e 5, Section 3.2.
12.
Exercise 12, Section 3.2 .
7.
Exercise 6, Section 3.2.
13.
Exercise 8, Section 3.6.
14.
3.8
Show that if the function F(x, y) in equation (l) of this sectio n does not explicit ly involve the variable y, then the RungeKutta formulas (7) thro ugh (12) reduce to a special case of Simpson's rule.
A Continuing Method The methods u ed in the previous sections of th is chapter may be call ed "starting" methods for finding approximations of olutions of the problem
l
= F(x, y);
x
= Xo, Y = Yo.
(1)
By tins we mean that no additional information is know n other than that given in prob lem (1) itself. Once an approxi mate value of YI has been obtained for XI = Xo + h, we have then used YI to compute Y2, and so on . We shall now describe a "continu ing" method developed by Milne (1890 197 1). 1 Suppose that we know the values of YII, YII _I, Y,,2, and YII  3. Then we can compute the values of FII , F,, _ I , F,,2, F,,  3, fro m eq uation (1) . Nextweapproximate y ' (x) by a cubic polynomial that passes through the fo ur points (x" , FII ), (XII  I, FII _ I ), (X,, 2 , F,,2)' and (X Il 3, F II  3 ). It can be proved that th is can be done and the polynomial thus obtained is un ique. Using this polynomial in place of y' (x) in the integra l YII+I  Y,/  3
=
l
X
n+ '
lex) dx,
(2)
Xn3
performin g the integration, and simplify ing gives an approx imation for result i (I) )111 + 1
=
411,
)111  3
+3
(2F"  F,, I + 2FII _2)·
)111+ I .
The
(3)
The detail of this derivation are discussed in Exercise 10. The problem of est im ating the error in our approx im ations and of designing programs for reducing or con'ecti ng for errors is, of course, crucial to any method we may use. In the method of Milne, (3) is called a predictor formula and the value Y';~ I obtained from it is used to find a corrected value for YII+I . I See, fo r example, W. E. Mil ne, Numerical Solutions of Differential Equations (New York: John Wil ey & Sons, Inc. , 1953), Chapters 3 and 4.
3.8
A Continuing Method
59
TABLE 3 .9 n
x" 0.0 0.1 0.2 0.3
0 2 3
YII
FII
1.00 1.11 1.25 1.42
1.00 1.22 1.52 1.92
A derivation of the correction formula is suggested in Exercise 11. The re ult is
m
YII+I =
h .
YIl1
.
+ '3(F"  1 + 4F" + F,, +I ) ,
(4)
where the value of F,,+ I is calculated by usi ng the YI~~ I obtained from the predictor formula. To illustrate the procedure outlined previously, we use Milne's method to find the value of Y at x = 0.4 for the probJem
x = 0, Y = 1. For starting points we take the values of Y I, Y2, and Y3, which were computed using the RungeKulla method. These numbers are presented in Table 3.9 . Now using the predictor formula (3), we obtain (I)
Y4
= Yo
4h
+ 3(2F3 
F2  2F 1 )
= 1.00 + 4(~.1) [2(1.92) 
1.52 + 2(1.22)]
= 1.63. Using this value to compute F4 and applying Lhe corrector formula (4), we bave (?)
Y4 = Y2
II
+ 3(F2 + 4F3 + F4 )
= J .25
0.1
+ 3[1.52 + 4(1.92) + 2.50]
= 1.64.
Exercises 1.
Continue the problem of tbis section to estimate values of Y at 0.5 and 0.6.
In Exercises 2 through 9, use the RungeKulla method to obtai n estimated values for YI, Y2, Y3 and then compl1te approximation for)'4 and Y5 by Milne's method .
60
Chapter 3
Num.erical Methods
2.
Exercise 1, Section 3.2 .
6. Exercise 7, Section 3.2.
3.
Exercise 3, Sectio n 3.2.
7.
4.
Exercise 4, Section 3.2.
8. Exercise 11, Section 3.2. 9. Exe rcise 13, Section 3.2.
5.
10.
Exerc ise 6, Section 3.2.
Exercise 9, Section 3.2.
Let \1 FII =
F" 
F,, _ I ,
2
\1 FII = \1(\1FI/) = \1(F"  F,,_ I) = F"  2FI/ _ 1 + F II  2, \13 FII
(a)
= \1(\12 FI/)'
Verify th at the graph of Y = FII
\1 FII
+ (x I !h
 XII)
+
\12 r~, ?
2 !h
\13 £"
(x  XI/)(X  XII  I)
+3 (x 3 !h 
 XII) (X  XI/ _ I)(X  XI/ 2 )
passes through the four poi nts (XI/ 3 . F,, 3 ), (X,, 2 , F,, 2 ), (X,, _ I , F,,_ I), and (x", F,,). (b)
11 .
3.9
Using the po lynomia l above as a replacement for the integrand in eq uation (2) above, derive Mi lne's formula (3).
Suppose that the va lue Y/~11 of equ ation (3) is used to estimate FII + 1. By substitution into the differential equation and the use of Simpson 's rule, show that a reca lcul ation of YII+I gives the res ul t of fo rmul a (4).
Computer Supplement Given the nature of its content, it would seem that th is section wo uld be th e pelfect pl ace for computer ass istance. Surprisingly, it is one of the most di fficult. The rea on is that the methods described are equally easy for the com puter. For example, the additional work required in moving fro m the Euler method to the RungeKutta method is alm ost unnoticeable excep t at the hi ghe t degrees of accuracy. As a result, most computer sy te ms have builtin sophisticated numerical methods that are beyond the scope of thi s book. In order to experim ent with the easier methods , it is therefore necessary to write our own programs. This can be done in a lmost any computer (or calculator) language . For illustration purposes we show a Maple program to replicate the results in the first two columns of Table 3.2 in Section 3.2. > f : =(x,y) >y A 2x A 2 ; > Xinit:=O;
3.9 >
Computer Supplemelll
61
Xfin al:=O.5 ;
> Yinit : =l ;
n:=10; h:=(X final  Xinit) I n; > x : =Xinit; > y : =Yinit; > for i from 1 to n do y : =y+h*f (x, y) ; x : =x+h; ad ; > Yfinal : =y ; Note two facts about the program: first we have chosen to assign a valLle to n, the number of steps, and then let the machine compute the step size h. This is the reverse of the process described in the section, and relieves the user of checking that the step size i a divisor of the interval length . The seco nd fact worth note is the order of the two commands inside the for loop. We need to compute the new y value before we increment the x value. As noted in Section 3.7, the equation in question can be solved to any degree of accuracy 0 that we can compare the resu lts of the program above to the actual va lues. We can then experiment with various step sizes, and with minimal work modify the program to implement other numerica l methods .
>
>
• Exercises 1.
Implement the ELller method given above in the programming language of your choice.
2.
Modify your program for 17 = 50 and n correct answer given in the text.
3.
Modify your program to solve some of the exercises in Section 3.2.
4.
Modify your program to implement the modified Euler method. Thi s will require adding a few steps to the for loop. Be caref ul to make your assignments in the right order. Compare the accuracy of the two Euler methods for the same n.
5.
Modify your program to imple ment the RungeKutta method and proceed as in Exercise 4.
= 100. Compare your results to the
Elementary Applicatio ns
4.1
4
Velocity of Escape from the Earth Ma ny p hysical problems invo lve differential equatio ns of o rder one. Co nsider the prob lem of determ ining the velocity of a partic le projected in a rad ial directio n outward fro m the earLh and acted upo n by only one force, the gravitatio nal attraction of the earth . We shall assume an initial velocity in a radi a l directi on 0 th at the motion of the part icle takes place entire ly on a line th rough the center of the earth. According to the Newtonian law of gravitation, the acceleration of the parti cle will be inversely proportional to the square of the distance fro m the particle to the center of the earth. Let r be th at va ri able d istance, and let R be the rad iu s of the earth. If t represents time, v the ve locity of the particle, a its accelerati o n, and k the con tant of propo rtionality in the Newtonian law, then dv k a = = . df r2 T he acceleration is negative because the veloc ity is decreas ing. Hence the constant k is positive. W hen r = R, then a =  g, the acceleration of grav ity atthe surface o f the earth . T hus
from which
a=
gR2 ? .
r
We wish to express the acceleration in terms of the ve locity and the d istance. We have a = dv / dt and v = dr / dt. Hence dv dr dv dv Q =  =   =V  , dt dldr dr so the differentia l equation for the velocity is now seen to be dv gR2 v  = . (I ) dr ,.2
62
4. J
Velocity of Escape from the Earth
63
The method of separation of variab les applies to equation (l) and leads at once to the set of solutions 2 2gR2 V =  +C. r
Suppose that the particle leaves the earth 's surface with the velocity Va . Then v = Va when r = R, fro m which the constant C is easily determined to be C = v5  2gR.
Thus, a particle projected in a radial direction outward from the earth's surface with an initial velocity Va will travel with a velocity V given by the equation
v2 =
2gR2 
r
+ v6
 2g R.
(2)
It is of considerable interest to determine whether the particle will escape from the earth. Now at the surface of the earth, at r = R, the velocity i positive, V = Va . An examination of the right member of equation (2) shows that the veloc ity of the particle will remain positive if, and only if,
v52gR::::0.
(3)
If the inequality (3) is satisfied, the velocity given by equation (2) will remain positive because it cannot vanish, is continuous, and is positive at r = R. On the other hand , if (3) is not satisfied, then 2g R < 0, and there will be a critical value of r fo r whi ch the right member of equation (2) i zero. That is , the particle would stop, the velocity would change from posit ive to negative, and the particle would return to the earth. A particle projected from the earth with a velocity Vo such that
v6 
Vo ::::J2g R
will escape from the earth. Hence the minimum such velocity of projection, Ve
= J2g R,
(4)
is called the velocity of escape . The radius of the earth is approximately R = 3960 miles. The acceleration of gravity at the surface of the earth is approximately g = 32.16 feet per second per second (ft/sec 2 ), or g = 6.09(10) 3 mile/sec 2 . For the earth, the velocity of escape is easily fo und to be Ve = 6.95 miles/sec. Of course, the gravitational pull of other celestial bodies, such as the moon, the su n, Mars , Venu , and so on , has been neglected in the idealized problem treated here. It is not difficult to see that such approximations are justified, since we are interested only in the critical initial velocity Ve . Whether the particle actually recedes from the earth forever or becomes, for inslance, a satellite of so me heavenly body is of no consequence in the pre ent problem. If in this study we happen to be thinking of the particle as an ideali zation of a ballistictype rocket, then other elements must be considered. Air resistance in the
64
Chapter 4
Elementary Applicatiolls
first few mi le may not be negligible. Methods for overco ming such d ifficulties are not su itab le topics for discussion here. It mu st be realized that the formula Ve = ,J2g R app lies eq uall y well for the velocity of escape from the other members of the solar sys te m, as long as Rand g are given their appropriate values.
I 4.2 I Newton's Law of Cooling Experiment has shown that under certain cond itions, a good approximation to the temperature of an object can be obtai ned by using Newton 's law of cooling: The temperature of a body changes at a rate that i proportional to the difference in temperature between the outside medium and the body itself. We shall ass ume here that the constant of proportionality is the same whether the temperature is increasing or decreasing. Suppose, for in ta nce, that a thermometer, which has been at the reading 70°F inside a house, is placed outside where the air temperature is 10°F. Three minutes later it is fo und tha t the thermometer read ing is 2YF We wish to predict the temperature reading at various later times . Let u (OF) represent the temperature of the thermometer at tim e f (m in ), the time being measured from the in tant the therm ometer is placed outs ide . We are given that when t = 0, u = 70 and when t = 3, LI = 25 . According to Newton's law, the time rate of change of temperature, du / d t , is proportional to the temperature differe nce (u  10) . Since the thermometer temperature is decreasing , it is convenient to choose (k) as the constan t of proportionality. Thus the u is to be determined fro m the differential equation
elu
dt = k(u
 10),
(1)
and the conditions th at when t = 0 ,
1.1
(2)
= 70
and
(3) when t = 3, LI = 25. We need to know the thermometer read i ng at two different times because there are two constants to be determ ined, k in equation (l) and the "arbi trary" constant that occurs in the solution of differential equation (l). From equation (1) it follow s at once th at u = 10 + Ce  k ,. Then condition (2) yields 70 = 10 + C, from whi ch C = 60, so we have
u
= LO + 60e k , .
The val ue of k wi ll be determined now by usi ng condition (3) . Putting and u = 25 into equation (4), we get 25 = 10 + 60e 3k , from which e 3k
= i,so k = ~ In 4.
(4) f
= 3
4.3
Simple Chemical Conversion
65
Thus the temperature is given by the equation u = 10+60exp (  ~t ln4).
(5)
Since In 4 = 1.39, equation (5) may be repl aced by u = 10 + 60exp (0.46t).
I 4.3 I
(6)
Simple Chemical Conversion It is known from the results of chemical experimentation that, in certain reactions in which a substa nce A is being converted into another substance, the time rate of change of the amount x of unconverted substance is proportional to A. Let the amount of unconverted substance be known at so me specified time; that is, let A = AO at t = O. Then the amount x at any time t > 0 is determined by the differential equation
dx (1) : = kx df and the condition that x = Xo when f = O. Since the amount A is decreasing as time increases, the constant of proportionality in eq uation (L) is taken to be (k) . From equation (1) it follows that
x = Ce  kt . But x
= xo when t =
O. Hence C
= Xo.
Thus we have the res ult
x = xoe 
kt
(2)
Let us now add another condition, which will enable us to determine k. Suppo e it is known that at the end of hal f a minute, at t = 30 (sec), twothirds of the original amount Xo has already been converted. Let us determine how much unconverted substance remains at t = 60 (sec). When twothirds of the sub tance has been converted, onethird remains unconverted. Hence x = txo when t = 30. Equation (2) now yields the relation .!.X 3 0 
x 0 e 30k ,
Jo
from which k is easily fo und to be In 3. Then with t measured in seco nds , the amount of unconverted substance is given by the equation x=xoexp(  Jotln3) .
(3)
At t = 60, x
= Xo exp (2In 3) = xO(3)  2 =
~xo .
• Exercises 1.
The radi us of the moon is roughly 1080 miles. The acceleration of grav ity at the surface of the moon is about 0.165g, where g is the acceleration of gravity at the su rface of the earth. Determine the velocity of escape for the moon.
66
Chapter 4
Elementary Applications
2.
Determine, to two significant figures , the velocity of escape for each of the celestial bodies listed in Table 4.1 . The data given are rough and g may be taken to be 6.1 (10) 3 mile/sec 2 .
3.
A thermometer reading 18°F is brought into a room where the temperature is 70°F; 1 min later the thermometer reading is 3 1°F. Determine the temperature reading as a function of time and, in particular, fi nd the temperature reading 5 minutes after the thermometer is first brought into the room.
4.
A thermometer reading 75°F is taken ou t where the temperature is 20°F. The reading is 30°F 4 min later. Find (a) the temperature reading 7 min after the thermometer was brought outside and (b) the time taken for the reading to drop from 75°F to within a half degree of the air temperature.
5.
At 1:00 P.M., a thermometer read ing 70°F is take n outside where the air temperature is 10°F (te n below zero). At 1:02 P.M. , the reading is 26°F. At 1:05 P.M., the thermometer is taken back indoors, where the air is at 70°F. What is the temperature readin g at I :09 P.M.?
6.
At 9 A.M., a the rmometer reading 70°F is taken outdoors, where the temperatu re is 15°F. At 9:05 A.M., the thel1110meter reading is 45°F. At 9 : 10 A.M. , the thell110meter is taken back indoors, where the temperature is fi xed at 70°F. Find (a) the read ing at 9:20 A.M. and (b) when the reading, to the nearest degree,will show the correct (70°F) indoor temperature.
7.
At 2:00 P.M. , a thermometer read ing 80°F is taken outside, where the air temperature is 20°F. At 2:03 P.M., the temperature read ing yielded by the thermometer is 42°F. Later, the thermometer is brought inside, where the air is at 80°F. At 2: 10 P.M., the reading is 71 °F. When was the therm omete r brought indoors?
8.
Suppose that a chemical reaction proceeds according to the law given in Section 4 .3. If half the substance A has been co nverted at the e nd of 10 sec, find when ninetenths of the substance will have been converted.
9.
The co nversion of a substance B follows the law used in Section 4. 3. If only a fourth of the substance has bee n converted at the end of 10 sec, find when njnetenths of the su bstance wi ll have been converted .
Venus Mars Jupiter Sun Ganymede
TABLE 4.1 Acceleration of Gravity at SUlface
Radius (miles)
Answer (111 ileslsec)
0.85g 0.38g 2.6g 28g 0. 12g
3,800 2, 100 43,000 432,000 1,780
6.3 3. 1 37 380 1.6
4.3
Simple Chemica! Conversion
67
10.
For a substance C, the time rate of conver ion is proportional to the square of the amount x of unconverted ubstance. Let k be the numerical value of the constant of proportionality and let tbe amount of unconverted substance be Xo at time t = O. Determine x for all t ~ O.
It.
For a substance D, the time rate of conversion is proportional to tbe square root of the amount x of unconverted substance. Let k be the numerica l va lue of the constant of proportionality. Show that the substance will disappea r in finite time and determine the time.
12.
Two substances, A and B, are being converted into a single compo und C. In the laboratory it has been shown that for these substances , the following law of conversion hold : The time rate of change of the amount x of compound C is proportional to the product of the amounts of unconverted substances A and B. Assume the units of measure so chosen that one unit of compound C is formed from the combination of one unit of A with one unit of B. If at time t = 0 th ere are a units of substance A , b units of substance B , and none of compound C present, how that the law of conversion may be expressed 'by the equation
dx  =k(ax)(b  x).
df Solve this equation with the given initial conditions.
13.
In the solution of Exercise 12, assume thatk > 0 andinvestigate the behavior of x as t ~ 00 .
14.
Radium decompose at a rate proportional to the quantity of radium present. Suppose it is found that in 25 years approximately 1.1 % of a certain quantilY of radium has decomposed. Determine approximately how long it wi 11 take for onehalf the original amount of radium to decompose.
15.
A certain radioactive substance has a halflife of 38 hr. Find how long it takes for 90% of the radioactivity to be dissipated.
16.
A bacterial population B is known to have a rate of growth proportional to B itself. If between noon and 2 P.M. the population triples, at what time, no controls being exerted , should B become 100 time what it was at noon ?
17 .
In the motion of an object through a certain medi urn (air at certain pressures is an example) , the medium furnishes a re isting force proportional to the square of the ve locity of the moving object. Suppose a body falls , du e to the action of g ravity, through the medium. Let t represent time, and v represent ve locity, positive downward. Let g be the usual constant acceleration of gravity, and let 11) be the weight of the body. U e Newton's law, force equals ma s times acceleration, to conclude that the differential equation of motion is 11) dv 2   = 11)  kv , g dt where kv 2 is the magnitude of the res ist ing force fu rni hed by the medium.
68
Chapter 4
ELel'llentCilY Applications
18.
Solve the differential equation of Exercise 17 , with the initial condition that v = Vo when t = O. . Introduce the constant a 2 = w / k to si mplify the formulas.
19.
There are mediums that resist motion through them with a force proportional to the first power of the ve locity. For such a medium, state and solve problems analogous t.o Exercises 17 and 18, except that for convenience a constant b = w / k may be introduced to replace the a 2 of Exercise 18. Show that b has the dimens ions of a velocity.
20.
Figure 4.1 shows a weight, w pounds (lb), sliding down an inclined plane that make an angle ex. with the horizontal. Assume that no force other than gravity is acting on the weight ; that is, there is no friction , no air resistance, and so 011 . At time t = O. let x = Xa and let the initia l velocity be Va . Determine x for t > O.
2 1.
A long, very smooth board is inclined at an angle of 10° w ith the horizontal. A weight starts from rest 10ft from the bottom of the board and slides downward under the action of gravity alone. Find how long it will take the weight to reac h the bottom of the board and determine the terminal speed.
22.
Add to the conditions of Exercise 20 a retarding force of magnitude kv, where v is the velocity. Determine V and x under the assumption th at the weight tarts from rest with x = Xo. Use the notation a = kg/w.
23.
A man, stand ing at 0 in Figure4.2, holds a rope oflength a to which a weight is attached, initially at Wo oThe man walks to the right dragging the weight after him. When the man is at M, the weight is at W. Find the differential eq uat ion of the path (called the tractrix) of the weight and sol ve the equation.
24 .
A tank co ntains 80 gallolls (gal) of pure water. A brine solution with 2lb/gal of salt en ters at 2 gal/min, and the wellstined mixture leaves at the same
Figure 4.1
4.4
Logistic G,vwth and th e Price of Coml1lodities
69
y
w (I
y
 +'  '  x
o
x
M
Figm"e 4.2
rate. Find (a) the amount of salt in the tank at any time and (b) the time at which the brine leaving will contain lib/gal of salt.
4.4
25.
For the tank in Exercise 24, determine the limiting value for the amount of salt in the tank after a long time. How much time mllst pass befo re the amount of salt in the tank reaches 80% of thi lim iting value?
26 .
A certain sum of money P draws interest compounded continuously. If al a certain time there are Po dollars ill the account, determine the tim e when the principal attains the value 2Po dollars, if the annual interest rate is (a) 2 % or (b) 4 %.
27.
A bank offers 5 % interest compounded cont inu ously ina sav in gs account. Determine (a) the amount of interest earned in 1 year on a deposit of $100 and (b) the equivalent rate if the compounding were done annuall y.
Logistic Growth and the Price of Commodities Numero us attempts have been made to develop mode ls to study the growth of popu lations. One means of obtaining a simple model for that study is to ass ume that the average birthrate per individual is a positive constant and that the average death rate per individual is proportional to the popu lation. If we let x (t) represent the population at time t, the assumptions above lead to the differential equation
1 dx   = b ax
x elt
'
(1 )
where b and a are positive constants. This equation is commonly called the logistic equation and the growth of population determined by it is ca ll ed logistic growth.
70
Chapter 4
Elementary Applications
The variab les in the logistic equation may be separated to obtai n
dx     = dt, x(b  ax) or
dx = bdt. ( ~x + _b a_) ax Integrating both sides gives us In
[_x_[
= bt
b  ax
+ c,
or X[ [ b ax
C br = ee .
(2)
To exped ite the study of eq uation (2), let us ass ume further that at t = 0 the population is the positive number Xo. T hen eq uation (2) may be written x Xo br e , b  ax b  axo and upon so lving for x, we have
(3)
It is interest ing to note that the population function obtained in equatio n (3) has a limiting val ue lim x(t) = lim r >oo
r>oo
· I1m
r >oo
bx ebr b  axo b 2 x ebr
0
+ axoebr
0
abxoebr
b a where we have used l'Hospital's rule to eval uate the limi t. We should also note that the logistic eq uation (1) will dicta te growth or decline in the pop ulation, depending upo n whether the in itia l population is less th an or greater than b / a. As a further example of an applicatio n in which a firstorder differen tial eq uation occurs, we consider an economic model of a certai n commodity market. We assume that the price P, the supply S, and the demand D of that commodity are fu nctions of time and that the rate of change of the price is proportional to the difference between th e demand and the supply. T hat is, dP
dt
= keD  S) .
(4)
4.4
Logistic Gro~vrh and th e Price of Commodifies
71
We assume further that the con ta nt k is positive so th at the p rice will increase if the demand exceeds the supply. D ifferent models of the commodity ma rke t w ill result, depending upo n the nature of the demand and suppl y functions tha t are indi cated. If, for example, we assume that
D = c  dP
and
(5)
S=a+bP,
where a, b, c, and d are posi tive con tants, we obta in a differential equation dP = k[(c  a)  (d
+ b)P] (6) dt that is linear in P. The ass umpLions (5) refl ect the tendency for the de mand to decrease as the price increases and the tendency for the suppl y to increase as the pri ce increases, both reasonable assumptions for many commodities. We should also as ume that 0 < P < cld, so that D is not negative. Eq uation (6) may be written dP
dt + ked + b) P =
k(c  a),
and o lved by multiplying by the integratin g factor obtai n
Pct) = c
e  k(d+ b)t
ek(d+b)1
(7) and in tegrating to
c a +.
I
d+b
If the pri ce at I = 0 is P = Po, we have C(l
CI
= Po   d + b'
so th at
ca pet) = ( Po  ) d +b
e k(d+b) 1
ca +.
d+b
(8)
Equation (8) shows that under the ass umpLi ons of (4) and (5) the price will tab il ize at a value (c  a)/(d + b) as t becomes large .
• Exercises 1.
A certain population is kn ow n to be grow ing at a rate given b y the logisti C eq uation dxld t = x(b  ax). Show that the maximum rate of growth will occur when the population is equal to half its eq uilibri um size, that is, whe n the popu lati on is bl2a.
2.
A bacterial population is know n to have a logi. ti c growth pattern with initi al population 1000 and an equilibrium population of 10,000. A cou nt shows th at at the end of 1 hr there are 2000 bacteria present. Determine the popula tion as a function of ti me.
72
Chapter 4
ElementCilY Applications
3.
For the population of Exercise 2, determine the time at which the population i increasing most rapidly and draw a sketch of the logistic curve.
4.
A college dormitory houses 100 tudents, each of whom is susceptible to a ce11ain virus infection. A simple model of epidemics assumes that during the course of an epidemic the rate of change with re pect to time of the number of infected students f is proportional to the number of infected students and is also proportional to the number of uninfected students, 100  f. (a)
If at time t = 0 a single student become. infected, show that the number of infected students at time t is given by 100e lOOkl
J    1OOkl 
 99+e
(b)
5.
'
If the constant of proportionality k has value 0.01 when t is measured in days, find the value of the rate of new cases f l (f) at the end of each day for the first 9 days.
Glucose is being fed intravenously into the bloodstream of a patient at a contant rate c grams per minute. At the same time, the patient's body converts the glucose and removes it from the bloodstream at a rate propol1ional to the amount of glucose present. If the constant of proportionality is k, how that as time increases, the amount of glucose in the bloodstream approaches an equilibrium value of c/ k.
6. The supply of food for a certain population is subject to a seasonal change that affects the growth rate of the population. The differential equation
dx  = cx(t) cost, dt where c is a positive constant, provides a si mple model for the seasonal growth of the population. Solve the differential equation in terms of an initial popu lation Xo and the constant c. Determine the maximum and the minimum populations and the time interval between maxima. 7.
Suppose that the human body dissipates a drug at a rate proportional to the amount y of drug present in the bloodstream at time t. At ti me t = 0 a fi rst injection of Yo grams of the drug is made into a body that was free from that drug prior to thal li me . (a)
Find the amount of residual drug in the bloodstream at the end of T hours .
(b)
If at time T a second injection of Yo grams is made, find the res idual amount of drug at the end of 2T hours .
4.5
Computer Supplement
73
(c)
If at the end of each time period of le ngth T , a n injection of Yo grams is made, find the residual amount of drug at the end of nT hours.
(d)
F ind the limiting value of the answer to part (c) as n approaches in fin ity.
8.
If the demand and supply functions for a commodity market are D = c  d P and S = a sin /3t, determine Pct ) and analyze its behavior as t increa es.
9.
An analysis of a certa i n commodity market reveal s that the demand a nd supply funcLions are given by D = c  d P and S = a + bP + q sin /31 , where a, b, c, d, q, and /3 are posit ive constants. Determ ine pet) and analy ze its behavior as t increases.
I 4.5 I Computer Supplement In Section 4.4 we discussed the logistic equation dx

(l) = x(bax). dl As noted in the text, thi s equation can be used to model the growth of a population x subject to some upper limit. In this model we a sume that when x is near 0 , the term ax 2 will be negligible and we will see the popu lation behaving like a solution to dx
 =bx. dt
We know from earlier work that this sol ution is exponential. On the other hand , whe n x grows to nearb/a, the term bax wi]] be nearOand thegrowLh w ill slow down. T his is easi ly demonstrated by choosing posi tive value' for the consta nts a and b and graphing several so lutions to (1). Figure 4 .3 was produced by the Maple command: >
DEplot(diff(x(t) ,t)=x* (3 2*x) ,x(t),t=  O.5 . . 2, { [ 0, . 5) , [0,1] , [0,2 ) , [0,2.5) } , x= 0 .5 . . 3 ) i
We see that the population will s tabilize at 3/2 or b / a. If a small distu rbance either inc rea es o r decreases the popu lation, it will return to thi s stable val ue. This is an example of a sustainable population .
• Exercises 1.
Use a computer to prod uce Figure 4.3.
2.
Nowa sume that there is a con tant level of harvesting of the population, so that the equation becomes dx
 =x(b  ax)  h. dt
74
Chapter 4
Elementary Applications x
Figure 4.3
For ma ll levels of harvesting, that is, when h is near zero, we would expect that the population wou ld not be greatly affected. Modify you r solution to the above by in erti ng a small value for h and plot the results. 3.
If the population is a food source fo r hu man, we wo uld want to maximize the harvesti ng without endangering the health of the population . Gradually increase h and see what happens. Wh at is the maximum harvesting leve l th at can safe ly be mai ntained?
4.
Next, su ppose that the harvesting i sea onal. For example, replace the h by h(si n (t) + 1). See what happens to the population for different va lues of h.
5.
Can you find a va lue for h and initial cond itions so that the popu lation survives for one "season", but dies alit in the second?
Additional Topics on Equations
5
of Order One I 5 . 1 I Integrating Factors Found by Inspection In Section 2.5 we found that any linear equation of order one can be solved with the aid of an integrating factor. In Section 5.2 there is some d iscussion of tests for the determination of integrating factors. At pre ent we are concerned with equations that are simple eno ugh to enable Ll S to find integrating factors by inspection. The ability to do this depends largely upon recognition of certain common exact differentials and upon experience. Following are four exact differentials that occur frequently:
d(x y ) = xdy
+ y dx,
(1)
d (;) = y dx
;Z x d y ,
(2)
d (X 
~) _ xdy  ydx
(
Y) =
d arctan 
X
'
(3)
x dy  y dx 2 2' X + y
(4)
X
2
Note the homogeneity of the coefficients of dx and dy in each of these differentials. A differential involving only one variable, sLlch as x 2 dx, is an exact differential.
EXAMPLES.} So lve the equation
ydx
+ (x +x 3 l
)dy = O.
(5)
Let us group the terms of like degree, writing the equation in the form
(ydx +xdy) +x 3 /dy = O.
75
76
Chapter 5
Additiollal Topi cs on Equations of Order Oll e
Now the com binati on (y elx tion, obtai ning
+ x dy) attracts attention, so we rewrite the equ a
d(xy)
+ x 3l
(6)
ely = O.
Since the differentia l of x y is prese nt in eq uation (6) , any fac tor that is a f unction of the product x y will not disturb the integrability of that term . But the other term contai ns the differential el y, and hence should co ntain a function of y alone. Therefore, let u divide by (xy )3 and write d(xy ) (x y )3
+ ely
= O.
y
The equation above is integrab le as it stands . A fami ly of sol utions is defined by
1
2x2 + In Iy l = y2
In
lei,
or
• EXAMPLE 5.2 Solve the equation y (x 3  y)dx x(x 3
+ y )dy = O.
(7)
Let us reg roup the terms of (7) to obtain x 3(yd x  x ely)  y (y d x + xdy) = O.
(8)
Recalling that
d(~)
Y = ydx ? Xd ,
we divide tbe term s of equatio n (8) throughout by
x3 d
(~)
l
to get
_ d(;Y ) = O.
(9)
Equation (9) will be made exact by introduc i ng a factor, if it can be fo und, to make the coeffi cient of d( xly ) a fu nction of xly and the coefficient of d(xy) a fu nction of (xy ). Some skill in obtaining suc h factors can be developed with a little practice. There is a stra ightforward attack on equation (9) that has its good poi nts. As um e that the integrati ng factor desired is xk yll, where k and n are to be determined. Applying that facto r, we obtain X k+3 y ll d
(~)
 xk y,,  Id(x y ) = O.
(10)
77
Integrating Factors Found by Inspection
5.1
Since the coefficient of d (x j y ) is to be a fu ncti on of the ratio x / y, the exponents of x and y in that coefficient must be numerically equal but of opposite sign. That is,
+3=
k
(11)
no
In a similar manner, from the coefficient of d(x y ) it follows that we must put
k = n  l.
(12)
=
From equations ( II) and (12) we conclude that k integrating factor is x  2 y 1 and (10) becomes
~d (~) y
Y
2, n
= I.
The desired
 d(xy) = 0, x 2 y2
of wh ich a set of solutions is given by
~ (~r + Xly = ~. Finally, we may write desired solutions of equation (7) as
x 3 +2y = exl .
•
EXAMPLE 5.3 Solve the equation 3x 2 y d x
+ ( y4 
x 3 )dy = 0.
Two terms in the coefficient of dx and d y are of degree three, and the other coefficie nt is not of degree three. Let us regroup the terms to get 2
(3 x y d x  x 3 d y )
+l
d y = 0,
or
Y d(x 3 )

3
x dy
+l
ely = O.
The fo rm of the first two terms now suggests the numerator in the differential of a q uotient, as in
d
(!!.) _ velu v u dv ?
.
V
Therefore, we divide each term of our equation by y 2 and obtain
y d(x 3 ) .x 3 ely y
  2 =
+y
2
d y = 0,
or
d
(X; )+ l d y
= O.
78
Chapter 5
Additional Topics
011
Equations oJ Order One
Hence a solution set of the original equation is x3
y3
y+]
C
=
3'
or 3x 3 + l = cy .
•
•
Exercises
Except when the exercise indicates otherwise, find a set of so lutions.
y(2xy + 1) elx  X ely = O. 3. 2. y(y3 x ) elx +x(y3+.y)ely =0. 4. 4 5. y(x4  y2) elx + x(x + y2) ely = O. 1.
6. 7.
8. 9. 10.
II. 12. 13. 14. 15. 16. 17.
18. 19. 20. 2 1. 22. 23. 24 . 25. 26. 27.
(x 3 y 3 + 1) dx
+ x 4i
ely = O. 2 2tds+s(2+s t)dt=0.
y(y2 + L)elx +x(y2  I) ely = O. Do Exercise 6 by another method. y(x 3  y5) dx  x(x 3 + y5) dy = O. y(x 2y2  J)elx +x(x 2y2 + I) ely = O. x4y' = _ x 3 y  csc(xy). y(x 2 y2  m) dx + x(x 2y2 + n) ely = O. y(2  3xy) dx  x ely = O. y(2x + y2) elx + x(y2  x) ely = O. ydx+2(y4x)dy=0. y(3x 3  X + y) dx + x 2(1  x 2) dy = O. 2x S y' = y(3X4 + y2) . i (l x 2)dx+x(x 2y+2x+y)ely =0. [1 + y tan (xY)l elx + x tan (xy) dy = O. y(x2  y2 + I)dx x(x 2  y2  I)dy = O. (x 3 +xy2 + y)dx + (y3 +x 2y +x)dy = O. y(x2 + y2 l)dx +x(x 2 + y2 + I )dy = O. (x 3 + xy2  y) dx + (y3 + x 2y + x) dy = O. y(x 3exy  y) dx + x(y + x 3exy ) ely = O. xy(y2 + J)elx + (x 2y2  2)dy = 0; whenx = J, Y = I. x(x 2  y2  x) dx  y(x2  y2) dy = 0; when x = 2, y = O. y(x2 + y)dx +x(x 2  2y)ely = 0; when x = 1, y = 2. y(x 3 y 3 + 2x2  y) elx + x 3 ( xy 3  2) dy = 0; when x = I , Y = I.
5.2
28. 29.
5.2
Th e Determinatioll of Integrating Faclors
79
+ ay) dx + (X" + y" + bx) dy = O. I (X" + y" + ay) dx + (Xl/y" +1 + ax) ely = O. I
(x" y"+1
The Determination of Integrating Factors Let us see what progress can be made on the problem of the determination of an integrating facto r for the equation
M dx
+ N dy
= O.
(1)
Suppose that u, possibly a func tion of both x and y, is to be an integrating facto r of (l). T hen the equation uM dx
+ £IN dy
(2)
= 0
mu st be exact. Therefore, by the result of Section 2.4,
o .
a
ay
ox
(uM) =  (u N).
Hence u must atisfy th e parti al differential equation
au ay
aM
u+M 
oy
aN ax
aU. ax
=u  + N  ,
or (3)
F urthermore, by reversing the argument above, it ca n be seen that if u satisfies equation (3), u i an integrating factor for equatio n (1). We have " reduced" the problem of solvin g the ordinary differenti al equation (l) to the problem of obtai ning a particular sol ution of the parti al differential equati on (3) . Not much ha been gai ned becau e we have developed no methods for attac kin g an equation sLlch as (3) . Therefore, we turn the problem back into the realm of ordinary differential eq uations by restricting u to be a function of on ly one vari able. First let £I be a function of x alone. Then ou/ay = 0 and becomes du/dx. Then (3) reduces to
au/ax
u
(aM _ ON) = Ndu , ay
ax
elx
or
~ N
(aM _ aN )ctx = ay ax
duo u
(4)
80
Chapter 5
Additional Topics on Equations of Order One
If the left member of equation (4) is a f unction of x alo ne, we can determine u at o nce. Indeed , if
~ N
(aM _ aN) = f(x), ay ax
(5)
th en the desired in tegrating factor is u
= exp [ / f (x) dx
l
By a simi lar argument, assu min g that u is a f unction of y alone, we are led to th e co nclu sion that if 1
M
(aM  aN) =g(y), ay
(6)
ax
th en an in tegrating factorfor equat ion (1) i
1
u = exp [  / g(y) dy
Our two re ults are ex pressed in the following rules:
(a)
(aM aN)
If 1  = fex), a f unction of x alone, then exp N ay ax an integrating factor for the equation Mdx
(b)
(j
+ Ndy =0.
f(x) dx) is
(1)
(aM aN)
If  1  = g(y), a f un ctio n of y alone, then exp M ay ax is an integratin g facto r for equation (1) .
C! g(y) dy)
It should be empb a ized that if neither of the preced ing criteria is satisfied, we can say o nl y that the equ ation does not have an integrating facto r that is a function of x or y alone. For exampl e, the student sho uld show that the criteria above fail in the case of Examp le 5.1 of Section 5.1 , even thoug h (xy) 3 is an integrating factor fo r the differe ntial eq uation.
EXAMPLE 5.4 Solve the equation (4xy
+ 3/  x)dx + x(x + 2y)dy =
O.
(7)
5.2
Here M = 4x y
+ 3y2 
aM

ay
X,
aN
ax
81
N = x 2 + 2xy, so
= 4x
 
Th e Deterlllillatioll of /lltegratil1g Factors
+ 6y 
(2x
+ 2y)
= 2x
+ 4y .
Hence 1 ( aM aN) 2x +4)1 2 Nay  ~ = x (x + 2 y)  :;.
Therefore, an integrating factor for equation (7) is
(f
exp 2
elX) ~ =exp(2 In[ xl) =x 2 .
Returning to the original eq uation (7), we insert th e in tegrati ng fac tor and obtain
(8) which we know must be an exact eq uation . The methods of Section 2.4 app ly. We are then Jed to put equation (8) in th e form
(4x 3 y dx
+ x4 ely) + (3x 2 /
elx
+ 2x 3 y dy)
 x 3 dx = 0,
from which the solution set
or
foll ows at once.
•
EXAMPLES.S Sol ve the eq uation
y (x + y + J) elx + x (x + 3y
+ 2) dy
= O.
First we form
aM
=x+2y+ I, ay
aN

ax
= 2x + 3y+2 .
Then we see that
aM

ay
aN
  =  x  y l ,
ax
(9)
82
Chapter 5
Additional Topics
011
Equatiolls of Order One
so
1 (aM aN) x+y+ 1 N ay  ax =  x(x + 3y + 2) is not a function of x alone. But
1 (aM M
aN)
a;:
ay 
x +y +1 I =  y(x + y + 1) = 
y'
Therefore, exp On Iyl) = Iyl is the desired integrating fac tor for (9). It fo llows that for y > 0, y by itself is an integrating factor of equation (9) and fo r y < 0, y is an integrating factor. In eith er case (9) becomes
(xl
+ i + l)dx + (x 2 y + 3xy2 + 2xy)dy =
0,
or
(xidx
+ x 2 y dy) + (idx + 3xidy) + (ldx + 2xy dy)
= O.
Then a set of so lutions of (9) is defined implicitly by
~x2l +xi +xi = ~c, or
+ 2y + 2)
xl(x
= c.
•
EXAMPLE 5.6 Solve the equation
y(x From aM lay
+ y)dx + (x +2y 
= x + 2y,
aN lax
=
~ ( aM _ aN) N
ay
ax
l)dy = O.
(10)
1, we conclude at once that
= x + 2y x
I 1
+ 2y 
= l.
Hence eX is an integrating factor for (10). Then
(xye X + leX) dx
+ (xe or + 2ye X 
eo'") dy = 0
an exact equation. Grouping the terms in the following manner,
[xye X dx
+ (xe
X

eo'") dy ] + (leX d x
+ 2ye Xdy)
leads us at once to the family of solutions defined by
eX(x  l) y
+ leX
= c,
= 0,
5.2
The Determination of Integrating Factors
83
or
+y 
y(x
1) = ce x.
•
• Exercises Solve each of the equations in Exercises I through 14.
I. 2. 3.
+ y2 + l)dx +x(x  2y)dy = O. 2y(x2  Y + x) dx + (x 2  2y) dy = O. (x 2
y(4x+y)dx2(x2_y)dy = O.
5.
4. 7.
(xy+ 1) dx + x(x + 4y  2) dy = 0. 6. y(8x  9y) dx + 2x(x  3y) dy = O.
8.
Do Exercise 7 by another method.
9.
y(y+2x2) dx2(x+y) dy=O. y2 dx
+ (3xy + i  I ) dy =
+ 1) dx + (x  y) d y = O. + y + 2) dx + (y2  x 2  4x  1) d y = O.
y(2x2  xy
10.
2y(x
11.
2(2y2 + 5xy  2y + 4) dx + x(2x + 2y  1) d y = O. 3(x 2 + i) dx + x(x 2 + 3y2 + 6y) dy = O.
12.
O.
+ 3x y
1) dy = O.
(2y2
14.
O. Euler's theorem (Exercise 36, Section 2.3) on homogeneous functions states that if F is a homogeneous function of degree k in x and y, then
15.
y(2x  y
 2y
+ 6x) dx + x(x + 2y 
13 .
+ 1) dx + x(3x 
4y
+ 3) dy =
aF
aF
ax
ay
x+y 
=kF.
Use Eu ler's theorem to prove the result that if M and N are homogeneous functions of the same degree, and if M x + Ny# 0 , then Mx + N y
is an integrating factor for the equation Mdx
16.
+ Ndy
= O.
(A)
In the result to be proved in Exercise 15 there is an exceptional case, namely, when Mx + N y = O. Solve equation (A) when Mx + Ny = O.
Use the integrating factor in the resu l.tofExercise 15 to solve each of the equations in Exercises 17 through 20.
17. 18.
xydx  (x 2 + 2y2) dy = O. v 2 dx +x(x + v)dv = O.
84
Chapter 5
5 .3
Additional Topics
011
Equations of Order One
19.
v(u 2 + v 2) du  u(u 2 + 2v2) dv = O.
20.
(x 2 + y2)dx  xydy = O.
21.
Apply the method of this section to the general linear equation of order one.
Substitution Suggested by the Equation An equation of the form Mdx
+ N dy =
0
may not yield at once (or at all) to the methods of Chapter 2. Even then the usefulness of those methods is not exhausted. It may be possible by some change of variables to transform the equation into a type that we know how to solve. A natural source of suggestions for useful transformations is the differential equation itself. If a particular function of one or both variables. tands out in the equation, it is worthwhile to examine the equation after that function has been introduced as a new variable. For instance, in the equation (x the combination (x
+ 2y 
1) dx
+ 3(x + 2y) dy =
0
+ 2y) occurs twice and thus attracts attention.
(1)
Hence we put
x +2y = v, and because no other function of x and y stands out, we retain either x or y for the other variable. The solution is completed in Example 5.7. In the equation
(1 +3xsiny)dx  x 2 cosydy=O,
(2)
the presence of both sin y and its differential cos y dy, and the fact that y appears in the equation in no other manner, leads us to put sin y = wand to obtain the differential equation in wand x. See Example 5.8.
EXAMPLE S.7 Solve the equation (x
+ 2y 
1) dx
+ 3(x + 2y) dy =
As suggested above, put x +2y = v. Then dx
= dv
 2dy
O.
(1)
5.3
Substitution Suggested by the Equation
85
and equation (1) becomes
(v l)(dv  2dy)
+ 3vdy =
0,
or
(vl)dv + (v+2)dy = 0. Now the variables can be separated. From the equation in the form v I   dv + dy v+2
= 0,
we get
(1_3_) v+2
dv
+ dy
=
0
and then
v  3 In Iv
But v = x
+ 21+ y + c =
O.
+ 2y, so our final result is x
+ 3y + c =
31n
Ix + 2y + 21 .
•
EXAMPLE 5.8 Solve the equation (1
Put sin y
= w.
+ 3x sin y) dx
Then cos y dy
= dw
 x 2 cos y dy = O. and (2) becomes
(\ + 3xw) dx 
x 2 dw
= 0,
an equation linear in w. From the standard form
3 dx dw  wdx = x x2 an integrating factor is seen to be
Application of the integrating factor yields the exact equation
x  3 dw  3x  4 w dx = x  5 dx,
(2)
86
Chapter 5
Additional Topics on Equations of Order One
for either x >
°
or x < 0, from which we get
or
4xw =
CX
4
1.

Hence (2) has the solution set
4x sin y
5.4
= cx 4 
1.
•
Bernoulli's Equation A wellknown equation that fits into the category of Section 5.3 is Bernoulli 's equation, y'
+ P(x)y = Q(X)Y".
(I)
If n = 1 in (1), the variables are separable, so we concentrate on the case n Equation (J) may be put in the form
i=
I.
(2) But the differential of y  II+ 1 is (1  n)y  II dy, so equation (2) may be simplified by putting
from which (I  n)y  II dy
Thus the equation in
= d z.
z and x is dz
+ (1 
n)p z dx
= (ln)Qdx ,
a linear equation in standard form. Hence any Bernoulli equation can be solved with the aid of the foregoing change of variable (unless n = 1, when no substitution is needed) . EXAMPLE 5.9 Solve the equation y(6/  x  I) dx
+ 2x dy =
O.
(3)
5.4
Bernoulli's Equation
87
First let us group the terms according to powers of y, writing
2x dy  y(x
+ I) dx + 6i dx
= O.
Now it can be seen that the equation is a Bernoulli equation, since it involves only terms containing, respectively, dy, y, and y" (n = 3 here). Therefore, we divide throughout by y3, obtaining
2xy 3 dy  y 2(x
+ 1) dx = 
6dx .
This equation is linear in y 2 , so we put y2 = v , obtain dv = _2y3 dy, and need to solve the equation
xdv
+ vex + 1)dx =
6dx,
or
dv
+ v(l + xI) dx
= 6x  1 dx.
(4)
Since exp (x
+ In Ix!) =
Ixle x
is an integrating factor for (4), the equation
xex dv
+ veX (x + l)dx =
6e x dx
is exact. Its solution set
together with v = y  2, leads us to the final result,
i(6 + ce  X) = x.
•
EXAMPLE 5.10 Solve the equation 6i dx  x(2x 3 + y) dy = O.
(5)
This is a Bernoulli equation with x as the dependent variable, so it can be treated in the manner used in Example 5.9. That method of attack is left for the exercises. Equation (5) can equally well be treated as follows. Note that if each member of (5) is multiplied by x 2 , the equation becomes (6)
88
Chapter 5
Additional Topics on Equations a/Order One
In (6), the variable x appears only in the combinations x 3 and its differential 3x 2 dx. Hence a reasonable choice of a new variable is w = x 3 . The equation in wand y is 2/dw  w(2w
+ y)ely =
0,
an equation with coefficients homogeneous of degree two in y and w. The further change of variable w = zy leads to the equation 2ydz  z (2 z  1) ely = 0, 4elz 2elz 2z  1  z.
ely _ 0

Y .
Therefore, we have
21n 12z  II  2ln Izl  In Iyl = In lei, or (2 z  1)2 = cYZ2.
But z = w/y = x 3 /y, so the solutions we seek are determined by (2x 3
_
y)2 = eyx6.
•
• Exercises In Exercises I th rough 20, solve the equation.
3.
+ 1) dx + (3x  2y + 3) ely = O. sin y(x + sin y) dx + 2x2 cos y dy = O. elyjdx = (9x + 4y + 1)2. 5. dy/dx
4.
y'=yxle 2x .
7. 8.
(3 sin y  5x) dx
9.
1.
2.
(3x  2y
6.
+ 2x2 cot y ely = o.
y' = 1 +6xexp(x  y).
11.
dv/du = (u  v)2  2(£1  v)  2. 2ydx+x(x 2 Iny  l)dy =0. (ke 2v  u) dL! = 2e 2v (e 2v + ku) dv.
12.
y' tan x sin 2y
13.
(x
14 .
y(x tan x
10.
+ 2y 
= sin 2 x + cos2 y . 1) dx  (x + 2y  5) dy
+ In y) dx + tan x dy
= O.
= O.
= sin (x
+ y) .
xydx+(x 2 3y)dy=0.
5.5
15.
Coefficients Linear ;11 the Two Variables
xy'  y = xkyll, where n =F 1 and k 2
16.
(3 tan x  2 cos y) sec x d x
17.
(x
+ 2y 
l)dx
+n
=F l.
+ tan x sin y d y
+ (2x + 4y 
89
= O.
3) dy = O. Solve by two methods.
18. Solve the equation 6idx x (2x 3 + y) dy = 0 of Example 5.10 above by treating it as a Bernoulli equation in the dependent variable x . 19 . 2x 3 y' = y(y2 + 3x 2) . Solve by two methods. 20.
cos y si n 2x d x
+ (cos2 Y 
cos2 x) dy = O.
21 . Solve the equation of Exercise 15 for the values of k and n not included there. In Exercises 22 through 27, find the particul ar sol ution requ ired.
1, y
= O.
4 (3x
23 .
y'
24.
2xyy'
25.
(y4  2xy) dx
26.
(2i when x = 1, Y = I. Solve by two methods. (x 2 +6y 2) dx  4xy dy = 0; when x = I, y = 1. Solve by three methods.
27.
5.5
+ y  2)dx  (3x + y)dy = 0; when x = = 2(3x + y)2  1; when x = 0 , y = 1.
22.
= y2 
2x 3 . Find the solution that passes through the point (1, 2).
+ 3x 2 dy = 0; x 3 ) dx + 3xi ely = 0;
when x
= 2, y =
I.
Coefficients Linear in the Two Variables Consider the equation (1)
in which the a's, b's, and c's are constants. We know already how to solve the special case in which C I = 0 and C2 = 0 because then the coefficients in (1) are each homogeneous and of degree one in x and y. It is reasonable, therefore, to attempt to reduce equation (1) to that situatio n. In connection with (1) consider the lines
+ b1y + C I = 0, a2x + b2 y + C2 = o. a lx
(2)
They may be parallel or they may intersect. There will not be two lines if al and b l are zero or if a 2 and b 2 are zero, but equation (1) will then be linear in one of its variables. If the Jines (2) intersect, let the point of intersection be (h, k). Then the translation (3)
90
Chapter 5
Additional Topics on Equations of Order One
will change the equations (2) into equations of lines through the origin of the uv coordinate system, namely,
a lu +b lv = 0, a2U +b2 v = O.
(4)
Therefore, since dx = du and dy = dv, the change of variables x y
= u +h, = v +k,
where (h, k) is the point of intersection of the lines (2), will transform the differential equation (1) into
(5) an equation that we know how to solve. If the lines (2) do not intersect, a constant k exists such that
so that equation (1) appears in the form
(6) The recurrence of the expression (aJx + b l y) in (6) suggests the introduction of a new variable w = alx + b J y . Then the new equation, in wand x or in wand y, is one with variables separable, since its coefficients contain only w and constants.
EXAMPLE 5.11 Solve the equation (x
+ 2y 
4) dx  (2x
+y 
The lines
x +2y 4 = 0 and
2x+y5=0 intersect at the point (2, I), Hence put x = u +2,
y=v+l.
5) dy = O.
(7)
5.5
Coefficients Linear in the Two Variables
91
Then equation (7) becomes
+ 2v) du
(u
 (2u
+ v) dv = 0,
(8)
which has coefficients homogeneous and of degree one in u and v. Therefore, let u = vz, which transfonns (8) into (Z + 2)(zdv
+ vdz) 
(2z
+ 1)dv =
0,
or
1) dv
(Z2 
Separation of the variables
+ v(z + 2) dz =
O.
v and z leads us to the equation
dv
+ (z + 2) dz
= O.
Z2  1
V
With the aid of partiaJ fractions, we can write the equation above in the form 2dv
3dz
dz
 +     =0. v zl z+1 Hence we get
21n Ivl + 31n Iz 
11  In Iz + 11
= In lei,
from which it follows that ~2(Z  1)3 = e(z
+ 1),
or (vz  v)3
= c(vz + v) .
Now vz = u, so a set of solutions appears as (u  v)3
= c(u + v).
But u = x  2 and v = y  I. Therefore, the desired result in terms of x and y is (x  y  1)3 = c(x
+y 
3).
For other methods of solution of equation (7), see Exercises 22 and 29 below.
•
EXAMPLE 5.]2 Solve the equation (2x
+ 3y 
1) dx
+ (2x + 3y + 2) dy
with the condition that y = 3 when x = 1.
= 0,
(9)
92
Chapter 5 Additional Topics on Equations of Order One
The lines
2x
+ 3y 
2x
+ 3y + 2 =
I= 0
and
0
are parallel. Therefore, we proceed, as we should have upon first glancing at the equation, to put
2x+3y=v . Then 2 dx = dv  3 d y, and equation (9) is transformed into
(v  I)(dv  3dy)
+ 2(v + 2)dy =
0,
or
(v  I) dv  (v  7) dy = O.
(10)
Equation (10) is easi ly solved, leading u ' to the relation
vy+c+6Inlv  71=0. Therefore, a so lution set of (9) is
2x
+ 2y + c =
 61n 12x
+ 3y 
7 1.
Buty = 3whenx = 1,soc =  8  6In4. Hence the particular solution required is given by
x
+y 
4 = 31n [i(2x
+ 3y 
• Exercises In Exercises 1 through 17. solve the eq uations .
1.
(y  2) dx  (x  y  1) dy = 0,
2.
(x4y9)dx+(4x+y 2)dy=0.
3. 4.
(2x  y) dx + (4x + y  6) dy = O. (x  4y  3) dx  (x  6y  5) dy = O.
5. 6. 7. 8.
(x+y  l)dx +(2x +2y+ l)dy=0 . (x  2)dx +4(x + y  l)dy = O. (x3y+2)dx+3(x+3y4)dy=0. (6x  3y +2)dx  (2x  y  I)dy = O.
7)].
•
5.5
9. 10.
11. 12. 13. 14.
15.
Coefficients Linear in the Two Variables
93
+ 4) dx  (2x  y + 1) dy = O. 4) dx + (x + 4y  5) dy = O. 1) dx  (2x + y  5) dy = O. (x  1) dx  (3x  2y  5) dy = O. (3x + 2y + 7) dx + (2x  y) dy = O. Solve by two methods. (2x + 3y  5) dx + (3x  y  2) dy = O. SoJve by two methods. 2 dx + (2x  y + 3) dy = O. Use a change of variables. (9x  4y
+ 3y (x + 2y (x
16.
Sol ve the equation of Exercise IS by using the fact that the equation is linear in x.
17.
(x  y
+ 2) dx + 3 dy =
O. Solve by two methods.
In Exercises 18 through 21 , obtain the particular solution indicated.
18.
(2x  3y + 4)dx
+ 3(x 
J)dy = 0; when x = 3, y = 2.
19. Solve the equation of Exercise 18 but with the condition that when x =  1, y = 2. 20. (x + y  4) dx  (3x  y  4) d y = 0; when x = 4, y = 1. 21. Solve the equation of Exercise 20 but with the condition that when x = 3, y =7. 22. Prove that the change of variables x = a,u
+ a 2V,
Y= u
+v
will transform the equation (a,x
+ b ,y + c ,) dx + (a2x + b2y + C2 ) dy =
0
(A)
into an equation in which the variables u and v are separable if a , and a2 are roots of the equation (B)
and if a2 # al . Note that this method of solution of (A) is not practical for us unless the roots of equation (B) are real and distinct. Solve Exercises 23 through 28 by the method indicated in Exercise 22.
23.
Do Exercise 4. As a check, the equation (B) for this case is
a2 so we may choose a) be

Sa
+6=
0,
= 2 and a2 = 3. The equation in u and v turT)S out to (v  1) du  2(u
+ 2) dv =
O.
94
Chapter 5
Additional Topics on Equations of Order One
24.
Do Exercise 3.
26.
Do Exercise 11 .
25.
Do Exercise 5.
27.
Do Exercise 12.
28.
Do Example 5.11 in this section.
29.
Prove that the change of variables
x = a]u + f3v.
y
= u+ v
will transform the equation
into an equation that is linear in the variable u if a] is a root of the equation (D) and if f3 is any number such that f3 =f:. al. Note that this method is not practical for us unless the roots of equation (D) are real; however, they need not be distinct as they had to be in the theorem of Exercise 22. The method of this exercise is particularly useful when the roots of (D) are equal. Solve Exercises 30 through 34 by the method indicated in Exercise 29.
5.6
f3 may be chosen to
30.
Do Exercise 10. The only possible a] is ( 2). Then be anything else.
31.
Do Exercise 6.
32.
Do Exercise 9.
33.
Do Exercise 12.
34.
Do Exercise 4. As seen in Exercise 23, the roots of the "a equation" are 2 and 3. If you choose a I = 2, for example, then you make f3 anything except 2. Of course, if you choose a, = 2 and f3 = 3. then you are reverting to the method of Exercise 22.
Solutions Involving Nonelementary Integrals In solving differential equations, we frequently are confronted with the need for integrating an expression that is not the differential of any elementary function. I I By an elementary function we mean a function studied in thc ordinary beginning calculus course. For example, polynomials, exponentials, logarithm. , trigonometric, and inverse trigonometric functions are elementary. AJI functions obtained from them by a fin ite number of applications of the elementary operations of addition, subtraction , multiplication, division, extraction of root " and raising to powers are elementary. Finally, we include such functions as sin sin x , in which the argument in a function previously classed as elementary is replaced by an elementary function.
5.6
Solutions Involving Nonelementary Integrals
95
Following is a short list of nonelementary integrals:
f f f
f f f
exp (_x 2 ) dx 2
sinx dx 2
cosx dx
x
e: dx sinx dx x
cosx dx x
f f f
x tanxdx dx Inx dx ~
Integrals involving the square root of a polynomial of degree greater than two are, in general, nonelementary. In special instances they may degenerate into elementary integrals. The following example presents two ways of dealing with problems in which nonelementary integrals arise.
EXAMPLE 5.13 Solve the equation y'  2xy = 1 with the initial condition that when x = 0, y = 1. The equation being linear in y, we write
dy  2xydx
= dx,
obtain the integrating factor exp( x 2 ), and prepare to solve exp(x2)dy  2xyexp(x2)dx = exp(x 2)dx.
(1)
The left member is, of course, the differential of y exp( _x 2 ) . But the right member is not the differential of any elementary function; that is, exp(  x 2 )dx is a nonelementary integral. Let us turn to power series for help. From the series
J
exp(x
2
00 (_1)lIx211 )=L ' 11 =0 n!
obtained in calculus, it follows that
f
2
exp(x )dx
00 (_1)lI x 211+ 1
= c+L
11=0 n!(2n
+ 1)
Thus the differential equation (I) has the general solution ?
y exp (x)
00 (_I)II X211 + 1
=c+ L
11=0 n!(2n
+ 1)
.
.
96
Chapter 5
Additional Topics on Equations of Order One
Since y
= 1 when x = 0, c may be found from 1 = c + O.
Therefore, the particular solution desired is 00
+L
y exp (  x 2 ) = I
(_I)II X211+1
11 =0 n!(2n
+ J)
.
(2)
An alternative procedure is the introduction of a definite integral. In calculus, the error function defined by eff x =
2 ,In
l
0
x
(3)
exp (_fJ2) dfJ
is sometimes studied. Since, from (3), d 2 2  erf x = exp ( x ), dx ,In
we may integrate the exact equation (1) as follows: y exp (_x 2 ) = ~,Jn erf x Since erf 0 = 0, the condition that y alternative to (2) we obtain
=
1 when x
+ c.
= 0 yields c =
(4)
1. Hence, as an
y exp ( _ x 2 ) = 1 + !,In elf x.
(5)
Equation (5) means the same as yexp(x 2 ) = 1
+ foX exp( 
fJ2)dfJ.
(6)
WIiting a sol ution in the form of (6) implies that the definite integral is to be evaluated by power se ries, approximate integration such as Simpson's rule, mechanical quadrature, or any other available tool. If it happens, as in this case, that the definite integral is itself a tabulated function, that is a great convenience, but it is not vital. The essential thing is to reduce the solution to a computable form.
•
• Exercises In each exercise, express the solution with the aid of power series or definite integrals.
1.
y' = y[1  exp ( _ x 2 »).
2.
(xy  sinx)dx
+x 2 dy =
O.
5.6
3.
y' = 1 4x 3y.
4.
(y cos 2 X
5.
(1
6.
[x exp
7.
x(2y

X
si n x) dx
+ xy) dx 
x dy
(~:) 
+ x) dx
+ sinx cosx dy =
= 0;
Y] dx
 dy =
Solutions Involving NOllelemelltary Integrals
97
O.
when x = 1, y = O.
+ x dy = 0; when x = 1, y = 2. 0; when x = 0, y = I .
• Miscellaneous Exercises In each exercise, find a set of solutions unless the statement of the exerc ise stipulates otherwise.
+ (2y 
= O.
J.
(y 2  3y  x) dx
2. 4.
+ y + l )dx +x(x  3y2  l )dy = (x + 3y  5) dx  (x  y  J) dy = O. (x 5  y2)dx + 2xydy = O.
5.
(2x+y4)dx+(x  3y+ 12)dy=0.
6.
y3 sec2 x dx  (I  2y2 tanx) dy = O. x 3 y dx + (3x 4  y3) dy = O.
3.
7.
3) dy
(y3
+ 7) dx + (x + 2y + 1) dy + (y4  3x 2) dy = O.
8.
(x  4y
9.
xy dx
O.
= O.
12.
+ 2y  1) elx  (2x + y  5) ely = (5x + 3e Y ) dx + 2xe Y ely = O. (3x + y  2) dx + (3x + y + 4) ely =
13.
(x  3y +4)elx +2(x  y  2) ely = O.
14.
(x 2) dx +4(x + y  1) ely = O.
16.
2(xy) elx+(3x y l ) ely = O.
15 .
y dx
= x(l + xy4) ely = O.
17.
ydx
10. 11.
(x
O. O.
18 . ely/dx = tanycotx  secycosx.
21. 22.
+ 3y  7) elx + (4x + 3y + 1) dy = O. (x + 4y + 3) dx  (2x  y  3) dy = O. (3x  3y  2)dx  (x  y + J)dy = O. (x  6y + 2) dx + 2(x + 2y + 2) ely = O.
23 .
(x  y  I ) dx  2(y  2) ely = O.
24.
(x  3y
19. 20.
(4x
+ 3) elx + (3x + y + 9) dy
= O.
25 .
(2x +4y  l )dx  (x +2y  3)dy = O.
26.
y(x 1)elx  (x 2  2x  2y)dy = O.
27.
+ 2y) elx + 2(x 4 dx + (x  y + 2)2 dy = O.
28 .
(6xy  3y2
y) ely = O.
+ x(x 2 y 
1) dy
= O.
98
Chapter 5
Additional Topics on Equations of Order One
29.
Solve in two ways the equation y' = ax
30 .
(alx
31. 32. 33. 34. 35. 36.
+ by + c;
+ ky + CI) dx + (kx + b2Y + C2) dy =
with b =f:. O.
O.
2
2xdv+v(2+v x )dx=0; whenx=l,v=!. (2x  5y
+ 12)dx + (7x
 4y
+ 15 )dy =
O.
[1 + (x + y)2 ]dx + [1 +x(x + y) ]dy = O. (x  2y  1) dx  (x  3) dy = O. Solve by two methods.
+ 1) dx
+ 2y 
4) dy = O. Solve by two methods. Find a change of variables that will reduce any equation of the form (2x  3y
 (3x
xy' = yf(xy)
37. 38.
to an eq uation in which the variables are separable. (x 4  4x 2 y2 y4) dx + 4x 3 y dy = 0; when x = J, y = 2. 4ydx + 3(2x  l) (dy + y 4 dx) = 0; when x = 1, y = 1.
39. y' = x  y + 2. Solve by two methods. 40.
5 .7
(x
+y 
2) dx  (x  4y  2) dy = O.
Computer Supplement The techniques developed in Chapter 5 for dealing with special cases are unnecessary when using a Computer Algebra System . Most of the equations in the chapter can be so lved directly using the elementary techniques in Section 2.7. Even if the solutions involve "nonelementary integrals," a CAS will often find a solution. For example, we can solve the equation y'  2xy = 1
given in the examp le in Section 5.6 by the si ngle Maple command >dsolve({diff(y(x),x)  2*x*y=l ,y(O)=1},y(x))
i
x2
y(x) =
e .fieli(x) 2
+e
x2
.
It is easy to see that this is the same as equation (5) of Section 5.6 .
• Exercises I.
Solve a selection of exercises fro m the chapter using a Computer Algebra System.
Linear Differential Equations
6.1
6
The General Linear Equation The general linear differential equation of order n is an equation that can be written d"y d"Iy dy bo(x) + b l (X)   I + ... + bl/ _ I (x) + bJ/(x)y = R(x). (1) dx" dx" dx If the value of the function R (x) is zero for all x, then the equation is calJed a homogeneous l linear differential equation . If the coefficient functions bo, ... , btl and the function R are continuous on an interval/and bo(x) is never zero on /, then the equation (1) is said to be normal on I. For example, the equation dy (x  1) + y sinx dx is a firstorder linear, nonhomogeneous, and normal equation on any interval that does not contain x = 1. On the other hand ,
=
d2y 3 dx 2
+ xy =
°
is a secondorder linear, homogeneous, and normal equation on any interval. We now prove that if YI and Y2 are so lutions of the homogeneous equation bo(x)y 1. We do not prove Theorem 6.2 in this book, but in Chapter 13 prove an existence and uniqueness theorem for firstorder equations in general.
EXAMPLE 6.1 Find the unique solution of the initial value problem
y"
+y =
0,
yeO) = 0,
y' (0)
= 1.
(2)
We observe that sin x and cos x are solutions of the differential equation in (2), so that for arbitrary c, and C2,
y = c , sinx
+ C2 cosx
is also a solution by the theorem of Section 6.l. Because of the initial co nditions in (2), we must choose c, and C2 so that sin 0 + C2 cos 0 = 0 and cos 0  C2 sin 0 = 1. This can be done in only one way, namely by choosi ng = 1 and C2 = 0. We find that the function sin x is a sol ution of the initial value problem (2). Moreover, since the problem satisfies the conditions req uired in Theorem 6.2, for any interval that co ntain s x = 0, sin x is the only so lution of the problem given in (2) .
c,
c, c,
EXAMPLE 6.2 Consider the ini tial value problem x 2 y"
+ 2xy' 
12y = 0,
y(l) = 4,
y'( I) = 5.
(3)
The differential equation is normal on either x > 0 or x < O. Since the initial conditions are stated for Xo = 1, we let I be the interval x > O. It is a simple matter to show that x 3 and x  4 are solutions of the differential equation in (3), so that for arbitrary c , and C2,
y = c,x 3
+ C2 X  4
is also a sol ution. The two initial conditions now require that
c, + C2 =
4
and
It follows that c, = 3 and C2 = 1 and therefore the function
y = 3x 3 + x 4 satisfies the in itial value problem for x > O.
102
Chapler 6
Linear Differential Equations
Now by Theorem 6.2 we can assert that the solution we have found is the only solution valid for x > 0.
•
• Exercises In Exercises 1 through 4, determi ne all intervals on which the equation is normal.
1.
2.
(x  l)y" + xl + y = sinx.  l) y" + 6y = eX.
(x 2
3.
x2ylfl+eXy=lnx.
4.
(cotX)Y"'
+y =
0.
In Exercises 5 through 8, determine the unique so lution of the initial value problem following the examples of this section.
5. y"  y = 0, yeO) = 4, leO) = 2. Use the fact that eX and e x are solutions of the differential equation. 6. y" + 4y = 0, yeO) = 2, y'(O) = 4. Use the fact that sin 2x and cos 2x are solutions of the differential equation. 7. y"  2y' + y = 0, yeO) = 7, y'(o) = 4. Use the fact that eX and xe x are solutions of the differential equation. 8. x 2y" + xy'  9y = 0, y(l) =  1, y'(l) = 15. Use the fact that x 3 and x  3 are solutions of the differential equation. 9. Establish the following important corollary to Theorem 6.2. If the differential equation is normal and homogeneous on I and Yo = YI = ... = YII_I = 0, then y = is the only solution.
°
6 .3
linear Independence Given the functions fl' such that
12, ... , '/;"
cdl (x)
if constants Cl,
C2,'"
+ c2h(x) + ... + cnf,,(x)
I
Cn,
= 0
not all zero, ex ist (1)
for all x in some interval a ~ x ~ h, then the functions fl' 12, ... , in are said to be linearly dependent on that interval. If no such relation exists, the functions are said to be linearly independent. That is, the functions fl' .h ... , f" are linearly independent on an interval when e uation (1) implies that
id ,l tl (0 he
I "II~tly f>/fhti('}l{ C I = C2 = ... = Cn = 0.
It should be clear that if the functions of a set are linearly dependent, at least one of them is a linear combination of the others; if they are linearly independent, none of them is a linear combination of the others.
6.4
6.4
The Wronskian
103
The Wronskian With the definitions of Section 6.3 in mind, we shall now obtain a sufficient condition that n functions be linearly independent on an interval a ~ x ~ b. Let us assume that each of the functions II, }2, . .. , J" is differentiable at least (n  1) times in the interval a ~ x ~ b. Then from the equation (I)
it follows by successive differentiation that
cd; + c2I; + ... + CTI/'; = 0, +" = 0 ,. C I j ." I + (.'2 f" 2 + ... + CTI J" CI
I I(II 
I)
+ (.'2 I(n  I) + .. . + G" j 'II(II 2
0
I) .
For any fixed value of x in the interval a ~ x ~ b, the nature of the solutions of these n linear equations in GI, (.'2, ... , G" will be determined by the value of the determinant
II (x) I(x)
/" (x) /';(x)
hex) I;(x)
W(x) =
(2) I?,I)(x)
Ii"  I)(X)
fY,  I)(x)
Indeed, if W(xo) ::f. 0 for some Xo on the interval a ~ x ~ b, it follows that CI = C2 = ... = C" = 0, and hence the funct ions II, ... , j;, are linearly independent on a ~ x ~ b. The function W(x) defined by equation (2) is called the Wronskian 2 of the n functions, II, ... , In. We have shown that if at one point on the interval the Wronskian is not zero, the functions are linearly independent on that interval. The converse of this statement is not true, as is exhibited in Exercise LO. If the n function involved are solutions of a homogeneous linear differential equation, the situation is simplified as is shown by Theorem 6.3. A proof of this theorem in the case n = 2 is suggested in Exercises 11 through 14.
Theorem 6.3 If on the interval a ~ x ~ b, bo(x) ::f. 0, bo, b l ,
.. . ,
b ll are continuous, and
YI, Y2, ... , Yn are solutions of the equation
boY (n)
+ b IY (n  I ) + . .. + b fIIY 1+ b flY
 0 ,
(3)
2 The Wronsk ian determinant is named after the Poli sh mathematician Hoene Wronski ( 1778  1853).
104
Chapter 6
Linear Differential Equatio/1.\'
then a necessary and sufficient condition that YI, ' .. , Yn be linearly independent is that the Wronskian of YI , ... , YII differ from zero at at least one point on the interval a ::::: x ::::: b.
EXAMPLE 6.3
Consider the set of functions cos ax, sin ax, sin (ax + b). The functions are linearly dependent on any interval sincesin (ax + b)cosbsin ax  sinbcos ax = 0 for all x. If we compute the Wronskian for this set of functions, we find that W (x) = 0 for all x . In itself this is not enough to determine the linear dependence of our set of functions. However, if we note that each of these functions is a solution of the differential equation
y''' + a 2 y' = 0, then Theorem 6.3 applies and the fact that W(x) = 0 for all x guarantees that the functions are linearly dependent on any interval.
•
EXAMPLE 6.4 One of the bestknown sets of n linearly independent functions of x is the set I, x, x 2 , ... ,X"  I. The linear independence of the powers of x follows at once from the fact that if CI, C2 , . . . ' CII are not all zero, the equation
can have, at most, (n  1) distinct roots and so the polynomial cannot vanish identically in any interval. See also Exercise J below.
•
• Exercises I.
Obtain the Wronskian of the functions
1, x, x 2 ,
... , X" 
I
forn > 1.
2.
Show that the functions eX, e 2x , e 3x are linearly independent.
3.
Show that the functions eX, cos x, sin x are linearly independent.
4.
By determining the constants C I , C2, C3 , C4, which are not all zero and are such that C I fl + c2h + c3iJ + c4f4 = 0 identically, show that the functions
f·
. I =X ,
are linearly dependent.
6.4
The Wronskian
105
5.
Show that cos (wt  (3), cos wt , sin wt are linearly dependent functions of t.
6. 7.
Show that 1, sin x, cos x are linearly independent.
8.
Show that if f and f' are continuous on a S x S band f (x) is not zero for all x on a S x S b, then f and xf are linearly independent on a S x S b.
9.
Show that if f, f', and f" are continuous on a S x S band i(x) is not zero for all x on a S x S b, then i, xi, and x 2 f are linearly independent on a S x S b. Let fl (x) = 1 + x 3 for x SO, fl (x) = 1 for x ~ 0;
10.
Show that 1, sin 2 x, cos 2 x are linearly dependent.
hex) = 1 for x sO, hex) = 1 + x 3 for x ~ 0;
f3(X) = 3 + x 3 for all x. Show that (a) i, 1', f" are continuous for all x for each of ii, 12, I,; (b) the Wronskian of fl ' 12, h is zero for all x; (c) II, .h h are linearly independent over the interval  I S x S I. In part (c) you must show that if Ctfl(X) + c2h(x) + c3h(x) = 0 for all x in  I S x S 1, then CI = C2 = C3 = O. Use x = ], 0, 1 successively to obtain three equations to solve for C I , C2, and C3.
11.
Given any interval a S x S b with Xo a fixed number in the interval and suppose y is a solution of the homogeneous equation y"
+ Py' + Qy = o.
(A)
Further, suppose that y(xo ) = y' (xo) = 0. Then use the existence and uniqueness theorem of Section 6.2 to prove that y(x) = 0 for every x in the interval a S x S b. 12.
Suppose that YI and Y2 are solutions of equation (A) of Exercise 11 and suppose the Wronskian of )I I and Y2 is identically zero on a S x S b. Show that for Xo in the interval a S x S b, there must exist constants c] and C2 not both zero such that
and
CI Y; (xo) 13.
+ C2Y~ (xo) =
O.
Consider the function defined by
where CI and C2 are the constants determined in Exercise 12. Show that this function is a solution of eq uation (A) above and that it fo ll ows that y(x) == 0 on a S x S b. (Use the results of Exercise 11.)
106
Chapter 6
Linear D(fferential Equations
14.
Combine the results of Exercises 11 through 13 to obtain a proof of the necessity condition of Theorem 6.3 in the case when n = 2. Also note that the sufficiency condition was established in the text of this section.
I 6 . 5 I General Solution of a Homogeneous Equation One of the basic results of the subject of linear differential equations is contained in Theorem 6.4. Theorem 6.4 Let {YI, Y2, ... , Ynl bea linearly independent set ofsolutions of the homogeneous linear equation
for x on the interval a :5 x :5 b. Suppose further that the equation is normal on a:::: x:::: h. If ¢ is any solution of equation (1), valid on a :::: x :::: b, there exist constants Cl, C2, . . . , clI such that
¢ = CIYI
+ C2Y2 + ... + C"YIl '
(2)
It is because of this theorem that we define the general solution of equation (L) to be
Y
= CIYI + C2Y2 + ... + Cn YI1,
(3)
where Cl, C2, ... , CII are arbitrary constants. In a sense each particular solution of the linear equation (1) is a special case (some choice of the c's) of the general solution (3). The basic ideas needed for a proof of this important theorem are exhibited here for an equation of order two. No additional complications occur for equations of higher order. Proof.
Consider the equation bo(x)y"
+ hi (x)y' + b2 (x)y = O.
(4)
Let Yl and Y2 be linearly independent solutions of equation (4) on the interval a :::: x :::: h. By Theorem 6.3, there exists a number Xo in the interval such that
w = I Y~ (xo) YI(xO)
Y;(x o) Yz (xo)
I # O.
It follows that the system of equations
+ C2Y2(XO) = ¢(xo), CIY;(XO) + C2 Y~(XO) = ¢'(xo), CIYI(XO)
(5)
6.6
has a unique solution CI
General Solution of a Nonhomogeneous Equation
= CI, C2 = C2.
107
That is,
+ C2Y2 (xo) = ¢ (xo), CI Y; (xo) + C2Y~ (xo) = ¢ I (xo).
CI YI (xo)
Now consider the function
(6) Because f is a Jinear combination of two solutions of equation (4) on the interval a S x S b, it is also a solution on that interval. Moreover, f(xo) = CIYI(XO)
fl (xo) = CI Y; (xo)
+ C2Y2(XO) , + C2Y~ (xo),
so that f (xo) = ¢ (xo) and f' (xo) = ¢ I (xo). It follows from the uniqueness theorem of Section 13.2 that f and ¢ are the same solution. That is,
¢ = CIYI
+ C2Y2,
which completes the proof of the theorem .
•
It is necessary to keep in mind that the discussion above used the fact that # 0 on the interval a S x S b. It is easy to see that the linear equation
bo(x)
xy'  2y
=0
has the general solution Y = cx 2 and also such particular solutions as Os x, x < O.
The solution YI is not a special case of the general solution. But in any interval throughout which bo(x) = x # 0, this particular solution is a special case of the general solution. It was, of course, made up by piecing together at x = 0 two parts, each drawn from the general solution.
6 .6
General Solution of a Nonhomogeneous Equation Let Yp be any particular solution (not necessarily involving any arbitrary constants) of the equation boy(n ) + b l /
n

I
)
+ ... + bn I y' + bny =
R(x)
(1)
and let Yc be a solution of the corresponding homogeneous equation bo/")
+ bl/n  I) + ... + b,, _ ly + blly = l
O.
(2)
108
Chapter 6
Linear Differential Equations
Then Y = Yc + yp
(3)
is a solution of equation ( 1). For, using the y of equation (3), we see that bOy(ll ) + ...
+ b,.,y = (bOY~") + ... + bnyc) + (bOY~') + ... + b"YfJ) = 0 + R(x) =
R(x).
If y" Y2, ... , y" are linearly independent solutions of equation (2), then Yc = C'YI
+ C2Y2 + ... + c"y",
(4)
in which the c's are arbitrary constants, is the general so lution of equation (2). The right member of equation (4) is called the complementary function for equation (1).
The general solution of the nonhomogeneous equation (I) is the sum of the complementary function and any particu lar solution. To justify this usage of the term general solution, we must show that if f is any solution of equation (1), then f == Yc + y" for some particular choice of the c" ... , c" . We note that since f and yp are both solutions of the nonhomogeneous equation (1), f  Yp is a solution of the homogeneous equation (2). Hence by Theorem 6.4 of Section 6.5,
f 
YfJ
== CIY' + C2Y2 + ... + CIIY"
for some particular choice of the show.
CI, ... , C".
This establishes what we wished to
y" = 4 .
(5)
EXAMPLE 6.5 Find the general sol ution of
We first observe that the functions 1 and x are linearly independent on any interval and are solutions of the homogeneous equation y" = O. Hence the complementary function for equation (5) is Yc = CI
+ C2 X .
On the other hand, the function 2x2 is a particular solution of equation (5). Hence the general solution of equation (5) is Y = c,
+ C2X + 2x 2.
•
EXAMPLE 6.6 Find the general solution of the equation y"  Y =4.
(6)
6.7
Differential Operators
109
It is easily seen that y = 4 is a solution of equation (6). Therefore, the y p in equation (3) may be taken to be ( 4). As we shall see later, the homogeneous equation
y"  y
=0
has as its general solution Thus the complementary function for equation (6) is C I eX + C2ex and a particular solution of (6) is y p =  4. Hence the general solution of equation (6) is y in which
I
6 .7
I
CI
= clex + C2ex 
4,
and C2 are arbitrary constants.
•
Differential Operators Let D denote differentiation with respect to x, D2 differentiation twice with respect to x , and so on; that is, for positive integral k ,
k D y
=
dk y  k ' dx
The expression
A = aoD" + aID" 
1
+ ... + all _ 1D + all
(1)
is called a differential operator of order n. It may be defined as that operator which, when applied to any function 3 y , yields the result
d" y d"  1y dy Ay =ao   +al   ] + ··· + all _ I  + a" y . dx" dx" dx
(2)
The coefficients ao, a i, ... , an in the operator A may be functions of x , but in this book most operators used will be those with constant coefficients. Two operators A and B are said to be equal if, and only if, the same result is produced when each acts upon the function y . That is, A = B if, and only if, Ay = By for all functions y possessing the derivatives necessary for the operations involved. The product AB of two operators A and B is defined as that operator wh ich produces the same result as is obtained by using the operator B followed by the operator A. Thus ABy = A(By ). The product of two differentia:! operators always exists and is a differential operator. For operators with constant 3 The function y is assumed to possess as many de rivati ves as may be required in whatever operations take place.
110
Chapler 6
Linear Differelltial Equations
coefficients, but not usually for those with variable coefficients, it is true that AB = BA. EXAMPLE 6.7 Let A = D + 2 and B = 3 D  1. Then dy By = (3 D  1) y = 3   Y dx
and A(8y) = (D
+ 2) (3;~
d2 y
= 3 dx 2 
dy dx
d 2y = 3 dx 2

y)
dy
+ 6 dx
 2y
dy
+ 5 dx  2y = (3D 2 + SD  2)y. I) = 3D 2 + SD  2.
Hence AB = (D + 2)(3D Now consider BA. Acting upon y, the operator BA yields B(Ay) = (3D  1)
(;~ + 2Y )
d2y = 3 dx 2
+ 6 dx
d 2y = 3 dx 2
dy
=
dy
dy  dx  2y
+ 5 dx  2y (3D 2 + SD  2)y.
Hence BA
= 3 D2 + 5 D 
2
= A B.
EXAMPLE 6.8 Let G = x D
+ 2, and H
= D  1. Then
G(Hy) = (xD
+ 2) (;~

y)
d2y dy dy = x 2  x  +2   2y dx dx dx =
X
d2y dx 2
+ (2 
dy x) dx  2y ,
•
6.8
The Fundamental Laws of Operation
111
so GH = xD2
+ (2 
x )D  2.
On the other hand, H(Gy) = (D  1)
=
(x dxdy + 2Y )
Y ~ ( x d +2Y ) dx dx
d 2y = x dx 2 d 2y = x 2 dx
dy
Y _ ( x d + 2Y ) dx dy
+ dx + 2 dx + (3 
dy
x) 
dx
dy  x dx  2y  2y ;
that is, HG = xD2 + (3  x )D  2.
It is worthy of notice that here we have two operators G and H (one of them with variable coefficients), whose product is dependent on the order of the factors . On this topic see also Exercises 17 through 22 in the next section.
•
The sum of two differential operators is obtained by expressing each in the form
aoD"
+ al D" 1 + . . + all _ I D + a"
and adding corresponding coefficients. For instance, if A = 3D 2

D
+x 
2
and then A
+B =
(3
+ x 2)D2 + 3D + x + 5.
Differential operators are linear operators; that is, if A is any differential operator, CI and C2 are constants, and II and h are any functions of x eac h possessing the required number of derivatives, then A(c i I I + c2h)
6.8
= CI All + czAh·
The Fundamental Laws of Operation Let A, B, and C be any differential operators as defined in Section 6.7. With the definitions of addition and multiplication above, it foll ows that differential operators satisfy the followin g:
112
Chapter 6
Linear Differen.tial Equations
(a)
The commutative Law of addition:
A+B=B+A . (b)
The associative law of addition:
(A + B) + C = A + (B + C) . (c)
The associative law of multiplication:
(AB)C = A(BC). (d) The distributive law of multiplication with respect to addition:
A(B (e)
+ C) = AB + AC.
If A and B are operators with constant coefficients, they also satisfy the commutative law of multiplication:
AB = BA. Therefore, differential operators with constant coefficients satisfy all the laws of the algebra of polynomials with respect to the operations of addition and multiplication. If m and n are any two positive integers, then
DIIID" = DII1 +II , a useful result that follows immediately from the definitions. Since for purposes of addition and multiplication the operators with constant coefficients behave just as algebraic polynomials behave, it is legitimate to use the tools of elementary algebra. In particular, synthetic division may be used to factor operators with constant coefficients .
• Exercises Perform the multip lications indicated in Exercises 1 through 4.
1.
2.
(4D + l)(D  2). (2D  3)(2D + 3).
3. 4.
(D + 2)(D2  2D + 5). (D  2)(D + 1)2.
In Exerc ises 5 throu gh 16, factor each of the operators .
5.
6. 7. 8.
2D2 + 3D  2. 2D25D12. D3  2D2  5D + 6. 4D 3 4D 2 11D+6.
9. 10.
D4  4D2. D J  3D 2 +4.
II. D3  21D +20. 12.
2D 3

D2  13D  6.
6.9
13. 14.
Some PlVperties of Differential Operators
2D4 + 11 D3 + 18D 2 + 4D  8. 8D 4 + 36D 3  66D 2 + 35D  6.
15 .
D4 + D3  2D2
16.
D3 11 D  20.
+ 4D 
113
24.
Perform the multiplications indicated ill Exercises 17 through 22.
I
6 .9
I
+ x) .
l7.
(D  x)(D
18.
(D
19.
D(xD  I).
+ x)(D 
x).
20.
(xD  l)D.
2l.
(xD
22.
+ 2)(xD  I). CxD  l) (xD + 2).
Some Properties of Differential Operators Since for constant m and positive integral k, (I)
it is easy to find the effecl that an operator has upon eJllx. Let f (D) be a po lynomial in D,
feD) = aoD"
+ alD
IJ 
1
+ ... + a,, _ 1D + all'
(2)
Then
so (3)
If m is a root of the equation f(m)
= 0, then in view of eq uation (3),
f(D)e
I1lX
= O.
Next consider the effect of the operator D  a on the product of e" x and a funclion y. We have
CD 
a)(e"Xy) = D(e(lX y )  ae"Xy = eClx Dy
CD 
a) 2(e" x y)
and
= CD  a) (eClX Dy) = e" x D2y.
Repeating the operation, we are led to
CD  at (e"I y) = e llx D" y.
(4)
114
Chap/er 6
Linear Differential Equations
Using the linearity of differential operators, we conclude that when feD) is a po lynomial in D with constant coefficients, then (5)
The relation (5) shows us how to shift an exponential factor from the left of a di Fferential operator to the right of the operator. This relation has many uses, some of which we examine in Chapter 7. EXAMPLE 6.9 Let feD) = 2D2
+ 5D 
12. Then the equation f(m)
= 0 is
2m 2 + 5m  12 = 0, or (m
+ 4) (2m 
3) = 0,
of which the roots are m I =  4 and m 2 = ~. With the aid of equation (3) it can be seen that
(2D 2
+ 5D 
12)e 4X = 0
and that (2D 2
+ 5D 
12) exp ( ~ x) = 0.
In other words, YI = e 4x and Y2 = exp ( ~ x) are solutions of (2D 2
+ 5D 
12)y = 0.
•
EXAMPLE 6.10 Show that for k = 0, 1, ... , (n  1). In equation (5) we let shift, we obtain
f
(D)
=
D" and y
= xk.
(6)
Then using the exponential
But D"x k = 0 for k = 0, 1,2, ... , n  I, which gives us equation (6) directly. The results obtained in equations (3), (5), and (6) are of fundamental importance to the so lving of linear differential equations with constant coefficients, which we consider in Chapter 7.
•
6.10
Computer Supplement
115
EXAMPLE 6.11 As an example of the llse of the exponential shift, we solve the differential equation (D
+ 3)4 y
= O.
(7)
First we multiply equation (7) by e3x to obtain e 3X (D
+ 3)4 y = O.
Applying the exponential shift as in equation (5) leads to D 4 (e 3x y) = O.
Integrating four times gives us e
3x
y=
CI
+ C2X + C3 x 2 + C4x3,
and finally, (8)
Note that each of the four functions e 3x , xe 3x , x 2 e 3x , and x 3 e 3x is a solution of equation (7). This, of course, is assured by the theorem of equation (6) of Example 6.10. If we now show that the four function are linearly independent, equation (8) gives the general solution of equation (7) . See Exercise 5 .
• Exercises In Exerc ises I thro ugh 4, use the ex ponential shift as in Example 6.11 to find the genera l
solution .
1.
(D  2) 3y = O.
3.
(2D  1) 2y = O.
2.
(D + l) 2y = O.
4.
(D+7)6 y =O.
5.
To show that the four functions in Examp le 6.11 are linearly independent on any interval, assume that they are linearly dependent and show that this leads to a contradiction of the results obta ined in Exercise 1 of Section 6.4.
6.
Prove that the set of functions
is a linearly independent set on any interval. See Exercise 5.
16. 101 Computer Supplement While most of the material in Chapter 6 is fairly theoretical and therefore not suited to computer implementation, there are two areas where a Computer Algebra System can be of help. The first of these is in factoring differential operators. For example, the operator D3  3D2 + 4 can be factored with the Maple command
116
Chapter 6
Linear Differential Equations
(D
+ 1)(D 
2)2
A second computer application is illustrated in showing that the four functions exp(3x), xexp(  3x), x 2 exp(3x), and x 3 exp(3x) in Example 6.11 of Section 6.9 are linearly independent. >y :=vector([exp (3*x) ,x*exp(  3*x), (x"2) *exp ( 3*x), (x " 3) *exp (3*x) l);
>Ans : =det(Wronskian(y,x)) ;
We can eas ily see that the Wronskian is never zero, and hence the functions are linearly independent.
• Exercises I.
Do a se lect ion of the facto ring exercise in Section 6.8 using a computer.
2.
Do Exercises 2, 3, 6, and 7 in Section 6.4.
Linear Equations with Constant Coefficients
7. 1
7
Introduction Several methods for solving differentia l equations with constant coefficients are presented in thi book. A classical technique is treated in this and the next chapter. Chapters 14 and 15 contain a development of the Laplace transform and its u e in solvi ng linear differential equations. In Chapter 12 we study matrix techniques for solvin g linear equations w ith constant coefficients. Each method has its advantages and its disadvantages. Each is theoretically ufficient: all are necessary for maximum efficiency.
7.2
The Auxiliary Equation : Distinct Roots Any linear homogeneous differentia l equation w ith constant coefficients,
d"y dx"
d"  ' y dx" 
ao +a l  ,
+ ... +a
dy +a"y =0, dx'
(L)
ll _ , 
may be written in the fonTI
f(D»),
= 0,
(2)
where feD) is a linear differential operator. As we saw in the preceding chapter, if In is any root of the algebraic equation f(m) = 0, then
f(D)e IllX = 0, which means simply that y = elllx is a sol ution of equation (2) . The equation f(nt) = 0
(3)
is called the auxiliary equation associated with (I) or (2). The auxiliary equation for (1) is of degree n . Let its roots be In" .. . , mil' If these roots are all real and distinct, then the n solutions YI
= exp (m,x),
Y2
= exp (m 2x ),
. .. ,
YII
= exp (mllx) 117
118
Chapter 7
Linear Equations with Constant Coefficients
are linearly independent and the general solution of (1) can be written at once. It is Y=
CI
+ C2 exp (m2X) + ... + Cit exp (mllx),
exp (m ix)
in which C I , C2, ... , C" are arbitrary constants. Repeated roots of the auxiliary equation will be treated in the next section. Imag inary roots will be avoided until Section 7.5, where the corresponding solutions will be put into a desirable form.
EXAMPLE 7.1 Solve the equation d3 y dx 3
d2y 4 dX2

dy
+ dx + 6y =
O.
First write the auxiliary equation m3

4m 2 + m
+ 6 = 0,
whose roots m =  1, 2, 3 may be obtained by synthetic division. Then the general sol ution is seen to be
y=
C I e
x
+ cze 2x + C3e3x.
•
EXAMPLE 7.2 Solve the equation
(3D 3 + 5D2  2D)y = O. The auxili ary equation is
3m 3 + 5m 2 and its roots are m = 0, 2, solution may be written Y

2m = 0
t. By using the fact that eOx
= CI + cze  2.1' + C3 exp (I3x ) .
EXAMPLE 7.3 Solve the equation dZx  Z4 x = 0 dt with the conditions that when t = 0, x = 0 and dx / df = 3.
1, the des ired
•
7.2
119
The Auxiliary Equation: Distinct Roots
The auxiliary equation is m 2 4 = 0, with roots m = 2,  2. Hence the general solution of the differential equation is x
=
c] e
21
+ C2e  2t.
It remains to enforce the conditions at t = O. Now dx  = 2c]e 21  2c?e 21. dt Thus the condition that x = 0 when t = 0 requires that
0=
CI
+ C2,
and the condition that dx / dt = 3 when t = 0 requires that 3
=
2c]  2C2.
From the simultaneous equations for C2 = Therefore,
t.
and
CI
x = t(e 21

we conclude that
C2
CI
1 and 4
e 21) ,
which can also be put in the form x = ~ sinh (2t).
•
• Exercises
In Exercises 1 through 22, find the general solution. When the operator D is used, itis implied that the independent variabl e is x .
(D 2 + 2D  3)y = O. 2. (D 2 + 2D)y = O. 3. (D 2 + D  6) y = O. 4. (D 2  5D + 6)y = O. 5. (D 3 + 3D 2  4D)y = O. 6. CD 3  3D 2  lOD)y = O. 7. (D 3 + 6D 2 + lID + 6)y = O. 8. (D 3 + 3D 2  4D  12)y = O. 9. (4D 3  7 D + 3)y = O. l.
17. 18. 19.
C4D4  8D 3  7D 2 + llD + 6)y (4D4  I6D 3 + 7D2 + 4D  2)y (4D 4 + 4D 3  13D 2 7D + 6)y
10.
13D  6)y = O. d x d 2x dx + 3 2 2=0.
(4D 3

3
11.
dt
dt
dt
d 3x
14.
dx  3  19 + 30x = O. dt dt (9D 3 7D+2)y=O. (4D 3  2ID  lO)y = O.
15. 16.
CD 3  14D + 8)y = O. (D 3  D2  4D  2)y = O.
12. 13.
= O. = O. = O.
120
Chapter 7
Linear Equations with Constan.t Coefficients
21.
(4D 5  8D 4  17 D3 + 12D2 + 9D)y = 0. (D 2  4aD + 3( 2 )y = 0; a real "# 0.
22.
[D 2
20 .

(a
+ b)D + ab]y =
0; a and b real and unequal.
In Exercises 23 and 24, find the particular solution indicated.
23. (D 2  2D  3)y = 0; when x = 0, y = 0, yt =  4. 24. (D 2

D  6) y
= 0;
when x = 0, y
In Exercises 25 through 29, find for x
25. (D 2  2D  3) y
when x
= 0,
y
= 4, y' = 0. = 0, y" = 2.

4D)y = 0; when x = 0, y = 0 , yt
27. (D 2

D  6)y = 0; when x = 0, y = 3, yt =  1.
28. CD2
+ 3D 
3

10)y
= 0;
2D2  5D + 6) y
= 1, y = e 3 .
I the y value for the particular solution required.
26. CD 3
29. CD
7 .3
= 0;
=
= 0, and when x
= 0, y = 0, and when x = 2 , y = l. when x = 0, y = 1, yt =  7, y" =  1.
when x
= 0;
The Auxiliary Equation : Repeated Roots Suppose that in the eq uation
f(D) y =
°
(1)
°
the operator feD) has repeated factors; that is, the auxiliary equation f(m) = has repeated roots. Then the method of the preceding section does not yield the general solution. Let the auxiliary equation have three equal roots m I = b, m 2 = b, m 3 = b. The correspo nding part of the solution yielded by the method of Section 7.2 is y =
y =
+ C2evx + C3evx, vx (CI + C2 + c3)e .
C Ie
Vx
(2)
Now (2) can be replaced by (3)
with C4 = CI + C2 + C3. Thus, corre ponding to the three roots under consideration , this method has yielded only the solution (3). The difficulty is present, of course, because the three solutions corresponding to the roots m I = m2 = m 3 = bare not linearly independent. What is needed is a method for obtaining n linearly independent solutions corresponding to n equal roots of the auxili ary equation. Suppose that the auxiliary eq uation f(m) = has the n roots
°
7.3
The Auxiliary Equalion: Repeated Roots
m , = m 2 = ... = mIl =
Then the operator feD) must have a factor independent y's for which (D  bty
121
b.
CD  b)". We wish to find n linearly
= o.
(4)
Turning to result (6) near the end of Section 6.9 and writing m = b, we find that
(D  b)" (x k e/}X) = 0
for
k = 0,1,2, ... , (n  1) .
(5)
The functions Yk = xk e/}x where k = 0, 1, 2, ... , (n 1) are linearly independent because, aside from the common factor e/}x , they contain only the respective powers xo, xl , x 2, ... , X"  I. (See Exercise 5 of Section 6.9.) The general solution of equation (4) is (6)
Furthermore, if feD) contains the factor (D 
bY', then the equation
f(D)y = 0
(1)
g(D)(D  b)"y = 0,
(7)
can be written
where g(D) contains all the factors of feD) except (D  b)" Then any solution of (D  b)"y = 0
(4)
is also a solution of (7) and therefore of (1). Now we are in a position to write the so lution of equation (1) whenever the auxiliary equation has only real roots. Each root of the auxiliary equation is either distinct from all the other roots or it is one of a set of equal roots. Conesponding to a root mi distinct from all others, there is the solution
Yi =
Ci
exp (mix),
(8)
and conesponding to n equal roots In" m 2, ... , m", each equal to b, there are solutions (9)
The collection of solutions in (8) and (9) has the proper number of elements, a number equal to the order of the differential equation, because there is one solution corresponding to each root of the auxiliary equation. The solutions thus obtained can be proved to be linearly independent.
122
Chapter 7 Linear Equations with Constanl Coefficients
EXAMPLE 7.4 Solve the equation (D 4  7D3 + 18D 2
20D + 8)y = O.

(10)
With the aid of synthetic division, it is easily seen that the auxiliary equation m 4  7m 3 + 18m 2
20m

+8=
0
has the roots m = 1, 2 , 2, 2. Then the general sol ution of equation (10) is
or
y=
Cte x
+
( C2
+
C3X
C4X 2) e 2r ' .
+
•
EXAMPLE 7.5 Solve the equation
d4 y d 3 y d2 y +2+= 0. dx 4 dx 3 dx 2 The auxiliary equation is m
4
+ 2m 3 + m 2 =
0,
with roots m = 0, 0,  1,  1. Hence the desired solution is
•
• Exercises In Exercises 1 through 20, fi nd the general solutj on.
1.
2. 3. 7. 8.
9. 10.
13.
( D 2  6D +9)y = 0 . (D 2 + 4D + 4) y = O. (4D 3 3
+ 4D2 + D)y =
4. 5.
O.
6.
8D 2 + 16D)y = O. (D4 + 6D 3 + 9D2)y = O. (D 3  3D 2 + 4)y = O. (D 3

(4D  3D + Oy = O. ( D4  3D 3  6D 2 + 28D  24)y = O. 11. (D s  D 3 )y = O. (D 3 + 3D 2 + 3D + l) y = O. 12. (D s  16D 3 )y = O. (D 3 + 6D 2 + 12D + 8)y = O. 4 2 (4D +4D3  3D  2D + l)y = O.
7.4
14. 15. 16. 17. 18. 19. 20.
A Definition of exp z forlmaginary
z
123
(4D 4  4D 3  23D 2 + 12D + 36)y = 0. (D 4 + 3D 3  6D 2  28D  24)y = 0. (27D4  18D 2 + 8D  ' l)y = 0. (4D 5  23D 3  33D 2  17D  3)y = 0. (4D 5  15D 3  5D2 + 15D + 9)y = 0. (D 4  5D2  6D  2)y = 0. (D 5  5D 4 + 7 D3 + D2  8D + 4)y = 0.
In Exercises 2 J through 26, find the particular olution indicated.
2l. (D 2 +4D+4)y = 0; when x = 0, y = 1, y' =1. 22. The equation of Exercise 21 with the condition that the graph of the solution pass through the points (0, 2) and (2, 0). 23. (D 3  3D  2)y = 0; when x = 0, y = 0, y' = 9, y" = 0. 24. (D 4 + 3D 3 + 2D2)y = 0; when x = 0, y = 0, y' = 4, y" =  6, y"l = 14. 25. The equation of Exercise 24 with the conditions that when x = 0, y = 0, y' = 3, y" =  5, y"' = 9. 26. CD 3 + D2  D  l)y = 0; when x = 0, y = 1, when x = 2, y = 0, and also with the condition as x 7 00, Y 7 0. In Exercises 27 through 29, find for x
= 2 the y value for the particular solution required.
27. (4D 2  4D + l)y = 0; when x = 0, y =  2, y' = 2. 28. (D 3 + 2D2)y = 0; when x = 0, y = 3, y' = 0, y" = 12. 29. CD 3 + 5D2 + 3D  9)y = 0; when x = 0, y = 1, when x = 1, y = 0, and also with the condition as x 7 00, Y 7 0.
7.4
A Definition of exp
z for
Imaginary
z
Since the auxiliary equation may have imaginary roots, we need to lay down a definition of exp z for imaginary z. Let z = 0: + if3 with 0: and f3 real. Since it is desirable to have the ordinary laws of exponents remain valid, it is wise to require that (l)
To eCL with 0: real , we attach the usual meaning. Now consider e i /3, f3 real. In calculus it is shown that for all real x eX = 1
X
x2 21
x3 31
x"
+  +  +  + ... +  + .. . I!
n!
(2)
124
Chapter 7
Linear Equations with Constant Coefficients
or
1::: xn 00
ex 
 n=O n."
If we tentatively put x = if3 in (2) as a definition of e i{3, we get ,. e {3
if3
j2f3 2
j3f33
i"f3"
I!
2!
3!
n!
= 1++  +  + ... ++ .. ..
(3)
Separating the even powers of f3 from the odd powers of f3 in (3) yields
+ ... , or (4)
Now i 2k = ( _ l)k, so we may write
or .
00
( 1/f32k
k=O
(2k)!
e'{3 =1:::
.
00
(_ l/f32k+1
+ /1:::   k=0 (2k + 1)'
(5)
But the series on the right in (5) are precisely those for cos f3 and sin f3 as developed in calculus. Hence we are led to the tentative result
ei {3
= co f3
+ i sin f3.
(6)
The student should realize that the manipulations above have no meaning in themselves at this stage (assuming that infinite series with complex terms are not a part of the content of elementary mathematics). What we have accomplished is this: The formal manipulations above have suggested a meaningful definition of exp (0:: + i(3), namely, exp (0::
+ i(3) =
e" (cos f3
+ j sin (3)
when ex and f3 are real.
(7)
7.5
The Auxiliary Equation: Imaginary Roots
125
Replacing f3 bye  f3) in (7) yields a result that is of value to us in the next section , exp (ex  if3) = e U (cos f3  i sin f3) . It is interesti ng and important that with the definition (7), the function eZ for comp lex z retains many of the properties possessed by the function eX for real x. Such matters are often studied in detail in books on complex variables. 1 Here we need in particul ar to know that if y = exp (a
+ ib) x ,
with a, b, and x real, then (Daib)y =O.
The result desired follows at once by differentiation, with respect to x, of the function
y
7 .5
= e"x (cos b.x + i sin bx) .
The Auxiliary Equation ; Imaginary Roots Consider a differential equation f(D)y = 0 for which the auxiliary equation fern) = 0 has real coefficients. From elementary algebra we know that if the auxi liary equation has any imaginary roots, those roots must occur in co njugate pairs. Thus if rn l=a +ib
is a root of the equation fern) = 0 , with a and b real and b
# 0, then
rn 2 = a  ib
is also a root of fern) = o. It must bekept in mind that thi s result is a consequence of the reality of the coefiicients in the equation fern) = O. Imaginary roots do not necess arily appear in pairs in an algebraic equation whose coefficients involve imaginaries . We can now construct in u able form solutions of f(D)y = 0
(1)
corresponding to im aginary roots of f(m) = O. For since f(m) is assumed to have real coefficients, any imaginary roots appear in conjugate pairs, 112 1
=a+ib
and
m 2=a  ib.
I For exampl e, R. V. Churchi ll and 1. W. Brown, Complex Variables and ApplicQlions. 6th ed. (New York : McGrawHill . 1996).
126
Chapler 7
Linear Equations with Constant Coe.tficients
Then, according to the preceding section , equation (1) is satisfied by
y=
c, exp [(a + ib)x] + C2 exp [(a
 ib) x ].
(2)
Taking x to be real along with a and b, we get from (2) the result
+ i sinbx) + cze"X(cos bx
y = c,e"X (cos bx
 i sinbx).
(3)
Now (3) may be written
y = (c , + C2)e"x cosbx
+ iCc,

C2)e"x
sinbx .
Finally, let c, + C2 = C3 and iCc,  C2) = C4, where C3 and C4 are new arbitrary constants. Then the equation (1) is seen to have the solutions (4)
corresponding to the two roots m, = a + ib and m 2 = a  ib (b "I 0) of the auxiliary equation. The reduction of solution (2) to the desirable form (4) has been done once and that is enough. Whenever a pair of conjugate imaginary roots of the auxiliary equation appears, we write down at once, in the form given on the right in equation (4), the particul ar solution corresponding to those two roots. EXAMPLE 7.6 Solve the equation (D 3

3D 2 + 9D
+ 13)y =
3m 2 + 9m
+
O.
For the auxili ary equation m3

13 = 0,
one root, m , = 1, is easily found. When the factor (m + 1) is removed by synthetic division, it is seen that the other two roots are solutions of the quadratic equation
m2

+ 13 = O. 2 + 3i and m3
4m
Those roots are found to be m2 = = 2  3i. The auxiliary equation has the roots rrt. = 1, 2 ± 3i. Hence the general solution of the differential equation is y =
c , e
x
+ C2e2x cos 3x + C3e2x sin 3x.
•
Repeated imaginary roots lead to solutions analogous to those brought in by repeated real roots. For instance, if the roots m = a ± i b occur three times, the corresponding six linearly independent solutions of the differential equation are
7.6 A Note on Hyperbolic Functions
127
those appearing in the expression (CI
+ C2X + c3x2)eQx cosbx + (C4 + CsX + c6x2)eax sin bx.
EXAMPLE 7.7 Solve the equation
+ 8D 2 + 16)y = O. The auxiliary equation m 4 + 8m 2 + 16 = 0 may be written (m 2 + 4)2 = 0, (D
4
so its roots are seen to be m = ±2i, ±2i. The toot ml = 2i and /11 2 = 2i occur twice each. Thinking of 2i as 0 + 2i and recalling that eOx = 1, we write the solution of the differential equation as
y =
(CI
+ C2 X ) cos 2x + (C3 + C4X) sin 2x.
•
In such exercises as those following Section 7.6, a fine check can be obtained by direct substitution of the result and its appropriate derivatives into the differential equation. The verification is palticularly effective because the operations performed in the check are so different fro m those performed in obtaining the solution.
7.6
A Note on Hyperbolic Functions Two particular linear combinations of exponential functions appear with such frequency in both pure and applied mathematics that it has been worthwhile to use special symbols for those combinations . The hyperbolic sine of x, written sinh x, is defined by eX _ e x
sinhx =   2
(1)
the hyperbolic cosine of x, written coshx, is defined by
+ e x
eX
coshx =   2 From the definitions of sinh x and cosh x it follows that sinh2 x = ! (e 2x and

2 + e 2x )
(2)
128
Chapter 7
Linear Equations with Constant Coefficients
so cosh2 X

sinh2 x = 1,
(3)
an identity similar to the wellknown identity cos 2 x + sin 2 x = 1 in trigonometry. Directly from the definition we find that
y = sinh u is equivalent to
Hence, if u is a function of x, then dy I /I  u du  = "?(e +e ),
dx

dx
that is, d du  sinh u = cosh u  . dx dx
(4)
The same method yields the result d du  cosh u = sinh u. dx dx
(5)
The graph of y = coshx and y = sinhx are exhibited in Figure 7.1. Note the important properties: (a)
cosh x ::: 1 for all real x .
(b)
The only real value of x for which sinh x = 0 is x =
(c)
cosh (x) = coshx; that is, coshx is an even function of x.
(d)
sinh (x) =  sinh x; sinh x is an odd function of x.
o.
The hyperbolic functions have no real period. Corresponding to the period
2rr possessed by the circular functions , there is a period 2rr i for the hyperbolic functions. The hyperbolic cosine curve is that in which a transmission line, cable, piece of string, watch chain, and so on, hangs between two points at which it is suspended. This result is obtained in Chapter 16. Since D2 cosh ax = a 2 cosh ax and D 2 sinh ax = a 2 sinh ax, it follows that both cosh ax and sinh ax are solutions of a
# O.
(6)
7.6
A NOle on Hyperbolic Functions
129
y
y
= cosh x
~~ x
Figure 7.1
F urthermore, the Wronskian of these two functions,
W
I
(x) =
co hax asinhax
sinhax acoshax
I=a ,
is not zero, so that cosh ax and sinh ax are linearly independent sol utions of equation (6). Hence the general sol utio n of (6) may be written
y
= C I cosh ax + C2 sinh ax
instead of using th e form
It is very often convenient to use th i alternative form for representin g the general solution of (6).
EXAMPLE 7.8 Find the solution of the problem (D 2

4)y = 0;
w hen x = 0, y = 0, y' = 2.
The general solution of the differe ntial equation (7) may be written
y=
CI
cos h 2x
+ C2 sinh 2x ,
from which
y' = 2cI sin h 2x
+ 2C2 cosh 2x.
(7)
130
Chapter 7 Linear Equations with COl/stant Coefficients
The initial conditions now require that
y
°= c,
and 2
= 2C2, so finally,
= sin h 2x.
Note that if we were to choose the alternative form
for the general solution of (7), we would obtain the same result with a little more fuss in detel111ining C3 and C4 . Indeed, one major reaso n for using the hyperbolic fu nctions is that cosh ax and sinh ax have values I and when x = 0, a fact that is particularly useful in solving initial value problems .
°
•
• Exercises Find the general sol ution except when the exercise stipulates otherwise.
1.
Verify directly that the relation (A)
satisfies the equation
[CD  a)2
2. 3. 4. S. 6. 12. 13. 14. 15. 16. 17. IS.
+ b2Jy =
0.
(D 2  2D + 5)y = O. (D 2  2D + 2)y = O. (D 2 + 9)y = O. (D 2  9)y = O. (D 2 + 6D + 13)y = O. (2D 4 + J ID J  4D2  69D
7. CD2  4D + 7)y = O. S. CD 3 + 2D2 + D + 2)y = O. 9. (D 4 + 2D 3 + 10D2)y = O. 10. (D 4 2D 3 +2D 2  2D+ l )y = O. 11. (D4 + lSD 2 + Sl)y = O. + 34)y = O. 4 6 (D + 9D + 24D2 + 16)y = O. (2D 3  D2 + 36D  lS)y = O. (D 2 l)y = 0; when x = 0, y = Yo, y' = O. (D 2 + J)y = 0; when x = 0, y = yo, y' = O. (D J +7D 2 + J9D + 13)y = 0; when x = 0, y = 0, y' = 2, y" = 12. (D 5 + D4  7D 3  llD2  SD  12)y = O.
20.
d 2x dx +k 2 x = 0, k real; when t = 0, x = 0,  = Va. dt dt J (D + D2 + 4D + 4)y = 0; when x = 0, y = 0, y' = 1, y" = S.
21.
czZx dt 2
19.
2
dx
+ 2b di + k
2
.
x = 0, k > b > 0, when
t =
0, x
= 0,
dx dt
= Vo.
7.6
A NOie on H yperbolic FUIlC/ioli s
131
• Miscellaneous Exercises Obtai n the general solution unless otherwise instructed.
(D 2 +3D)y=0. 10. (4D 3 2 1D  IO )y=0. 2. (9D 4 + 6D 3 + D2)y = O. 11. (4D 3  7 D + 3)y = O. 3. (D 2 + D  6)y = O. 12. (D J  14D + 8)y = O. 4. (D J + 2D2 + D + 2)y = O. 13. (8D 3  4D2  2D + l )y = O. 2 3 S. (D  3D + 4)y = O. 14. (D4 + D3  4D2  4D)y = O. 3 6. (D  2D2  3D)y = O. IS . (D 42D 3+SD 28D+4)y = O. 7. (4D 3  3D + l )y = O. 16. (D 4 + 2D2 + I)y = O. 8. (D J + 3D 2  4D  12»)1 = O. 17. (D 4 + SD 2 + 4)y = O. 3 2 9. (D + 3D + 3D + 1))1 = O. 18. (D 4 + 3D J  40)y = O. 19. (D 4  11D 3 + 36D 2  16D  64)y = O. 20. (D 2 + 2D + S) y = O. 21. (D 4 + 4D 3 + 2D2  8D  8)y = O. 22. (4D4  24D 3 + 3SD 2 + 60  9»)1 = O. 23. (4D 4 + 20D 3 + 350 2 + 2SD + 6)y = O. 24. (04  7D3+ IID 2 +SD  14)y=0 . 25. (D 3 +S D 2 +7D+3)y=0. 33 . (D 4 D 3  3D 2 +D+2)y=0. 3 26. (D  2D2 + D  2)y = O. 34. (D 3  2D2  3D + 10) y = o. 27. (D J  D2 + D  I)y = O. 35. (D s + D4  6D 3 )y = O. 28. (D J +40 2 +SD)y =0. 36. (4D 3 +28D 2 +6ID+37)y=0. 29. (D4  13D 2 + 36)y = O. 37 . (4DJ+ 12D2+ 13D + 10)y = o. 30. (D4SDJ+SD2 +S0 6)y = O. 38. (18D 3  33D 2 +20D 4)y = O. 31. (4D 3 + 8D 2  11 D + 3»)1 = O. 39. (D s 2D 3  2D 23D2)y = O. 32. (D J + D2  160  16)y = O. 40. (D 42D 3 +2D 22D+ 1).'1 = O. 41. (D 5  ISD 3 + lOD 2 + 60D  72)y = O. 42. (4D4  ISD 2 + SD + 6) y = O. 43. (D 4 + 3D 3  6D 2  28D  24)y = O. 44 . (4D 4  4D 3  23 D 2 + 12D + 36)y = O. 45 . (4D 5  23D J  33D 2  17 D  3)y = O. 1.
46. 47.
(D 2  D  6) y = 0; when x = 0, y = 2, )I' = l. (D4 + 6D 3 + 9D 2 »)I = 0; when x = 0, y = 0 , y' = 0, y" = 6, and as x ~ 00, y' ~ 1. For this particular solution, find the value of y when x =1.
132
Chapter 7
Linear Equations with Constant Coefficienls
48. 49. 50.
51.
(D 3 + 6D 2 + 12D + 8)y = 0; when x = 0, y = 1, y' =  2, y" = 2. (D s + D4  9D 3  13D 2 + 8D + 12) y = O. (4D 5 + 4D4  9D 3  llD2 + D + 3)y = O. (D s + D4  7 D3  J ID2  8D  12)y = O.
I 7.7 I Computer Supplement The techniques described in the Computer Supplement to Chapter 2 extend easily to higherorder equations. We can ill ustrate this with Example 7.6 in Section 7.5 (D 3

3D 2 +9D
+ 13»), =
O.
If we add the initial conditions, yeO) = 1, y' (0) = 2, y" (0) = 3, Maple solves the problem with the commands >di f f(y (x) ,x$3 )3* di f f(y (x) ,x$2) +9*diff(y(x) ,x)+13*y(x)=O; d3 d2 dx 3 y(x)  3 dx 2 y(x)
d
+ 9 dx y(x) + 13 y(x)
= 0
>dso lve({ " ,y(O) = l, D (y ) (O) =2,D(D(y)) (O ) =3 },y(x));
4 e x y (x) =  9
+
S e2 Xcos(3x) 9
+
4e 2 x sin(3x) 9
We can also use Maple to plot the resulting sol ution with the command >p l ot (rhs(" ) , x=2 .. 2); See Figure 7.2.
y
20 15 10 5 2
1
0 5  10
Figure 7.2
7.7
• Exercises 1.
Solve a variety of problems from the chapter.
2.
Use a plotting routine to display your results.
Computer Supplement
133
Nonhomogeneous Equations: Undetermined Coefficients 8.1
8
Construction of a Homogeneous Equation from a Specific Solution In Section 6.6 we saw that the general solution of the equation
(boD"
+ b l D,,  I + ... + b,, _ 1D + b,,)y =
R(x)
(I)
is
y = yc + y", where Ye, the complementary f unction , is the general solution of the homogeneous equation (boD" + b l D,I  I + ... + b,I _ 1D + b,Jy = 0 (2) and Y" is any particular so lution of the original equation ( 1). Various methods for gett ing a solution of (1) when the bo, b I , ... , b" are constants will be presented. In preparation for the method of undetermined coefficients it is wise to obtain proficiency in writing a homogeneous differential equation of which a given function of proper form is a solution. Recall that in solving homogeneous equations with constant coefficients, a term such as cle"'" occurred only when the auxiliary equation fern) = 0 had a root In = a, and then the operator feD) had a factor (D  a). In like manner, C2xeGX appeared only when feD) contained the factor (D  a)2, c3x2eGx only when feD) contained (D  a) 3, and so on. Such terms as ce llX cos bx or ce"x sin bx correspond to roots m = a ± ib, or to a factor [(D  a)2 + b 2].
EXAMPLES.l Find a homogeneous linear equation, with constant coefficients, that has as a particular solution
y = 7e 3x
+ 2x .
First note that the coefficie nts (7 and 2) are quite irrelevant for the present problem as long as they are not zero. We shall obtain an equation satisfied by y = cle 3x + C2X no matter what the constants CI and C2 may be.
134
8.1
Construction of a Homogeneous EquatiOIl
135
A term CI e3x occurs along with a root m = 3 of the auxiliary equatio n. The term C2X will appear if the auxiliary equation has rn = 0, 0, that is, a double root m = O. We have recognized that the equation
D 2( D  3)y = 0, or
(D 3

3D2)y = 0,
has y = c le 3x + C2X + C3 as its ge neral sol ution , and therefore that it also has y = 7e 3x + 2x a a particular solution.
•
EXAMPLE 8.2 Find a homogeneous linear equation with real, con tant coefficients that is satisfied by
y = 6 +3xe x

(3)
cosx.
The ten11 6 is assoc iated with m = 0, the term
3xex
with a double root 111 = In = 0 ± i. Hence
1, 1, and the term ( cos x ) with th e pair of imaginary roots
the auxiliary equation is
m(m  1)\m 2 + I) = 0, or
m5

2m 4
+ 2m 3 
2m 2
+m
= O.
Therefore, the function in (3) i a solution of the differential eq uation
(D 5

2D4
+ 2D 3 
2D2
+ D) y
= O.
(4)
That is, from the ge neral solution
y
= C I + (C2 + c3x)e< + C4 cos X + C5 si n x
of equatio n (4), the relation (3) follows by an appropriate choice of the constants : CI = 6, C2 = 0, C3 = 3, C4 =  1, C5 = O.
•
EXAMPLE 8.3 Find a homogeneous linear equation with real, consta nt coefficients th at is satisfied by
y = 4xe x sin 2x. The desired equati on must have its auxiliary equation with roots
m = I ± 2i, 1 ± 2i.
136
Chapter 8
Nonhomogeneous Equations: Undetermined Coefficients
The roots m must be
=
1 ±2i correspo nd to factors (m _ 1)2 + 4, so the auxiliary equation
or 4m3
+ 14m 2 
20m
+ 2S =
O.
Hence the desired equation is (D 4  4D 3
+ 14D2 
20D
+ 2S) y
= O.
m4

•
Note that in all such problems, a correct (but undesirable) solution may be obtained by inserting additi onal roots of the auxiliary equation .
• Exercises In Exercises I thro ugh 14, ob tain in factored form a linear differential eq uatio n with rea l, constant coefficients that is ati s fied by the given fu nction .
4.
= 4e 2x + 3ex . y = 7  2x + ~e4X. y =  2x + ~e4X. y = x 2  S sin 3x.
S.
y
1. 2. 3.
6. 7.
y
= 2e x cos 3x . 2x
y
=
y
= 2e
3e
sin 3x . 3x
cos x .
8.
y
= e x sin 2x.
9.
y
10.
y
11.
y
= xe  X sin 2x + 3e  x cos 2x. = sin2x + 3cos2x. = cos kx.
12.
y
13.
y
= x si n 2x. = 4sinhx.
14.
y
= 2 cosh 2x
 sinh 2x .
In Exercises 15 through 34, list the roots of the auxili ary equ atio n for a homogeneous linear eq uation with rea l, constant coef'ficients th at has the given fu nctio n as a particular soluti on.
17.
= 3xe 2x . = x 2 e x + 4e y = e cos4x.
18.
y
3e  Xcos 4x
19.
y = x(e 2x + 4) . y =4+2x 2 e 3x .
IS . Y 16. y
20. 21.
X
•
X
22.
y y
23.
y
24.
y
=
+ lS e
= xe = xex + Sex. = 4cos2x. = 4 cos 2x  3 s in 2x . X
•
x
sin 4x .
2S.
y
= xcos2x.
26.
y
= e 2x cos 3x .
27.
y = x cos 2x  3 sin 2x.
28 .
y
29.
y
30. 31.
y y
32.
y = x 2 sinx.
33 .
y = x 2 sinx
34.
y = 8 cos 4x + sin 3x .
= e 2x (cos 3x + sin 3x). = sin 3 x = ~(3 sinx  sin 3x). = cos2 x . = x 2  X + eX(x + cosx). + x cosx.
8.2
8 . '2
Solution ofa Nonhomogeneous Equation
137
Solution of a Nonhomogeneous Equation Before proceeding to the theoretical basis and the actua l working technique of the useful method of undetermined coefficients, let us examine the underlying ideas as applied to a simple numerical example. Consider the equation D2(D  l)y
= 3e + sinx. X
(1)
The complementary function may be determined at once from the roots m
= 0, 0,
1
(2)
of the auxiliary equation. The complementary function is
Yc = c,
+ C2X + C3 ex .
(3)
Since the general solution of (1) is
Y=Yc+YP' where Yc is as given in (3) and Yp i any particu lar so lution of (I), all that remains for us to do is to find a particular solution of (I). The righthand member of (1), R(x) = 3e x
+ sinx,
(4)
is a particular solution of a homogeneous linear differential equation whose auxiliary equation has the roots
m' = I, ±i.
(5)
Therefore, the function R is a particu lar so lution of the equation (D  1) (D 2 + l)R = O.
(6)
We wish to convert (1) into a homogeneous linear differential equation with constant coefficients, because we know how to solve any such equation. But, by (6), the operator (D  I) (D 2 + I) will annihilate the right member of (1) . Therefore we apply that operator to both sides of equation (1) and get (D  1)(D 2
+ l)D2(D 
I) y = O.
(7)
Any solution of (1) must be a particu lar so lution of (7). The general sol ution of (7) can be written at once from the roots of its auxiliary equation, those roots being the values m = 0, 0, 1 from (2) and the values m' = 1, ±i from (5). Thus the general solution of (7) is
y =
C,
+ C2X + C3e x + C4xe x + Cs cosx + C6 sin x.
(8)
But the desired general solution of (1) is y = Yc
where
+ Yp,
(9)
138
Chapter 8 Nonhomogeneous Equations: Vndetennined Coefficients
the Cl, c2, C3 being arbitrary constants as in (8). Thus there must exist a particular solution of (1) containing at most the remaining terms in (8). Using different letters as coefficients to emphasize that they are not arbitrary, we conclude that (1) has a particular solution Y" = A xe
x
+ B cosx + C sinx.
(10)
We now have only to determine the numerical coefficients A, B, C by direct use of the original equation
D2(D  l) y = 3e x
+ sil1X.
(1)
From (10) it follows that
Dyp = A(xe X X
D2yp = A(x e D3yp
=
A(x e
X
+ eX)  B sinx + C cos x, + 2e B COsx  C sinx , + 3e + B sinx  C cosx . X
X
) 
)
Substitution of y,J into (1 ) then yields A ex
+ (B + C)sinx + (B
 C)cos x = 3e x
+ sinx.
(11)
Because (11) is to be an identity and because eX , sinx, and cosx are linearly independent, the corresponding coefficients in the two members of (11) must be equal; that is,
A=3
B+C=1 B  C = O.
4, c = 4.
Therefore, A = 3, B = solution of equation (1) is
yp = 3xe'
t·
Returning to (10) , we find that a particular . + 2:l cosx + 2l SIOX.
The general solution of the original equation,
D 2(D  l) y
= 3ex + sil1X,
(1)
is therefore obtained by adding to the complementary function the yp fou nd above: (12) A caref ul analysis of the ideas behind the process used shows that to arrive at the solution (12), we need to perform only the following steps: (a)
From (1) find the values of In and
(b)
From the values of
In
In '
as exhibited in (2) and (5).
and m ' write Yc and yp as in (3) and (10).
8.3
8.3
The Method of Undetermined Coefficients
139
(c)
Substitute Yp into (1), equate corresponding coefficients, and obtain the numerical values of the coefficients in Yp'
Cd)
Write the general solution of (1).
The Method of Undetermined Coefficients Let us examine the general problem of the type treated in the preceding section. Let feD) be a polynomial in the operator D. Consider the equation f(D)y
= R(x).
Let the roots of the auxiliary equation f(m) =
°
(1)
be
In = mJ, In2, ... , In".
(2)
The general solution of (1) is (3)
Y = Yc+YP'
where Yc can be obtained at once from the values of m in (2) and where Y = YP is any particular solution (yet to be obtained) of (1). Now suppose that the right member R(x) of (1) is itself a particular solution of some homogeneous linear d~fferenlial equation with constant coefficients,
(4)
g(D)R = 0, whose auxiliary equation has the roots
(5)
Recall that the values of m' in (5) can be obtained by inspection from R(x). The differential equation g(D)f(D)y =
°
(6)
has as the roots of its auxiliary equation the values of m from (2) and m' from (5). Hence the general solution of (6) contains the Yc of (3) and so is of the form Y=Yc+Yq' But also any particular solution of (1) must satisfy (6). Now, if f(D)(y c + Yq) = R(x), then f(D)Yq = R(x) because f(D)y c = O. Then deleting the Yc from the general solution of (6) leaves a function Yq that for some numerical values of its coefficients must satisfy (1); that is, the coefficients in Yq can be determined so that Yq = Yp· The determination of those nume11cal coefficients may be accomplished as in the following examples.
140
Chapter 8
Nonhomogeneous Equations: Undetermined Coefficiems
It must be kept in mind that the method of this section is applicable when, and only when, the right member of the equation is itselfa particular solution of some homogeneous linear differential equation with constant coefficients.
EXAMPLE 8.4 Solve the equation (D 2
+D
2)y = 2x  40 cos 2x
(7)
Here we have
m = 1,2
and m'
= 0,0,
±2i.
Therefore, we may write x
 2x
Yc = cle +C2 e , y p = A + B x + C cos 2x
+ E sin 2x ,
in which C I and C2 are arbitrary constants, whereas A, B, C, and E are to be determined numerically, so that yp will satisfy equation (7). Since D yp = B  2C si n 2x
+ 2E cos 2x
and D2yp = 4C cos 2x  4Esin2x,
direct substitution of yp into (7) yields  4C cos2x  4E si n 2x
+B
2C sin2x
+ 2E cos2x 
2A
 2Bx  2C cos 2x  2E sin 2x = 2x  40cos2x.
(8)
But (8) is to be an identity in x, so we must equate coefficients of each of the set of linearly independent function s cos 2x , sin 2x, x, and 1 appearing in the identity. Thus it follows that  6C + 2E = 40,  6E  2C = 0,
= 2, 2A = 0.
2B
B
8.3
The Method oj Undetermined Coefficients
141
The equations above determine A, B, C, and E. Indeed, they lead to
!,
A =
B =  1,
C =6, E = 2.
Since the general solution of (7) is y = Yc + Yp, we can now write the desired result,
• EXAMPLE8.S Solve the equation (D 2
+ l)y
= sinx .
(9)
At once m = ±i and m' = ±i. Therefore,
cosx + C2 sin x, yp = Ax cos x + Bx sinx. Yc =
CI
Now Y;~ =
A(x cosx  2sinx)
+ B( x si n x + 2cosx),
so the requirement that Yp is to satisfy equation (9) yields
2Asinx +2Bcosx
= sin x,
from which A =  ~ and B = 0. The general sol ution of (9) is y =
CI
cosx
+ C2 sin x 
!x cosx.
•
EXAMPLE 8.6 Determine y so that it will satisfy the equation
y'll  y'
= 4ex + 3e2x
with the conditions that when x = 0, y = 0, y' = 1, and y" = 2. First we note that m = 0, 1, 1 and m ' =  1, 2. Thus Yc
= C I + C2ex + C3 e  x ,
Yp = Axe x
+ Be2x .
(10)
142
Chapter 8
NOllhomogeneous Equations: Undetermined Coefficiellts
Now Y;J = A(xe X + eX ) +2 B e 2x , Y;~ = A(xe X  2e X ) + 4Be 2x ,
+ 3e + 8B e 2x .
y;~' = A(xe X
X
)
Then x 2x p  y'p = 2Ae + 6Be , Y"'
so that from (10) we may condude that A = 2 and B The general solution of (10) is therefore
Y=
CI
= ~.
+ C2ex + C3ex + 2xe  x + !e2x .
(11)
We must determine C I, C2, C3 so (11) will satisfy the conditions that when x = 0, y = 0, y' = I, and y" = 2. From ( I I) it follows that (12) and (13)
°
We put x = in eac h of (11), (12), and (13) to get the equations for the determination of C I, C2, and C3 . These are
°
=
 I =
2= from which
CI
= ~,
C2
+ C2 + C3 + ~, C2  C3 + 3, C2 + C3  2, C1
= 0, C3 = 4 . Therefore, the final result is
y =
_¥ +4e 
x
+2xe x
+ ~e2x.
•
An important point, sometimes overlooked by students, is that it is the general sol ution, lhe y of ( I I ), that must be made to satisfy the initi al conditions .
• Exercises In Exerc ises I through 35 , obtain the general solution.
1.
(D 2 + D )y=cosx.
2.
(D 2

6D
+ 9)y
= eX.
3. 4.
(D 2 + 3D + 2)y = 12x 2 . ( D 2 + 3D+2)y=1+3x+x 2 .
8.3
The Method of Undetermined Coefficients
CD2 + 9)y = Sex  162x. 11 . CD2 + 9)y = SeX  162x 2. 12. y"3y'4y = 30e x . 13. 4x y"  3y'  4y = 30e . 14. 2 2x 9. (D  4) y = e + 2. 15. 10. (D 2  D  2)y = 6x + 6e x . 16. X 2 17 . (D  l) y = e (2s in x + 4cosx). 5. 6. 7. S.
IS. 19. 20. 2].
(D 2

3
CD (D 3 (D 3 +
y"  4 y' + 3Y = 20 cos x. y" 4y'+3y = 2cosx+4sinx. y" + 2 y' + y = 7 + 75 sin 2x . (D 2 + 4D + 5) y = SOx + l3e 3x . (D 2 + l)y = cos x . (D 2  4D + 4)y = e2x
l) y = Sx ex . D) y = x . D2 + D  1) y = 4sin x. D2  4D  4)y = 3e x  4x  6.
I)y = 7x 2 . 23 . (D  I)y = e x . 24. (D 2  l)y = 10 sin 2 x. Use the identity sin 2 x = 22 .
( D4
143

4
2
!(l  cos2x).
2
25.
(D + l )y = 12cos x.
26. 27. 2S. 29.
(D 2 +4)y=4sin 2 x. y"  3 y'  4Y = 16x  50 cos 2x. CD 3  3D  2)y = 100sin 2x. y" + 4y' + 3y = lS e 2x + e x .
34. 35.
CD 3 + D  10) y = 2ge4x . CD 3 + D2  4D  4)y = Sx + 8 + 6e ·I .
30. 3 1.
y"  y = eX  4.
32. 33.
y" + 6y' + 13y = 60cosx + 26 . (D 3 3D 2+4)y = 6 + S0cos2x .
y"  y'  2y
= 6x + 6e x .
In Exercises 36 th rough 44, find the particular solution ind icated.
36. 37 . 3S. 39. 40.
41.
CD2 + 1) y = 10e2.x; when x = 0, y = 0, y' = O. CD 2  4) y=28x; whenx=O,y= O,y' =S. CD2 + 3D)y =  ISx; when x = 0, y = 0, y' = 5. (D 2 + 4D + 5) y = 1Oe 3x ; when x = 0, y = 4, y' = 0. d 2x dx dx dt 2 + 4 dt + 5x = 10; when t = 0, x = 0, dt = 0. x + 4x + 5x = S sin t; when f = 0, x = 0, x = 0. Note that the notation
dx/dt and x = d 2 x/dt 2 is common when the independent variable is time. 42. y" + 9y = S1x2 + 14cos4x; when x = 0, y = 0, y' = 3. 43. (D 3 +4D 2+9D+ 10) y = 24e x ; when x = 0, y = 0, y' = 4, y" = 10. 44. y" + 2y' + 5 y = Se  x ; when x = 0, y = 0, y' = 8.
x=
144
Chapler 8
Nonhomogeneous Equatiolls: Undetermined Coefficients In Exercises 45 through 48, obtain , from the particular so lution indicated, the va lue of y and the va lue of y' at x = 2.
45. 46.
47. 48. 49. 50.
y" + 2/ + y = x; atx = 0, y = 3 , and at x = 1, y =  l. y" + 2y' + y = x; at x = 0, y = 2, / = 2. 4y" + y = 2; atx = Jr, Y = 0, / = 1. 2y"  5/  3y = 9x 2  1; atx = 0, y = 1, y' = o. (D2+ D)y = x+ 1; whenx = 0, y = 1, and when x = 1, y = ~. Compute the value of y at x = 4. 3 2 (D + l) y = x ; when x = 0, y = 0, and when x = Jr, Y = O. Show that this boundary value problem has no solution.
51.
52. 53.
8.4
(D 2 + l)y = 2cosx; when x = 0, y = 0, and when x = Jr, Y = O. S how that this boundary va lue problem has an unlimited number of so lu tions and obtain them. For the equation (D 3 + D 2)y = 4, find the solution whose graph has at the origin a po int of inflection with a horizontal tangent line. For the eq uation (D 2  D) y = 2  2x ,find a particular solutio n that has at some point (to be determined) on the x axis an inflection point with a horizontal tangent line.
Solution by Inspection It is freq uently easy to obta in a pariicular olution of a nonhomogeneous equation (boD"
+ b l D,,  I + ... + b,,_1 D + b,,)y =
by inspection. For example, if R (x) is a constant Ro and if b"
#
R(x)
(1)
0,
Ro yp= 
(2)
b" is a solution of
(boD"
+ b l D,,I + .. . + b,, _ 1D + b,,)y =
R o,
b" =f:. 0, Ro constant,
(3)
because all derivatives of yp are zero, so
(boD"
+ b l D" I + ... + b,, _ ID + bll)yp = b" Ro =
Ro.
b"
Suppose th at b" = 0 in equation (3). Let D ky be the lowes tordered derivative that actually appears in the di fferential equation. Then the equation may be written
(boD"
+ ... + b,,  k Dk) y =
Ro,
b,, k =f:. 0 , Ro constant.
(4)
8.4
Solution by In spection
145
Now Dk xk = k!, a constant, so that all higher derivatives of xk are zero. Thus it becomes evident that (4) has a soluti on
ROX k
(5)
YP=k1b ' . nk
EXAMPLES.7 Solve the equation (D 2
3D

+ 2)y
= 16.
(6)
By the methods of Chapter 7 we obtain the complementary function,
By in spection a particular solution of the original equation is
Y" =
Jf
= 8.
Hence th e general solution of (6) is
• EXAMPLES.S Solve the equation
d 5Y

dx
5
d 3Y + 4 3 =7, dx
From the auxi li ary equatio n m 5 + 4m 3 =
°
we get In = 0, 0, 0, ±2i . Hence
A particular solution of (7) is
7x 3 7x 3 Yp = 3 ! ·4 = 24' As a check, note that
(7)
146
Chapter 8
Nonhomogeneous Equations: Undetermined Coefficients
The general solution of equation (7) is Y= in which the
CI
+ C2X + C3x2 + ;p,x 3 + C4 cos 2x + C5 sin2x,
CI, ... , C5
are arbitrary constants.
Examination of (D 2
+ 4)y
= sin 3x
• (8)
leads us to earch for a sol ution proportional to s in 3x because if y is proportional to si n 3x, so is D2y . Indeed, from y = A sin 3x
(9)
we get
D2 y =  9A sin 3x, so (9) is a so lution of (8) if
(9+4)A = 1 A =
 k.
Thu (8) has the general solution
y =
C 1 cos
2x
+ (.'2 si n 2x  ksin 3x,
a result that can be obtained mentally. For equation (8), the general method of undetermined coefficients leads us to write
m = ±2i ,
m' = ± 3i,
a nd so to write YP = Asin3x + Bcos3x.
(10)
When the yp of (10) is substituted into (8), it is fo und, of course, that A =~
B =0.
In contrast, consider the equation (D 2 + 4D +4)y = sin3x.
(11)
Here any attempt to fi nd a solution propoltional to in 3x is doomed to failure because although D2y will also be proportional to sin 3x, the term D y w ill involve cos 3x. There is no other term on either side of (11) to compensate for this cosine term , so no sol ution of the form y = A sin 3x is possible. For this eq uation, m =  2, 2, m' = ±3i, and in the particular solution
8.4
Yp = A sin3x
Solution by Inspection
147
+ B cos 3x,
it must turn o ut that B 1= O. No labor has been saved by th e inspection . In more complicated situations, such as
(D 2 + 4)y = x sin 3x  2cos 3x , the method of inspectio n will save no work. For the equation (12)
we see, since (D 2
+ 4)e 5x
= 2ge 5x , that
is a sol ution. Finally, note that if Yl is a solution of
fCD)y
= RJ(x)
and Y2 is a solution of
f ( D )y = R 2 (x) , then YP = Yl
+ Y2
is a solution of
f(D) y
= R J(x) + R2(x) .
It follows readily that the task of obtaining a particular solution of
f(D)y = R(x) may be split into parts by treating separate terms of R(x) independently, if convenient. See the examples beJow. T his is the basis of the "method of superposi tion," which plays a useful role in applied mathematics.
EXAMPLE 8.9 Find a particular sol ution of
(D 2 Since (D 2


9) y = 3ex
+x 
sin4x.
9)e X = Sex, we see by inspection that 3 x
YJ =  ge
(13)
148
Chaplel' 8
Nonhomogeneous Equaliol1S: Undele rmilled Coefficiellls
is a particular solution of
In a similar man ner, we see that Y2 = 
§x satisfies
(D 2  9)Y2 =
X
and that . 4 Y3 = 25I S1l1 X
satisfies
( D 2  9)Y3 = sin4x. Hence Yp
3x = ge 
1.+1·4 ij"x 25S lI1 X
is a sol uti on of equation (13).
•
EXAMPLE 8.10 Fi nd a particular solution of (D 2
At once we see that YI =
+ 4)y =
sin x
+ sin 2x.
(14)
4sin x is a sol uti o n of 2
( D +4)YI = sinx.
Then we seek a sol ution of (D
2
+ 4)Y2 = si n 2x
by the method of undetermined coeffic ients. Because we put Y2 = Ax sin 2x
(15) In
= ±2i and tn' = ±2i,
+ B x cos2x
into (15) and determine that
4Acos2x  4 B sin2x = sin 2x ,
i.
from which A = 0, B = Thus a partic ul ar sol ution of (14) is
Yp =
4sin x 
ixcos2x .
•
Solution by Inspection
8.4
149
EXAMPLE 8.11 Find a particu lcu' solution of
(D 2
+ a 2 )y =
(16)
cosbx.
If b # a, then a particular solution of the form y = A cos bx will exist. It follows from (16) that
(_ b 2 A and A = (a 2

b2 ) 
I.
+ a 2 A) cos bx
= cos bx
A particular solution of (16) is
y = (a 2
b2 )  1 cos bx.

If b = a, then eq uation (16) becomes (D
2
+ a 2 )y =
(17)
cos ax,
and no fu nction of the form A cos ax is a particulcu' solution, si nce the operator D2 + a 2 will ann ihil ate the f unction A cos ax . However, a solution of th e for m Ax cos ax + Bx sin ax exists. Upon substitution into (17) we requi re that
2aA sin ax
+ 2a B cos ax =
an equation that is satisfied only if A = 0 and B =
x
cos ax,
do . Therefore,
.
y =  sm ax 2a
(18)
a particu lar solution of ( 17).
•
We have seen in this ex.ample an important distincti on between the cases b # a and b = a . In a phys ical app lication considered in Chapter 10, the prese nce of a sol utio n of the form given in (18) res ults in a p henomeno n ca lled resonance. At this point we need only notice that th e sol ution in (18) will be oscillatory in character, but the amplitudes of the oscillat ion will become in creasingly large as x increases.
Exercises I.
Show th at if b
#
a, then (D
2
+ a 2 )y =
has a particular solution y = (a 2 2.

sin bx
b 2 )  1 sin bx .
Show that the equation (D2
+ a 2 )y =
si n ax
has no solution of the form y = A sin ax , with A constant. Find a particular solution of the equation.
150
ChapTer 8
NOllhomogeJIeous Equations: Undet ermined Coefficients In Exercises 3 through 50, find a particular solution by inspection. Verify your solution.
3.
(D 2 + 4)y
4. 5.
(D 2 + 9) y (D 2 + 4D +4)y
= 8.
29.
6.
(D 2 +2D3)y=6.
30.
7.
(D 3 3D+2)y=7.
31.
S.
(D 4 + 4D2 + 4) y = 20. (D 2 + 4D) y = 12.
32.
(D 3  9D) y
34.
9. 10. 1l. 12. 13. 14.
IS. 16. 17. IS. 19. 20. 21. 22. 23. 24. 25. 26.
1 8.5
I
= =
27 .
12. 18.
28 .
33.
= 27. (D + 5D)y = IS . (D 3 + D) y = S. (D4  4D2)y = 24. (D 4 + D 2)y =  12. (D s  D 3 )y = 24. (D s  9D 3 )y = 27. (D 2 +4)y = 6sinx. (D 2 + 4)y = lOcos 3x. (D 2 + 4)y = 8x + 1  15e x . (D 2 + D)y = 6 + 3e 2x . (D 2 + 3D  4)y = 18e 2x . (D 2 + 2D + 5) y = 4ex  10. (D 2 l) y = 2e 3x . (D 2  I)y = 2x + 3. (D 2  l)y = cos 2x. 3
35 . 36. 37. 38. 39.
(D 2 +2D +I )y = 12e x . (D 2 + 2D + l )y = 7e 2x . (D 2  2D + l )y = 12e x . (D 2  2D + l)y = 6e 2x . (D 2  2D  3)y = eX . (D 2  2D  3)y = e 2x . (4D 2 + 1)y= 12 si n x.
42. 43.
(D 3  l) y
44.
(D 3
45.
(D 3  D)y
41.
46. 47. 48. 49. 50.
(D  l )y = in2x .
= eX + 3x . = 5e  3x . =  2x + cos 2x. 2 (D + 1) y = 4e  2x . (D 2 + l) y = 10 sin 4x. (D 2 + l)y = 6e 3x .
(4D 2 + I) y =  12cos x. (4D 2 + 4D + l)y = IS e x  5. (4D 2 + 4D + l)y = 7e  x + 2.
40.
2
(D 2 + l)y (D 2 + l)y (D 2 + 1) y
= e x .
l)y=43x 2.
= e 2x . (D4 + 4)y = 5e 2x . (D 4 +4)y = 6sin2x . (D 4 + 4)y = co 2x. (D 3  D)y = 5sin2x. (D 3  D)y = 5cos2x.
Computer Supplement The algebraic manipulations needed to solve a differential equation usin g the method of undetermined coefficients are easily perfonned using a Computer Algebra System. As an example, consider Example S.4 of Section 8.3. (D 2
+
D  2) y
= 2x
 40 cos 2x.
The following Maple session mimics the steps performed by hand in the section. >y(x) :=A+B*x+C * cos(2*x)+F*sin{2*x); y(x) := A + Bx
+ Ccos(2x) +
Fsin(2x)
8.5
CompHter Supplement
151
>yp(x) : =diff(y(x) ,x); yp(x) := B  2C sin (2x)
+ 2 F cos (2x)
>ypp (x) : =diff (yp (x) ,x) ; ypp(x) := 4 C cos(2x)  4 F sin(2x)
>ypp (x)+yp(x ) 2*y(x ) 2*x+40*cos( 2 *x ) =O;
6 C cos(2x)  6 F sin(2 x) + B  2 C sin (2x ) +2 F cos (2 x)  2 A  2 B X  2 X + 40 cos (2 x) = 0 >co l lec t (", [x , cos (2*x) , sin (2 *x) ) ) ;
(2 B2) x + (6 C + 2 F + 40) cos(2 x) + ( 6 F  2 C) sin(2x) + B  2 A = 0 >so l ve({  2*B 2=O,  6*C +2*F+40=O ,  6*F 2*C=O,B2*A=O}, {A,B , C,F}) ; [C = 6, A = ]/2, B = 1, F = 2)
We have only to substitu te the resulting coefficients back into the origi nal sol ution to obtain the desired result.
Exercises 1.
Use a computer to solve a selection of exercises from the chapter.
Variation of Parameters
9. 1
9
Introduction In Chapter 8 we so lved the nonhomogeneous linear equation with constant coefficients
(boD"
+ hi D
I1

1 + ...
+ b,, _1D + b,,)y = R (x)
(1)
by the method of undetermined coefficient . We saw that this method would be applicable only for a certain class of differential eq uations: those for which R (x) itself was a solution of a homogeneo us linear equation with constant coeffi cients. In this chapter we study two method that carry no such restrictions. In fact, much of what we do wi ll be app li cable to linear eq uations with variable coefficients. We begin with a procedure by D' Alembert that is often call ed the method of reduction of order.
9 . :2
Reduction of Order Consider the general secondorder linear equation
y"+py'+qy =R.
(1)
Suppose that we know a solution y = YJ of the correspo ndi ng homogeneoLl s equation
y"
+ py' + qy = O.
(2)
Then the introduction of a new dependent variable v by the substituti on Y
= YIV
(3)
will lead to a solution of eq uation (1) in the following way. From (3) it follows that
y' = YIV' + y;v, Y" = YI v" + 2y; V i
152
+ Y;'v,
9.2
Reduction of Order
153
so substitution of (3) into (1) yields
y,v"
+ 2y;v' + y;'v + py ,v' + py;v + qy ,v =
R,
+ (2y; + PY I)v' + (y;' + py ; + qy ,)v =
R.
or
YIV"
(4)
But y = y, is a solution of (2). That is,
y;'
+ py; + qy,
= 0
and equation (4) reduces to
y,v" + (2y; Now let v' =
W,
+ py,)v' =
R.
(5)
R,
(6)
so equation (5) becomes y,w'
+ (2y; + py,)w =
a linear equation of first order in w. By the usual method (integratin g facto r) we can find w from (6). Then we can get v from v' = w by an integration. Fi nally, y = y,v. Note that the method is not restricted to equations with constant coefficients. It depends only upon our knowing a single nonze ro solution of equation (2) . For practical purposes, the method depends also upon our being able to effect the integrations.
EXAMPLE 9.1 Solve the equati on
y" _ y = eX.
(7)
The co mplementary function of (7) is
Ye =
CI eX
+ C2e  x .
We shaIJ take the particular solution eX and use the method of reduction of order by setting
Then
and yl! = vex
+ 2v' eX + v" eX .
Substituting into equation (7) gives
v"
+ 2v' =
I.
(8)
154
Chapter 9
Variation of Parameters
Equation (8) is a firs torder linear equation in the variabl e v' . Applying the integrating factor e2x yields
Thus
e2x v' =
!e
2x
+ c,
(9)
where c is an arbitrary constant. Equation (9) readily gives
! + ce
=
VI
2x
,
and hence V
=
 2x
CIe
+ C2 + 2:I x ,
where Cl and C2 are arbitrary constants. Remembering that y = vex, we finally have Y = c le
x
+ C2e x + !xex .
•
Of co ur e, the sol ution to equation (7) could have been obtained by the method of undetermined coefficients. Let us now solve a problem not solvable by that method. EXAMPLE 9.2 Solve the equation (D 2
+ l )y =
cscx.
(10)
+ C2 sinx.
(11)
The complementary function is
Yc =
CI COSX
We may use any special ca e of (J 1) a the YI in the theory above. Let us then pu t
y=vsi nx. We find that
y' =
VI
si n x
y" = vI! sin x
+ 2v
+ v cosx
and l
cos X

v sin x .
The equatio n for v is
v" sin x
+ 2v' cosx = cscx ,
or
v"
+ 2v
l
cotx = csc2 x.
(12)
9.2
Put
Reduction of Order
155
= w; then equation (12) becomes
Vi
+ 2w cotx =
Wi
csc 2 x,
fo r whi ch an integrating factor is sin 2 x. Thus sin 2 x dw + 2w si n x cosx dx = dx
(13)
is exact. From (13) we get . ?
W SIl1 X
= x,
and if we seek only a particular so lution, we have 2 W = x csc X, or Vi
Hence
v=
= X csc 2 x.
f
X
2 csc x dx,
or
v =  x cotx
+ In I sinxl,
a result obtained using integration by parts. Now
y=vsi n x, so the particu lar solution which we so ught is yp = x cosx
+ sin x In I sinxl.
Finally, the complete solution of (10) is seen to be Y=
CI COSX
+ c2s in x 
x cosx
+ sin x In I si n xl·
•
• Exercises U e the met hod of reduction of order to solve the equation s in Exerc i e. I through 8.
( D 2  4D
4.
(D 2 +4)y=sinx .
(D 2
2.
(D
2
5. 7.
+ l) y = secx. (D + l )y = sec 3 x. Use y = v sinx. (D 2 + 2D + l )y = (eX  1) 2 .
8.
(D 2
6.


1) y =
5D
X 
l.
+ 6»)' =
2e
x
.
(D 2 2

3D + 2)y = (1 + e 2x )1/2.
+ 4)y =
3.
1.
eX .
156
Chapter 9
Variation of Parameters
9.
Use the substitution y = v cosx to solve the equation of Example 9.2.
= ve x
10.
Use y
11.
( D + 1) Y = csc 3 x . Take a hint from Exercise 6. Verify th at y = eX is a solutio n of the eq uation
12.
to solve the equation of Example 9.l.
2
(x  l )y"  xy'
+y =
O.
Use this fact to find the general solution of (x  I) y"  x y'
13.
+y
= I.
Observe that y = x is a particu lar soluti on of the equati on 2X 2 y" +xy'
 Y= 0
and fi nd the general sol ution . For w hat values of x is the solution valid? 14.
In Chapter 19 we shall study Bessel's differential equat ion of index zero
xl' + y' + xy = 0. Suppose that one solution of this eq uatio n is given the name lo(x). S how that a seco nd solution takes the form Jo(X)
15.
dx / x [Jo(x)]
2.
One so lution of the Legendre d iffere ntial equation
(1  x 2 )y"  2xy'
+ 2y
= 0
is y = x . Find a seco nd sol ution . I
9.3
I
Variation of Parameters In Secti o n 9.2 we saw that if Yl is a solution of the homogeneous equation
y"
+ p(x)y' + q(x)y =
0,
(1)
we can use it to determine the genera l sol ution of the nonhomogeneous eq uation
y"
+ p(x)y' + q(x)y
= R(x).
(2)
In using th e method of reduction of order, we proceeded as follows. Because is a solu tion of 0), the fu nction C1YI is also a sol ution for an arbitrary constant C I· We re pl aced the constant C l by a function vex) and considered the possibility of the existe nce of a solution of equation (2) of the form v . Yl. Th is led us to a firstorder linear equation in the variable Vi that we were able to solve.
Y1
9.3
Variation of Parameters
157
Suppose now that we know the general solution of the homogeneous equation (l). That is, su ppose that
Yc =
CIYI
+ C2Y2
(3)
is a solution of (1), where YI and Y2 are linearly independent on an interval a < x < b. Let us see what happens if we replace both of the constants in (3) with functions of x . That is, we consider
(4) and try to determine A(x) and B(x) so that AYI + BY2 is a soluti on of equation (2). Note that we are involved with two unknown functions A(x) and B(x) and that we have on ly insisted that these functions satisfy one condition: the function in (4) is to be a solution of equati on (2). We may therefore expect to impose a second condition on A(x) and B(x) in some way which would be to our advantage. Indeed , if we imply impose the conditi on B(x) == 0, we will be dealing with the method of reduction of order. Actually, we impo e a somewhat different co ndition on A and B. From (4) it follows that (5) Rather than becoming involved with derivatives of A and B of higher order than the first, we now choose some particular func ti on for the expression A'Y I
+ B'Y2.
Technically, we could let this function be sin x, eX, or any other suitable function. For simplicity we choo e
A' y,
+ B'Y2 =
(6)
0.
It then follows from (5) that y"
= Ay;' + By; + A'y; + B' y; .
(7)
Because y was to be a soluti on of (2), we substitute from (4), (5), and (7) into equation (2) to obtain
A (y;'
+ py; + q y ,) + B(y; + py; + qY2) + A' y; + B' y;
= R(x) .
But y, and Y2 are sol ution of the homogeneous eq uation (1), so that finally
A' y ; + B' y; = R(x).
(8)
Equations (6) and (8) now give us two equations that we wish to so lve for A' and B' . This so lution exists provided that the determinant
I~; ~~I
158
Chapter 9
Variatioll of Parallleters
does not vanish. But this determinant is precisely the Wronskian of the functions y, and Y2, and we presumed that these two functions were linearly independent on the interval a < x < b . Therefore, the Wronskian does not vanish on that interval and we can find A' and B'. By integration we can now find A and B . Once A and 8 are known, equation (4) gives the desired y. This argument can easily be extended to equations of order higher than two, but no essentially new ideas appear. Moreover, there is nothing in the method that prohibits the linear d(fferentia.l equation involved from having variable coefficients.
EXAMPLE 9.3 Solve the equation (D 2
+ I)y = secx tan x .
(9)
Of course,
Ye =
c,cosx +
C2
sin x.
Let us seek a particular sol ution by variation of parameter. Put
y = A cosx
+ B sinx ,
( 10)
from which
y' = A sinx + 8 cosx + A' cosx + 8' sin x. Next set
A' cosx
+ 8' sinx =
0,
(I 1)
so that
y' =  A sin x
+ B cosx.
Then
y" = A cosx  8 sin x  A' sin x
+ 8 ' cosx.
(12)
Next we eliminate y by combining equations ( 10) and (12) with the original equation (9) . Thus we get the re lation A'sinx
+ 8'cosx =
secxtanx.
( 13)
From (13) and (11), A' is easily eliminated. The result is
8' = tanx. so that 8 = In I secxl.
( 14)
9.3
Varia/ion of Parameters
159
in whi ch the arbitrary constant has been disregarded because we are seeking only a particular solution to add to our previously detenruned complementary function Ye' From equations (13) and (11) it also follows that
AI =  sin x ec x tan x , or
A' =  tan 2 x. Then
A= 
!
2
tan x d x =
!
2
(1  sec x) dx,
so that
A = x  tanx,
(15)
again disregarding the arbitrary constant. Returning to equation (10) with the known A from (15) and the known B from (14) , we write down the particular solution
Yp
=
(x  tanx) cosx
+ sinx In I secx l,
or YP = x cosx  sinx
+ sin x In I secxl ·
Then the general solution of (9) is
y
= c, cos x + C3 sin x + x cosx + sin x In I secxl ,
(16)
where the term ( sin x) in YP has been absorbed in the complementary function term C3 sin x, since C3 is an arbitrary constant. The solution (16) can, as usual, be verified by direct substitution into the original differential equation .
•
EXAMPLE 9.4 Solve the equation (D 2  3D+2) y =   X 1 + eHere
(17)
160
Chapter 9
Variation of Param eters
so we put ( 18) Because
we impose the condition ( 19)
Then (20) from which it follows that (21 ) Combining ( 18), (20), (2 1), and the origin al equaLion (17), we find that , ,_
1
, ? ,_
A e" + 2B e =   x I
+e
Elimination of B' from eq uations (19) and (22) yie lds
, ,_ 1 Ae' =    1 + e x '
e x 1+ e X
A' =    Then
Similarly,
B'e 2x = _ _ _ 1+ e x
'
so that
or
Then, from (18), Yp = eX In (l
+ e X)
 eX
+ e2x In ( I + e X).
(22)
Solution of y"
9.4
+ .Y = f(x)
161
The term (eX) in y" can be absorbed into the complementary function. The general solution of equation (17) is
y
9.4
Solution of y"
= C3 ex + C2 e2x + (eX + e2x ) In (1 + e X).
+y =
•
f(x)
Consider next the equation
(D 2
+ l)y
(1)
= f(x),
in which all that we require of f(x) is that it be integrable in the interval on which we seek a solution. For instance, f(x) may be any continuous function or any funct ion with only a finite number of finite discontinuities on the interval a ~ x ~ b. The method of valiation of parameters wi ll now be appl ied to the solution of (1). Put y
= A cos x + B si n x .
(2)
Then
y' =Asinx+Bco x +A'cosx +B' inx , and if we choo e
+ B' sin x
A' cosx
= 0,
(3)
we obtain
y" = A cosx  B sinx  A' sinx
+ 8 ' cosx .
(4)
From (1), (2), and (4) it follows that
+ B' cosx =
 A' sinx
f(x) .
(5)
Equations (3) and (5) may be solved for A' and B', yielding A' = 
lex) sinx
and
B' = f(x) cosx.
We may now write A =
1x 1
f(f3) sin f3 d[3,
(6)
1I
x
B =
f([3) cos f3 df3,
(7)
162
Chapter 9
Variation of Parameters
for any x in a ::s x ::s b. It is here that we use the integrability of f (x) on the interval a ::s x ::s b . The A and B of (6) and (7) may be inserted in (2) to give us the particular so lution yp =  cos x
=
l'
l'
f (f3) si n f3 df3
+ sin x
a
l'
f (fJ) cos f3 dfJ
a
f(f3)(si n xcosf3cosxsin fJ)df3.
Hence we have
y" =
l
(8)
x
f(f3) sin (x  f3) d{3,
(9)
and we can now write the general sol ution of equation (1):
1 x
y=
CI
cosx
+ C2 sinx +
f({3) sin (x  {3) d{3.
( 0)
a
• Exercises In Exercises I through 18, use variatio n of parameters.
3.
(D 2  l )y (D 2 + l )y (D 2 + l )y
4.
(D 2 + 2D
l.
2.
5.
10. II.
= cscx. + 2)y = e  x cscx . (D 2 + 1)y = sec 3 x.
6. (D 2 + l)y 7.
= eX + 1. = cscx cotx.
= sec4 x.
+ l)y = tan x . (D + I)y = tan2 x . (D 2 + 1) y = sec x csc x. CD2
12.
(D 2 + 1) y = sec 2 x csc x . CD2  2D+ l )y = e2x (e'+l)2. (D 2 3D+2)y = e2Xj Cl +e h ).
14.
(D 2  3D + 2)y = cos (e  X ). (D 2  l)y = 2(1  e 2x ) 1/2.
15.
(D 2
16.
CD  l)(D  2)(D  3)y
13.

l )y
= e 2x sin ex. = eX.
17. y"l  y' = x. 8. 9. J8. y'" + y' = tan x. 19. Observe th atx and eX are solutions of the homogeneolls equatio n associated with 2
(1  X)y"
+ xy' 
Y = 2(x  l}e x .
Use this fact to so lve the nonhomogeneo us equation. 20.
Solve the equation
y" _ y = eX by the method of variat io n of parameters, but instead of setting A' YJ +B' Y2 = k, for constant k.
oas in equation (6) of Sec tion 9.3, choose A' Yl + B' Y2 = 2 1.
App ly the sugge tion of Exercise 20 to Exerc ise 5.
9.4
22.
Solution of yO
+y
= f(x)
163
Let YI and Y2 be solutions of the homogeneous equation associated with Y"
+ p(x)y' + q(x) y =
f(x).
(A)
Let W(x) be the Wronskian of YI and Y2, and assume that W(x) =f:. 0 on the interval a < x < b. Show that a particular so lut ion of equation (A) is given by
l
Yp 
a
X
fCtn[ Y, CB )Y2(X)  YI(x )Y2CB)] df3 W(f3 ) .
(B)
23 . The conditions of Exercise 22 imply that (C)
and
y{
+ Py~ + qY2 = O.
CD)
If we multiply equation (C) by Y2 and equation (D) by YI and then subtract the two equations, we obtain
(Y2 Y;'  YIY{ ) + p(Y2Y;  Y IY~ ) = O. From this equation show that the Wronskian of YI and Y2 can be written
w(x)
= c exp ( 
f
(E)
P dx ),
where c is constant. Equation (E) is known as Abel's formula. 24.
Use Abel's formula to show that if W (xo) = 0 for some Xo on the interva l b, then W(x) == 0 for all a < x < b.
a < x
0 such that
I ~~I ~ K, fo r every point in T. Since (x, )' *) is in T, it follows that I/( x,
)' 1) 
I(x, Y2)1 =
1
~~ (x, )'*)I'IYI  Y21,
I/(x , YI)  I(x , Y2)1 ~ K IYI  )'21,
(1)
for every pair of poi nts (x , YI) and (x, Y2) in T. T he inequality (1) is call ed a Lipschitz condition for the f unction I . We have shown that under the hypotheses of our existence theo rem, the L ipschi tz co ndition (1) holds for every pair of points (x, )'1) and (x, Y2) in T . In the proof in Section 13.4 we shall actually use the Li psch itz condition rather th an the hypothesized continuity of af/ ay . Th us, we could restate the existence theorem in terms of condition (1) in stead of assuming that aflay is continuous in T .
11 3.41 A Proof of the Existence Theorem One hypothesis of the ex istence theorem of Section 13.2 is that I is co ntinuous in the rectangle T. It follows that f mu st be bounded in T. Let M > 0 be a num ber such th at I/(x. y)1 ~ M for every point in T . We now take h to be the small er of the two num bers a and b 1M, and de ll ne the rectangle R to be the set of points (x, y) fo r which
Ix 
xo l ~ h
and
Iy 
Yol ~ b .
Clearly, R is a subset of T . As indicated in Section 13.2, we now cons ider the sequence of f unction s Yl/(x) = Yo
+
1' '0
and prove the fo ll ow ing lemma.
J(t, YIII (t)) dt
(1)
13.4 A Proof of the Existence Theorem
Lemma 13.1
247
If lx  xo l .:::: hi then Iy,,(x)  Yol .:::: b, forn =1 ,2,3, .... The proof of this lemma wi ll be accomplished by induction. First of all, if Ix  xo l .:::: h, we have
IYl (x)  yol =
1 1~r f(t, Yo) dt l
.:::: M
1
lor dt l
.:::: M ix  xol .:::: Mh .:::: b. If we now as sume that for Ix  xo l .:::: h, IYk(X)  yol .:::: b, it follows that the point [x , Yk(X)] is in R so that If(x, Yk(x)) 1 .:::: M. Thus
IYk+ l (x)  yol .::::
11~( f(t, Yk(t)) dt l
.:::: M
1
l or dtl
.:::: Mh .:::: b.
By induction we can now assert the validity of the lem ma. Lemma 13.1 may be stated in a slightly different way: If Ix  xol .:::: h, then the points [x, y,,(x)], n = 0 , 1,2, .. . , are in R . The Lipschitz condition of Sectio n 13.3 may now be Ll sed to deduce the fo llowing lemma.
Lemma 13.2
!f lx  xol .:::: hi then If(x , Yl/(x))  f(x , YII  I(x)) 1 .:::: KIYI/(x)  Y,, _ I(x)l, forn = 1,2,3, .... We are now in a position to give an inductive proof of still another lemma.
Lemma 13.3
Ifl x  xol .:::: h, then Iy,, (x)  YII  l(x)1 .:::: for n = 1,2,3, ....
M K"  l lx  xo l" M K"  l h" n! < n!
248
Chapter 13
The Existence and Ul1 iqueness of So{UlioHS
For the case n = 1, we have f ro m the proof of Lemma 13 .1, IYI (x)  yo l
s
M ix  xo l·
Ass uming that IYII  I (x)  YII  2(X) I S
M K II  2 Ix  xo ill  I .:..:.~
(2)
(n  I)!
we mu st now show that
M KII  1Ix  xo lll IYII(x)  YII  I (x)1 S '  
n!
We will prove th is for the case Xo S x S Xo IYII(x)  Y,,  I (x) 1
+ h.
From Lemma 13.2 we have
= 11~r [f(t, YIII (t)) S
10'
S K
f(t, YIl  2(t ))] dt l
If(r, YII I (t»  f(t, YII  2(t»
1:
I dt
IYII  I (t)  YII  2(t) I dt .
U ing the hypothesis (2), we conclude that II
Iy//(x) 
YII  I (x) 1 S
MK 
1
1
(n  1).
1" (t 
XO)// I
dt,
Xo
or
(3) For the case Xo  h S x S xu, the same type of argum ent wi ll yield the same resul t. T he proof of Le mm a 13.3 is thus complete. To uti lize the res ults of Lemma 13.3, we now compare the two infi nite series 00
L
00
[Y//(x)  Y//I (x) ]
,, =1
and
L
11 = 1
M KII 1h ll n!
T he second of these series is an absolutely convergent seri es. M oreover, by Lemma 13.3, tbe second series dominates the firs t series. Hence, by the Weierstras M test the series 00
L [YII(x) 11=1
YII  I (x) ]
(4)
/3.4
A Proof of the Existence Theorem
249
converges absolutely and uniform ly on the interval Ix  Xo I .::: h. If we consider the kth partial sum of the series (4) k
L [Yn(X)  Y,,  I (x) ] = [YI (x)  Yo(x) ]
+ [Y2(X) 
YI (x)]
+ ...
11= 1
we see that k
L[YII(x)  Y,, I (x)] = Yk(X). 11 =1
That is, the statemenL that the series (4) converges absolutely and uniformly is equiva lent to the stateme nt that the sequence YII(X) converges uniformly on the interval Ix  xo l .::: h .
If we now define ¢(X) = lim YII(X) ,,t oo
and recall from the definition of the equence YII (x) th at each YII (x) is continuo us on Ix  xo l .::: h, iLfo ll ows (since the co nvergence is uniform) that ¢(x) is also con tinuo us and ¢(x)
=
lim YIl(X) = Yo
JI ~
+
lim
117 00
l
x
f(t, Y,,I (t» d t .
Xo
Because of the continui ty of f and the un iform convergence of the sequence y" (x), we may interchange the order of the two limiting processes to show that ¢ (x) is a so lution of the integral eq uali on
(5) It follows immediately upon differentiation of eq uat ion (5) that ¢ (x) i a sol utio n of the differential equation d y/dx = f(x, y) on the interval Ix  xol .::: h. Furthermore, it is clear from eq ualion (5) that ¢(xo) = Yo . Finally, since we have how n in Lemma 13.1 that IYI/(x)  Yo l .::: b for each n and for Ix  xo l .::: h, it follows that the same inequality must hold for ¢(x) = lim ll 7oo YI/(x) . That is , if Ix  xo l h, then I¢(x)  yol b. Thus we have co mpl eted the proof of parts (a), (b), and (c) of the existe nce theorem of Section 13.2.
s
s
250
Chapter J3
The Existence alld Uniqueness of Solwions
113.51 A Proof of the Uniqueness Theorem We mu t now show that the function 0, f3 + 00. Thus
W' +
0 as
f3
(2)
1
x
f3 x ,
yields
e fi f3  df3.
+ 0 , and, because
1
x is fixed,
(3)
e fJ f3x
+ 0 as
(4)
Theorem 14.10 Forx > 0 , r(x+ I) =XI(X). Suppose that n is a positive integer. lleration of Theorem 14.10 gives us f(n
+ 1) = nr(n) = n(n  l )r(/1  1)
= n(17  1)(17  2) .. ·2 . I . 1(1) = n! r(1) .
But by definition,
Theorem 14.11 Forpositive integral n, 1(11
+ I) =
n!.
°°
In the integral for I(X + 1) in (2), let us put f3 = sf with s > and t as the new variable of integration. This yields , since t + 0 as f3 + and t + 00 as
f3
+ 00,
rex + 1) = which is valid for x
+I
1
1
00
e S1sXtXsdt =
sx+ 1
eS1rr dt,
> O. We thu s obtain
r(x+l) = sx+l
00
r oo eSlrXdt,
Jo
s > 0, x > 1,
(5)
14.10
Periodic Functions
269
which in our Laplace transform notation says that L{t X } =
rex + 1) sx+1
If in (6) we putx =
s > 0, x >  1.
'
(6)
!, we get L{t 
I/2 }
rc!) s
=
1/ 2 .
But we already know that L(t 
I/2 }
= (n/s) 1/ 2 . Hence
(7)
114.10 1 Periodic Functions Suppose that the function F(t) is periodic with period w:
+ w) = F(t).
F(t
(1)
The function is completely determined by ( I ) once the nature of F(t) throu ghout one period, 0 ::: t < w, is given. If FCt) has a transform, L(F(t)} =
10
00
e· 2, graph F(r) and F'(t). Find L{F(t)}. Find L{F'( r)} in two ways.
10.
11.
_ ?
2s s 15
8s
S
(7r) 1/2 
,
s
>
S
>
o.
LU 2 sin kt}.
For the function H(t ) = t + 1, = 6, repeat Exercise 9.
o ::: t ::: 2, t > 2,
Define a triangularwave function T (f , c) by
T(t, c) = t , = 2c  t,
0::: t ::: c,
c < t < 2c;
T (t+2c, c) = T(t , c) . Sketch T(t , c) and find its Laplace transform. 12.
Show that th e delivative of the function T(t, c) of Exercise 11 is, except at certain points, the function Q(r , c) of Example 14.8, Section 14.10. Obtain L{T(t , c)} from L{Q(t , c)}.
13.
Find Lflsinkt l}.
14.
Find L{I cos kt I} .
14.10
Periodic FW lclions
273
15 . Define the function G (t) by
o~
G(t) = e', G(t
+ c)
r < c, t ::: O.
= G(t),
Sketch the graph of G(t) and find its Laplace transform. 16.
De fi ne th e funcLion Set) by
o~
S(O = I  t,
S(t
17.
+ 1) =
set),
I :::
O.
Sketch the graph of S(t) and find iLS Laplace transform. Sketch a halfwave rect ificat ion of the fu nctio n sin wt, as descri bed below, and find its transform.
o~
F (t) = sin wt,
F
rc
I ~ ,
w
n 2rc  < t < ; W W
=0,
18. 19.
t < I,
(1 + 2:)
= F(t) .
Find L {F(t) } where F(t) = r fo r 0 < r < wand F(t + w) = F(t). Prove that if L{ F(t)} = f(s) and if F(t) / I is of clas A,
1
00
L{F (r) / ,) =
f(f3)df3 .
Inverse Transforms
15
115.1 1 Definition of an Inverse Transform Suppose that the function F(t) is to be determ ined from a differential equation with initial conditions. The Laplace operator L is used to transform the ori ginal problem into a new problem from which the transform f (s) is to be fou nd. If the Laplace tran sformation i to be effective, the new problem mu st be simpler than the ori ginal problem . We first find j(s) and then must obtain F(r) fro m f(s). It is therefore desirable to deve lop methods for findi ng the object functio n F(t) when its tra nsform j(s) is known. If L{F(t)} = j(s),
(1)
we say that F(r) is an inverse Laplace trallJjorl17, or an inverse transform, of j(s) and we wri te F(t) = L  I (f(s) }.
(2)
Since (1) means that
10
00
e sl FCt) dt = j(s),
(3)
it fo Hows at once that an inverse transform is not unique. For example, if FI (t) and F2 (t) are identical except at a discrete set of po ints and differ at these point , the value of the integral in (3) is the same for the two functions; their transforms are identical. Let us employ the term nulljul1ction fo r any function N(t) for which
['0
Jo N(t)dt =
0
(4)
for every positive to . Lerch's theorem (not proved here) states that if L{FI (t)} = L{6(t)}, then F I (t)  F2 (t) = NCt). That is , an inverse Laplace transform is unique except for the addition of an arbitrary null function. The only continuous null function is the zero function . If an j(s) has a continuous inverse FCI), then F(t) is the only co ntinuous inverse of j(s) . If
274
15. ]
Definition of an In verse Tral'1~fo/'m
275
f (s) has an inverse F I (t) co ntinuous over a specifi ed closed interva l, every inverse that is also continuous over that interval is identical with FI (t) on that interval. Essentially, inverses of the same f (s) differ at most at their points of di scontinuity. In applications, fa ilure of uniqueness caused by add itio n of a null function is not vital, because the effect of that null function on physical properties of the solution are null. In the proble ms we treat, the inverse F (t) is required either to be continuous for r ::: 0 or to be sectionally continuous with values of F(t) at the points of discontinuity specified by each probl em. The F(r) is then unique. A crude but sometim es effective method for fi nding inverse Laplace transforms is to consu'lI ct a table of transforms (see the table at th e end of thi chapter) and then to u e it in reverse to find inverse transforms. We know from Exerc ise I , Section 14 .3, that s (5) L{cosk!} = 2 2' S
+k
Therefore,
L_ I
{
?
s
S } + k? = co
kr.
(6)
We shall refi ne the method above, and actuall y make it quite powerful, by developing theorems by which a given f(s) may be expanded into co mpo nent parts whose inve rses are known (found in the table). Other theorems wi ll pennit us to write f(s) in alternative forms that yield the desired inverse. T he most fundamental of such theorems is the one that states that the inverse transformation is a I i near operation.
Theorem 15.1 1fci and C2 are constants, L  I {cdl (s)
+ c2h
(s») = e lL  J U l (s) } + c2L  I {j2 (s) }.
Next let us prove a simple but extre mely useful theorem o n the manipu lati on of inverse transform . From
1
00
f(s) = we obtain
1
e SI F(t) dt,
(7)
00
f(s  a) =
=
10
e  (s  a)1
F(t) d t
e"I [eal
F(t) ] dt .
00
Thus, f rom L  I U(s») = F(t) it follows that L  I U(s  a)J = eal F(t), or
(8) Equation (8) may be rewritten with the exponent ial transferred to the other side of the equation. We thus obtain the fo ll owing result.
276
Chaprer 15
In verse Transforills
Theorem 15.2 L  I {f(s)} = e ol L  I {f(s  a)}. EXAMPLE 15.1 Find L 
{? 15 } . First complete the square in the denominator, s + 4s + 13
I
LI{
}=L + 15 + 9 } . Since we know that L I { k 2 } = sin kt , we proceed as foll ows: s + k 15
s2 + 4s
I
+ 13
{
2)2
(s
?
L I { ,\"2
15 }_5L
+ 4s + 13
I {

(s
3 }_5e
+ 2) 2 + 9
2I

L 1
{_3_} + s2
9
21
= 5e sin 3t ,
in which we have Ll sed Theorem 15.2.
•
EXAMPLE 15.2 Evaluate L 
I
{
2
s
S + 1 } . We write + 6s + 25
L
s+1
I {
s2
+ 6s + 25
}L
I {

(s
s+ 1
+ 3)2 + 16
}
.
Then L
I
L2:. ~ ~
25 }
= e31L  I = e
31
{ s~ ~ ~6}
[L L2~ 16}  ~ L L2: 16}]  I
 I
= e 3r (cos 4t  ~ sin 41) .
• Exercises In Exerc ises I through 10, obtain L  1 (f(s) } from the given f(s).
1. s2
2.
+ 2s + 10·
1 4s + 8 3s S2 + 4s + 13
4.
1
5. S2
S2 
3.
s .\"2+6s+13
+ 4s + 4 S
6. S2
+ 4s + 4
•
/5.2
7.
8.
2s  3 S2  4s + 8' 3s + I S2 + 6s + 13 '
s2
Show that for n a nonnegative .integer,
12.
Show that for m >  I,
L 1
{ + I} (s
a)II1+1
tille  a!
  

r(m
+ 1)'
Show that L
14.
(s + 4) 3'
10.
11.
13.
277
+3
2s
9.
Partial Fractions
I
{ I } _eI (s+a) 2+ b 2
=
b
ar
sin bt
Show that
L1{ (s + a)s? + b 2 }= ~ear(bCosbtasinbt). b 15.
For a > 0, show that fro m L  I (f(s)} = F(t) it follows that L  I (f(as)} =
~F (~) .
16. For a > 0, show that from L  I (f(s)} = F(t) it follows that 1 L  I {f(as + b)}=~exp
115. '21
(bt) (t) ;; F
~
.
Partial Fractions In using the Laplace transform to solve differential equations, we often need to obtain the inverse tran sform of a rational fraction N(s) D(s)
0)
The numerator and denominator in (1) are polynomials in s and the degree of D(s) is larger than the degree of N(s). The fraction (1) has the partial fraction expansion used in calc ul us. Because of the linearity of the inverse operator L  I, the partial fractions expansion of (1) permits us to replace a complicated problem in obtaining an inverse transform with a set of simpler problems.
278
Chapter l5
Inverse Transforms
EXAMPLE 15.3 Obtain L 
I {
6
S2 
s3
+ 4s 2 + 3..1'
} , Since the denominator is a product of distinci
linear factors , we know that constants A, B , C exist such that
6 s3+4s +3s
s2  6
S2 
::::= 2
ABC
s(s+ 1)(s + 3)
= 
s
+  + s+ 1
..1'+3'
Multiplying each term by the lowest common denominator, we obtain the identity
..1'2  6
= A(s + 1)(s + 3) + Bs (s + 3) + Cs(s + 1) ,
from which we need to determine A, B , and C , Using the values successively in (2), we get
s = 0:
6 = A(l)(3) ,
s=  1: s=  3:
5 = B(  1)(2) ,
!. Therefore,
6 2 + 4s + 3s
 2
..1' 2 
:_::__ = _ ,\'3
I
= 0, 1 ,3
3 = C(3)(2) ,
from which A = 2, B = ~ , C =
Since L 
..I'
(2)
{~} s
= I and L 
I
s
~
+ _2! _. s+I s+3
+_
2_
{ _ I__ } = e CII , we get the desired result,
s
+a
• EXAMPLE 15.4
5s3  6s  3 } Obtain L  1 { 3 ? ' Since the denominator contains repeated linear s (s + 1)facto rs, we must assume partial frac tions of the form shown: 5s 3  6s  3 ..1'3(..1'+ 1)2
AI s
:::,. = 
A2
A3
BI
8
+ s 2 + S3 +  + (s+ 21)2 ' s+1
(3)
Corresponding to a denominator factor (x  y y, we mu st in general ass ume r partial fractions of the form AI xy +
A2
(Xy)2
AI'
+ .. ,+(x y ),
/5.2
Partial FraClions
279
From (3) we get
5s 3  6s  3 = A IS2(S + 1)2 + A2S(S + 1) 2 + A 3(S
+ 1)2 + B Is\s + 1) + B2 s 3 ,
(4)
which must be an identity in s. To get the necessary five equations for the determination of A I , A 2 , A3, B I , B2 , two elementcu'y methods are popular. Specific values ofs can be used in (4), or the coefficient oflike powers ofs in the two members of (4) may be equated. We employ whatever combination of these methods yields simp le equations to be solved for A I , A 2, ... , B2 . From (4) we obtai n S
3 = A3(1),  2 = B2(  I ),
= 0:
s = I :
coeff. of S4: coeff. of s3: coeff. of s :
O=AI+B I ,
+ A2 + BI + B2 , 6 = A2 + 2A 3· above yield Al = 3, A2 = 0, A3 = 5 = 2A l
The equations Therefore, we find that
 3, BI
=
 3, B2
=
2.
• EXAMPLE 15.5 Obtain L  1
{
?
s(s
16
+ 4)
2
} ' Since quadratic factors require the corresponding
partial fractions to have li near numerators, we start with an expansion of the form
16 A = s(s2 +4)2 s
c:,::
+
BIs + C I 8 2 +4
B 2 s +C2 + ...,.;;....,:;(8 2 +4)2'
From the identity
16 = A(S2 +4) 2 + (Bls
+C
I
it is not difficult to find the values A = I , BI = We thus obtain
L
1
Lcs ! 21
4)2 } = L 
I
+ (B 2s + C2)S, I , B2 = 4, C I = 0, Cz = O.
)S(S2 +4)
{~

s2:
4  (S 2
= I  cos 2t  t sin 2t .
~4)2 }
•
280
Chapter 15
Inverse Transforms
• Exercises In eac h exerci se, find an inverse transform of the given
1. 2.
3. 7.
8. 9.
10.
S2
+ as
4.
+1 s(s + 1)2'
5.
4s +4 S2(S  2) .
s+2 s2  6s + 8 2S2 + 5s  4 S3
+ s2 
6.
2s
f (s). 2S2
1 s3(s2
+ 1) '
5s  2 S2(S
+ 2)(s
 1)
1 (s2
+ a 2)(s2 + b2) ,
a 2 =f:. b 2, ab =f:. O.
s a 2 =f:. 2 (S2 + ( ) (s 2 + b2)' S2 a 2 =f:. (S 2 + a 2) (s2 + b2) ,
b2, ab =f:. O. b2, ab =f:. O.
115.31 Initial Value Problems Because of Theorem 14.6 of Section 14.7, the Laplace operator will transform a linear differential equation with con tant coefficients into an algebraic equation in the transformed f uncti on. If upon solving this algebraic equation for the transformed fu nction we are able to obtain the inverse transform, we may have a solution of the original differential equation . Several examples will now be treated in detail so that we can get some feeling for the advantages and disadvantages of the transform method. One fact is apparent from the nature of the tran form of derivatives : This method is most readily applied if the appropriate initial conditions are given along with the differential equation. If they are not, the algebra is more complicated.
EXAMPLE 15.6 Solve the initial value problem y"(t)
+ yet) = 0;
yeO) = 0, y' (0) = I.
(1)
Applying the Laplace transform to both ides of the differential equation gives
L{ y"
+ y}
= 0,
and because of the linearity of the transform
L{y" } + L{y} = 0.
15.3
Initial Value Problems
281
An application of Theorem 14.6 now yie lds
s2L (y(t) }  1 + L( y(t)} = 0, an equation that may eas il y be solved for L{y(t»). We have
I
L{ y(t») =
S2
+ 1.
(2)
We know that sin t is a function that satisfies (2), and it is a simple matter to verify that sin t is the sol ution of (l).
•
EXAMPLE 15.7 Solve the problem y"(t)
+ f32 y (t)
= A si n wt;
yeO) = 1, y' (0) = O.
(3)
Here A , f3, ware constants. Because (3 = 0 would make the problem one of elementary calculus and becau e a change of sign of f3 or w would not alter the character of the problem, we may assume that f3 and ware positive. Let
L( y( t )} = ues). Then L( y'(t) } = sues)  1,
L(y"(t) } = slues)  05 · 1  0, and appli cation of the operator L transforms the problem (3) into 2
?
su(05)  S + f3 u (05) =
05
2
Aw
+w
2'
from which u(05) =
s
Aw
+ 2 . 05 2 + f3 2 (05 + (32) (05 2 + ( 2)
(4)
We need the inverse transform of the ri ght member of (4). The form of that inverse depends upon whether f3 and ware equal or unequal. If w =I f3 ,
282
Chapter /5
Inverse Tramfonns
Now yet) = L  I lues)}, so for w =I f3,
yet) = cos f3t
A
+ f3(f3 2 _
(
2)
(f3 sin wt  w sin f3t).
(5)
If w = f3, the transform (4) becomes
(6) We know from equation (8) of Section 14.8 that L
1{ 2 1 2 2} = ~(Sin f3tf3tcOSf3t) (s + f3 ) 2f3
Hence, for w = 13,
y et) = cos f3t
A
+ 213 2 (si n f3t
(7)
 f3t cos f3t).
It is a imple matter to show that this function is indeed the solution of the given initial value problem.
•
Note that the initi al conditions were ati sfied automati cally by this method when Theorem 14.6 was applied. We get not the general solution with arbitrary constants still to be determi ned but that particular solution which satisfies the desired initial conditio ns. The tran form method also gives us some insight into the reason that the sol ution takes different forms according to whether wand 13 are equal or unequal.
EXAMPLE 15.8 Solve the problem X" (t)
Let L{x(t)}
+ 2x' (t) + x(t) =
x(O)
3te  ';
= 4 , x' (O) = 2.
(8)
= yes). Then the operator L converts (8) into
s2y (s)4s2+2[sy(s)
.
 4]+y(s) =
3 (s
+ 1)
2'
or
4s y es) = (s
+ 10
3
+ 1)2 + (s + 1)4 '
(9)
15.3
Initial Valu e ProbLem s
283
We may wli te 4 (s
yes) =
+ 1) + 6
(05
+ 1)2
3
+ (s + 1)4 '
or
4 yes) =
6
3
s + 1 + (s+ 1)2 + (05+
1)4'
Empl oying the inverse transform, we obtain x(t) = (4 + 6t
+ ~ t 3 )e  l.
(10)
•
Again the knowledge of ini tial conditions co ntributed to the efficiency of our method. In obtai ni ng and in using equatio n (9), those terms that came from the ini tial values x (0) and x' (0) were not combined wi th the term that came from the transform of the ri ght member of the differential equ ation. To com bi ne such terms rarely simplifies and freq uently co mplicates the task of obtaining the inverse transform . From the solu tion (10) the student ho ul d obtain the derivatives xl (t) = (2  6t
x" (t) = (8
+ ~ t2 
+ 9t 
~ t 3 ) e l ,
3t 2 + ~t 3 )e l,
and thus verify th at the x of (10) satisfies both the d ifferent ial equation and the initial conditio ns of the problem (8). Such verification not o nly checks our work but also removes any need to justify temporary assumptio n abo ut the right to use the Laplace tran form theore ms on the function x(t) during the time that the function is ti11 unkn ow n.
EXAMPLE 15.9 Solve the problem w " (x)
+ 2W' (X) + w(x ) =
x;
w(O)
=
3, w(l)
= 1 .
(1 1)
In this example the bo undary co ndi tions are not both of the ini tial cond ition type. Us ing x rather tha n t as the independent variable, let L{ w(x ) } = g(s).
We know w(O) = 3, but we also need w"(x) . Hence we put
W i (0)
(12)
in order to write the transfo rm of
w' (O) = B
and hope to determine B later by using the co ndi tio n that w (l ) =  1.
(13)
284
Chapter J5
Inverse Transform s
The transformed problem is s2g(s)  s(  3)  B
+ 2 [sg(s) 
(3) ]
+ g(s)
1
= 2'
s
from which 3(s
+ 1) + B3 (s + 1)2
g(s) =
1
+ s2(s + 1)2
(14)
But by the usual partial fractions expansion, 1
2
]
S
S2
2
1
    =  +  +   +   +
S 2 (S
1) 2
S
+ 1
(s + 1)2 ,
so
1
2
1
g(s) =      s2 s s+l
+
B2 , (s + 1)2
(15)
from which we obtain
w(x) = x  2  e x
+ (B 
2)xe  x .
(16)
We have yet to impose the condition th at w( I ) =  1. From (16) with x = 1, we get
 I
=
1  2  e
I
+ (B 
2) e l ,
so B = 3. Thus our final result is
w(x)
=x 
2  e x + xe x .
(17)
•
The problem in Examp le 15.9 may be solved effi ciently by the methods of Chapter 8. See also Exercises 23 through 44 .
• Exercises In Exercises I through 22, solve the problem by th e Laplace transf orm method . Veri fy that your so lution satisfies the differential equation and the i niti al conditions.
y' = e l ; y eO) = 2. 3. y' + y l 2. / = 2e ; yeO) = 1. 4. y'  y 2 5. y" + a y = 0; yeO) = 1, y' eO) = O. 6. y" + a 2 y = 0; yeO) = 0, /(0) = a. 7. y"  3y' + 2y = e31; yeO) = y' (O) = O. 1.
= e y eO) = O. = e I ; yeO) = 1. 21
;
15.3
+y =
e t ; yeO) = y'(O) =
285
o.
8.
y"
9.
y"  2y' = 4; yeO) = 0, y'(O) = 4.
+ y' 
initial Value Problems
10.
y"
11.
x"(t)  4x'(t)
12.
x"(t)
13.
y"(t)  yet) = 4cos t; yeO) = 0, y'(O) = l.
14.
y"(t)  6y'(t)
2y = 4; yeO) = 2, y'(0) = 3.
+ x(t)
+ 4x(t)
= 4e2t ; x(O) = 1, x'(O) =  4.
= 6 sin 2t; x(O) = 3, x'(O) = 1.
+ 9y(t)
= 6t 2 e31; y eO) = y'(O) = O.
15 . x"(r) + 4x(t) = t +4; x(O) = 1, x'(O) = O. 16.
x"Ct)  2x' (t) = 6  4t; x(O) = 2, x'(O) = O.
17.
x" (t) + .x(t) = 4e' ; x(O) = 1, x'CO) = 3. x"(t) + x' (t)  2x(t) = 6; x(O) = 1, x'(O) = l.
18. ]9. 20. 21. 22.
+ 9y(x) = 40ex ; yeO) = 5, y'(0) =  2. y"(x ) + y(x) = 4e yeO) = 0, y' (0) = o. x"Ct) + 3x'(t) + 2x(t) = 4t 2 ; x(O) = 0, x'(O) = O. x"(r)  4.x'(t) + 4x(t) = 4cos 2t; x(O) = 2, x'(O) = 5. y"(x )
X
;
In Exercises 23 through 42, use the Laplace transform method with the rea lization th at these exercises were not constructed wi th the Lap lace transform techn ique in mind. Compare your work with that done in solving the same problems by the methods of Chapter 8. T he exerc ise numbers refer to the exercises in Section 8.3 .
23 . Exercise 1.
33 . Exercise 37 .
24.
Exercise 2.
34.
Exercise 38.
25.
Exercise 3.
35.
Exercise 39.
26.
Exercise 11.
36.
Exercise 40.
27 . Exercise 14.
37.
Exerci se 41.
28.
38.
Exercise 42.
Exerc ise 20.
29.
Exercise 21.
39. Exercise 43.
30.
Exercise 22 .
40.
Exercise 44.
3l.
Exercise 23 .
41.
Exercise 45.
32. 43.
Exerci se 36. Solve the problem
42.
Exerci se 46.
x" (t)  4x ' (t)
44.
+ 4x(t)
= e2t ;
x'(O) = 0 , x( l) = O.
Solve the problem x"(t )
+ 4x(t) =
 8t 2 ;
x(O) = 3, x(irr) = O.
286
Chapter J5
Inverse Transforms
115.41 A Step Function Applications frequently deal with situations that change abruptly at specifi ed times. We need a notation fo r a function that will suppress a given term up to a certain value of t and insert that term for all larger 1. The fu nction we are about to introduce leads us to a poweIful tool for constructing inverse transforms . Let us define function aCt) by
aCt) = 0,
t < 0,
= 1,
t ::: 0.
(1)
The graph of aCt) is hown in Figure J S.l. T he defin iti on (1) says that aCt) is zero when the argument is negative and aCt) is uni ty when the argument is positive or zero. It follows that
a (t  c) = 0,
t < c,
= I. ,
t :::
(2)
c.
The a fu nction permits easy designation of the result of translating the graph of F(t). If the graph of
y = F(t) ,
(3)
t ::: 0,
is as shown in Figure 15 .2, the graph of
y = a(t  c)F(t  c),
(4)
t ::: c,
is that hown in Figure 15.3. Furthermore, if F(t) is defined for  c ::; t < 0, then F(t  c) is defined for t < c and the y of (4) is zero for t < c beca use of the negative argumen t in a(t  c). Notice that the values of F(t) for negative t have no bearing o n this resul t becau e eac h value is multiplied by zero (from the a); only the existence of F fo r negative arguments is needed.
°: :
°: ;
a(f )
o Figure 15.1
15.4 A Step FUllction
287
y
o Figure 15.2 y
0\
c
Figure 15.3
The Laplace transfonn of aCt  c) F(t  c) is related to that of F(t). Consider
1
00
L{a(t  c)FCt  c) } =
e  S1a(t  c)F(t  c)dt .
Since aCt  c) = 0 for 0 ::: t < c and aCt  c) = ] for t
1
~
c, we get
00
L{a(t  c)F(t  c)} =
e Sf F(t  c) dt.
Now put t  c = v in the integral to obtain
1
00
L{a(t  c)F(t  c)} =
e  S (C+ V )
1
F(v) dv
00
= e  CS
e S V F(v) dv.
288
Chapler 15
Inverse Tral1~fonns
Since a definite integral is independent of the variab le of integration,
1
1
00
00
e
SU
F(v) elv
=
e  S ( F(t) elt
=
L {F(t)}
=
f(s).
Therefore, we have shown that
L{a(t  c) F(t  c)} = e CS L{F(t)} = e CS f(s).
(5)
Theorem t 5.3 ff L  1{f(s ) } = F(t ) , ifc ::: 0, a nd!f F(t) be assigned values (no matter which ones)for  c .::: t < 0, L  1 (e CS f(s)} = F(t  c)a(t  c ).
(6)
EXAMPLE 15.10 Find L{ y (t ) } where (Figure ]5.4) y et) = t 2 , = 6,
0
2.
Here, d irect LIse of the definition of a transform yields
Although the integration s above are not difficult, we prefer to use the a fu nction. Since aCt  2) = 0 for t < 2 and aCt  2) = 1 for t ::: 2 , we build the y et) in the following way. The crude trial
y 6
4
Figure 15.4
15.4
A Step FUllction
289
works for 0 < t < 2, but we wish to knock out the t 2 when t > 2. Hence we wri te
Thi s gives t 2 for t < 2 and zero fo r t > 2. Then we add the term 6a(t  2) and fi nally arri ve at
y et) = t 2
t 2 a(t  2)

+ 6a(t 
(7)
2).
The y of (7) is the y of our example and, of course, it can be written at o nce after a li ttle practice with the a function. Unfortunate ly, the y of (7) is not yet in the best form fo r o ur purpose. T he theorem we wish to use g ives LI S
L{F(t  c)a(t  c)} = e cs /(s). Therefore, we must have the coefficien t of aCt  2) expressed as a functio n of (t  2) . Since _ t2
y et) = t 2

+6 =
4t
_ (t 2 
+ 4) 
4(1  2)
+ 2,
(t  2) 2a(t  2)  4(t  2)a(t  2)
+ 2a(1
 2),
(8)
from which it foll ows at once that
• EXAMPLE 15.11 Find and sketc h a f un ction get ) fo r which get) = L 
3 I
{ S 
4e·\· S2
4e  3s
}
+  . S2
We know that L  I {4 j s 2} = 4t . By Theorem 15.3 we then get L 1 {
74e  S } = 4 (t 
1)a (t  I)
and L
1
4 e  3s } {~
=4(t3)a(t  3) .
290
Chapter 15
Inverse Trallsforms
We may therefore write
get) = 3  4(t  1)a(t  1) + 4(/  3)a(t  3) .
(9)
To write get) without the a function, consider first the interval 0:::t < 1
in which aCt  1) = 0 and aCt  3) = O. We find g(t) = 3,
(10)
O:::t < l .
For 1 ::: t < 3, aCt  1) = 1 and aCt  3) = O. Hence
get) For t
~
3, aCt  1)
=3
4(t  1)
= 1 and aCt 
get) = 3  4(t  I)
=7 
3)
4t,
1 ::: t < 3.
(11)
= 1, so
+ 4(/
 3) = 5 ,
t ~ 3.
(12)
Equations (10) , (1 1), and (12) are equivalent to equation (9). The graph of get) is show n in Figure 15.5 .
•
EXAMPLE 15.12 Solve the problem x"(t)
+ 4x(t)
= 1/I(t);
x(O) = 1, x' (O) = 0,
g(t)
3
o
5
Figure 15.5
(13)
A Step FUllction
15.4
291
in which 1jJ(t) is defined by 1jJ(t) = 4r ,
O::st::Sl ,
(14)
t > 1.
=4,
We seek, of course, a so lution valid in the ra nge / ::: 0 in which the function 1jJ(t) is deflned. In this problem another phase of the power of the Laplace tran fo rm method begins to emerge. T he fact that the function 1jJ(t) in the differential eq uation has discontinuou derivatives ma ke. li se of the classica l method of undetermi ned coefficient somewhat awkward, but such di scontinu it ies do not inte rfere at all with the sim pli city of the Laplace transform method. In attacking this problem, let us put L(x(r)} = h (s). We need to obtain L{1jJ(t)} . In terms of the a fu nction we may wri te, from (14), 1jJ(t) = 4t  4(t  I )a(t  I),
( 15)
t ::: O.
From (15) it follows that
4
4e .I'
L{l/f(t)} =    . s2 s2 Therefore, the application of the operator L tra nsfo rms problem (13) in to
4 4e s s2h(s)  s  0 + 4h(s) =    , .1'2 s2 from wh ich
(16) Now
4 o (16) becomes
h(s) = _ s_·_ . S2 + 4
+ ~ __ s2 .1' 2 + 4 1_,
__
(~ s2
1_)
__
.1' 2
+4
e s
.
( J7)
Since x(t) = L \ {h(s)}, we obtain the desired solution,
xCt) = cos 2t
+t 
! sin 2t 
[(t  1) 
! sin 2(t 
1)]a(t  1).
(18)
It is easy to ve ri fy our solution . Fro m (18) it follows that
+ 1  cos2t  [1  cos2(t  l)]a(/  1), 4 cos 2t + 2 sin 2t  2 sin 2(t  l)a(t  1).
x'(t) =  2sin 2t x" (t) =
(19)
(20)
292
Chapter 15
Illverse T/,{/II.~fonns
Therefore, x (0) = I and X"(t)
X l (0)
= 0, as desired. Also, from (18) and (20), we get
+ 4x(t) = 4r 
4(r  I )a(t  1)
= t(t) ,
t ::: O.
Exercises
•
In Exerci ses I through 7, sketch the graph of the given fun ction for f ::::: O.
I.
aU  c) .
2.
a(t  I) + 2a(t  2)  3a(t4) .
4. 5.
3.
(t  3)a(t  3).
6.
7.
t
2

2
r a(t
in (t  Jr) . aCt  Jr). (t  3)2a(t  3). / 2  (t  1)2a(t  1) .
 2).
I n Exerci ses 8 through 15, express F(t ) in terms of the Ci function and find LI F(I)I.
8.
F(r) = 3,
9.
F(t) = 4, 0 < 1 < 2, = 21  I , I > 2.
= I,
10. II.
16.
F(t) = [ 2, = t  I,
12.
O< t < l, t > I.
0< t < 2, 2 < t < 3,
t > 3. 0 < t < 2, I > 2.
= 7, F(t) = e t , =0,
13.
o < t < 2, 14. F(t) = 3, t > 2. F(r) = 12, 0 < 1 < I, I < t < 2, = 3, 15. F(t) = 0, I > 2. Find and sketch an inverse tran sform of e '! 5e 3s F(t) = [ 2 ,
0 2Jr.
= sin 3r, = 0, = sin 3r , =0,
O Jr.
s s
4s
17.
Eva lu ateL  1{ e
18.
If F (t) is to be continuous for t ::: 0 and
(s
+ 2)
3}' F(t) = L 
19.
e  3s I
{ (s
+ 1)
} 3
'
eval uate F(2 ), F(5), F(7) . If F(t) is to be conti nuous for t ::: 0 and F(/) = L
_I
eval uate F(I), F(3) , F(5) .
{(l  e
2s
)(l  3e 2S )
s
2
} '
J5.4 A 51ep Flillcrion
20.
293
Prove that 1jf(t, c) = L(I)"a(t  nc) is th e same function as was used
,,=0 in Example 14 .8, Section 14.10. Note that for any spec ific t, the series is finite ; no question of convergence is involved.
21.
Obtain the transform of the halfwave rectification F (t) of sin l by writing F(t) = sint 1jf(t, n)
in terms of the 1jf of Exercise 20. Use the fact th at (_I)" sint =
in (f 1771").
Check your result w ith that of Exerc ise 17, Section 14.10. In Exercises 22 through 25, solve the problem usi ng the Lap lace transform. Verify th at yo ur solution satisfies the differential equ ation and the initial cond itions.
22.
Xl/ (f)
+ x(t ) =
F(t); x(O) = 0, x ' (O) = 0 , in which F(t) = 4,
= t 23 .
xl/(I)
+ x(t) = H (t);
0
+ 2,
x(O) = L, x ' (O)
f
= 0,
= 2t  5 ,
25 .
2,
> 2.
in wh ich
0 ~ f ~ 4,
H(t )=3 ,
24.
~ f ~
t > 4.
x " (t ) +x(t) = G(l); x(O) = 0, x ' (O) = 1, in which
x"(t)
+ 4x(t) =
G(O= l ,
0 ~ t ~ n / 2,
= 0,
f > n/2.
M(t); x(O) = x'(O) = 0 , in which M(t) = sint  aCt  2n) sin (t  2n).
26.
Compute y(in) and y (2 + in) for the function y (x) th at satisfies the initial value problem y l/ (x)
27.
+ y(x)
= (x  2)a(x  2);
yeO) = 0, y' (O) = O.
Compute x(l) and x (4) for the function x(t) th at satisfies the initi al value problem X" (f )
+ 2x' (t) + x(t)
= 2 + (f  3)a(t  3);
x(O) = 2, x ' (O) = I .
294
Chapter 15
III verse Tran.Iorms
115 .51 A Convolution Theorem We now seek a fo rmu la for th e inverse transform of a product of transforms. Given
L  I (f(s)} = F(t),
L  I {g(s) } = G(t),
(1)
in which F(t) and G(t) are ass umed to be f unctions of class A, we shall obtain a formula for
L  I {f(s)g(s)} .
(2 )
Since f(s) is the transform of F (t) , we may write
10
f(s) =
00
e Sf F(t) dt .
( 3)
Si nce g (s) is the transform of G (t),
(4) in whi ch, to avoid confusion, we have used f3 (rather than t) as the variable of integration in the definite integral. By eq uat ion (4), we have
1 00
f(s)g(s) =
e s{3 f(s)G(fJ) df3 .
(5)
O n the right in (5) we encounterti1e produ ct e s {3 f(s) . By Theorem 15.3, Section 15.4 we know that from L  I (f(s)} = F(t)
(6)
L  I {e s{3 !(s) } = F(t  fJ)a(t  fJ) ,
(7)
it follows th at
in which a is the step function d iscussed in Section 15.4 . Equation (7) means that
1
00
e ·\·/3 f(s) =
eS'F(t  fJ)a (t  (3)dt.
(8)
With the aid of (8) we may put eq uation (5) in the form
/(s)g(s) =
10
00
1
00
e SI G(fJ)F(t  (3)a(t  (3) dt dfJ.
(9)
/ 5.5
Since o:(t  (3) = 0 fo r 0 < t < fJ and 0:(1  (3) may be rewritten as
A COl/vo/Cllion TheoreJ/l
= I for f
295
? f3, eq uat ion (9)
00
10 t OO e .IIG({3)F(l (3)d t dfJ.
f(s)g(s) =
(10)
In ( 10), the integration in the tf3 plane cover the haded reg ion shown in Figure 15.6. The elements are su mm ed from t = fJ to 1 = 00 and th en from f3 = 0 to (3 =
00 .
In advanced calculu s it is shown that because FC/) and G(t) are fu nctions of class A, it is legitimate to interchange the order of integrati on on the right in equation CI 0). From Figure 15.6 we see that in the new order of integration, the elements are to be ummed from f3 = 0 to fJ = f and then from t = 0 to f = 00 . We th us obtain f(s)g(s) =
10 10
1
e s l G(f3)F(t  fJ) dfJ dl,
or f(.')g(s) =
10
00
e s l
[10
1
G(fJ)F(t  (3)d fJ ] dt.
( I I)
Since the right member of (I I ) is precisely the Laplace tran sfo rm of
10
1
G(fJ)F(1  (3)d{3 ,
we have arrived at the desired result, which is called the convolution theo rem for the Laplace tra nsform.
{3
o Figure 15.6
296
Chapter 15
Theorem 15.4
Inverse Trallsforms
ff L I (f(s)} = F(t), of class A, then
If L 
I {g(s)}
= G(t), and
l'
L  I {f(s)g(s») =
if F(t) and G(t) are/unctions
G([3)F(t  (3) d[3.
(12)
It is easy to show that the right member of eq uation (12) is also a function of class A. Of course, F and G are interchangeable in (12) because f and g enter (12) symmetrically. We may replace (12) by
l'
L I (f(s)g(s») =
F([3)G(t  (3) df3,
(13)
a result which also follow from (12) by a change of variable of integration.
EXAMPLE 15.13 Evaluate L  I {f(s)/s}. Let L  I {f(s)} = F(t). Since
we use Theorem lS.4 to conclude that
L
I {
/;s) } =
l'
F(f3) d(3.
• EXAMPLE 15.14 Solve the problem x"(t)
+ k 2 x (t)
= F(t);
x(O) = A, x/CO) = B.
(14)
Here k, A, B are constants and F(t) is a function whose Laplace transform exist . Let
L{x(t»)
= u(s),
L{F(t»)
= f(s).
Then the Laplace operator transforms problem (14) into
s2u(s)  As  B u~) =
As S2
+ k 2u(s)
= /(s),
+B /(s) + k 2 + S2 + k 2 .
(15)
To get the inverse transform of the last term in (IS), we use the convolution theorem. Thus we arrive at
x(l) = A cos kt
+ B k
sin kt
+ 1 k
i' 0
F(t  [3) sin k(3 d[3,
J5.5
or
x(t) = A coskt
BIll +
+
k
sin kt
k
A Conlio/llIion Theorelll
F(f3) sink(t  (3) df3 .
0
297
(16)
•
Verification of the so lution (16) is simpl e. Once the check has been performed, the need fo r the assumption that F(t) has a Laplace tran sform is removed. It does not matter what method we use to get a solution (with certain exceptions naturally imposed during college examinations) .i f th e va lidity of the res ult can be verified from the result itself.
•
Exercises
(n Exercises I through 3, fi nd the Laplace transform or the given convo lution integral.
1.
fal (t 
2.
10
3.
11
(3) sin 3f3 df3.
1
e (/!3J
sin f3df3.
(t  (3) 3 e!3 df3.
In Exerc ise' 4 through 7, find an inver e transform of the given j(.I") u ing the convolution theorem .
4.
5. 8.
1 s(s
+ 2)
+ 2x'(t) + x(t) = F(t);
+ 1)2'
x(O)
= 0, x/C O) = O.
Solve the problem 2
= H (t);
yeO)
= 0, y'(0) = o.
Solve the problem yll(t)
11.
I (s2
Solve the problem
y"(r )  k y(t)
10.
S2(S  2) .
7.
x ll (t)
9.
4
6.
s(s2 + k 2)'
+ 4y'(t) + l3 y(t) = F(t);
yeO)
= 0,
y' (0)
= O.
Solve the problem x"(t)
+ 6x'(t) + 9x(t)
= F(t) ;
x(O) = A , x/CO) = B.
298
Chapter J5
Inverse Trallsforms
115.61 Special Integral Equations A differential equation may be loosely described a one that contains a derivative of a dependent variable; the equat ion contains a dependent variable under a de11vative sign. An equation that contains a dependent variable under an integral sign is called an in tegral equation. Because of the convolution theorem , the Laplace transfo rm is an excellent tool for solving a very special class of integral equations. We know from Theorem 15.4 that if L{F(t)} = j(s)
and L{G(t)} = g(s),
then
L
{lot F(f3)G(t 
(3) df3 } = j(s)g(s).
(1)
The relation (l) suggests the use of the Laplace transform in sol ving equation s that con tain co nvo lution integrals.
EXAMPLE ]5.15 Find F(t ) from the integral equation F(t) = 4t  3
lot F(f3) s.in (t 
(3) df3 .
(2)
The integral in (2) is in precisely the right form to permit the use of the convol ution theorem . Let L(F(t)} = j(s) .
Then, because L{sin t} =
1 2' S 1
+
appli cation of Theorem 15.4 yields
L
{lot F(f3) sin (t 
(3) d f3 } =
s{·~\'
Therefore, the Laplace operator co nverts eq uation (2) into j(s) = ~ _ 3j(s) . s2 .1'2 + I
(3)
/5.6
Special Integral EquatiollS
299
We need to obtain f(s) fro m (3) and then F(t) from f(s). From (3) we get
(1+
S2
~ 1) f (s) = ~ ,
or f(s)
4(S2 + 1) 1 3 = S2(S2+4) = S2 + . s2+4
Therefore, F(t) = L _ I
{
1+ s 3} +4
?
2
s
,
or
F(t)=t+~sin2t.
(4)
That the F(t) of (4) is a solution of equation (2) may be verified directly. Such a check is frequently tedious. We shall show th at for the F of (4), the righthand side of equation (2) reduces to the lefthand side of (2) . Since
RHS
= 4t 
3 fal (f3
+~
in 2f3) in (t  f3) df3,
we integrate by parts and obtain the result RHS = 4t  3 [(f3
+ ~ sin 2f3) cos (t 
f3)
I
+ 3 fal (I + 3 co
2f3) cos (t  f3) df3,
from which RHS = 4t  3(t
+ ~ sin 2t) + 3 fal cos (t 
fJ) df3
+ 9 fa' cos 2f3 cos (t 
fJ) dfJ,
or RHS = t 
~ sin 2t 
3 [ si n (t  ,8)
I
+ ~ fa' [cos (t + f3) + cos (t
 3f3)] df3.
300
Chapter J5 Inverse Tral1:,jorms
Thi s leads us to the result RHS = r 
=t 
~ sin 2t + 3 sin t + ~ [ sin (t + ,8) 1 sin (t ~ sin 2t
+ 3 sin t + ~ sin 21 + ~ sin 2t 
3,8)
~ sin t
I
+ ~ sin t ,
or RHS = t
+ ~ sin2t =
F(t) = LHS,
as desired. It is important to realize that the original equation F(r) = 4t 
310
•
1
F(,8) sin (t  ,8)el,8
(2)
could equally we ll have been encountered in the equivalent form F(t) = 4t 
310
1
F(t  ,8) sin,8 d,8.
An essential ingredient for the success of the method being used is that the integral involved be in exactly the convolution integral form. We must have zero to the independent variable as the li mits of integration and an integrand that is the product of a function of the variab le of integration by a function of the difference between the independent variable and the variable of integrat ion. The fact that integrals of that form appear with significant frequency in physical problems keeps the topic of this section from being re legated to the role of a mathematical parlor game. EXAMPLE 15.16 Sol ve the equation g(x) =
~x2
lox
(x  y)g(y) ely.
(5)
Again the integ ral involved is one of the convolution type with x play in g the role of the independent variable. Let the Laplace transform of g(x) be some as yet unknown function h(z): L{g(x)} = h(z) .
(6)
Since L{ ~ x 2 } = l iz? and L{x} = I/ z2 , we may apply the operator L throughout (5) and obtain h(z)
1
h(z)
= Z3  , Z2
15.6
Special Integ ral Equations
301
fro m which
or
1 h(z) = Z(Z2
+ 1
Z2
+ 1)
Z2
=  Z(Z::2+1)
Z
Z
Z2
+ 1.
Then g (x) = L 
J {
I  Z  }, Z Z2 + 1
or g(x) = 1  cosx.
(7)
Verification of (7) is simple. For the right member of (5) we get RHS =
!x
2

= !x 2 = =
•
!x !x
2 2
L'
(x  y) (J 
[ex y)(y 
[!/ +

0

!x 2 
cos x
cos y) ely
Sin y)I  L ' (y  sin y)dy
cos y
I
+ 1=
1  cosx = LHS.
Exercises
I n Exercises
J
•
through 4, olve the given equation and verify your so lution.
1.
F(r) = 1 +
211
2.
F(t) = 1 +
l'
3.
F(t) = t
4.
F(t) = 4t 2
+
F (t  (3)e 2f3 d{3.
F({3) sin (t 
11 11
/3) d{3.
F(t  (3)e  f3 d{3. F(t  (3)e  f3 d{3.
In Exerc i e 5 through 8, solve the given equation. If sufficient time is available, verify your solution .
302
Chap/er 15
Inverse Tral1~fo r/l1S
5.
F(I) = 13 +
[' F(fJ) sin (t 
Jo
fJ) dfJ·
1
6. F(t) = 8/ 2  31 F (fJ) sin (t  fJ) dfJ· 7.
F(r) = t 2

8.
F(t ) = I
+
211 F(t  fJ) sinh 2fJ dfJ.
211
F(t  fJ) co fJ dfJ ·
[n Exercises 9 through 12, solve the given equati on.
9.
H(f) =ge2/21 f H(t  fJ)cosfJdfJ.
i +
1 Y
H(x) sin (y  x ) dx.
[0 .
H( y ) =
I I.
g(x) = e x 
12.
f y (t) = 61 +41 (fJ  t)2y (fJ)dfJ.
13.
Solve the foll owing equation for F (t ) with the condition that F(O) = 4:
21\
g(fJ) co (x  fJ) dfJ ·
F' (r) =
14.
I
+ ['
Jo
Solve the following equation for F (t) with the condition that F (0) = 0: F ' (t) = sint
[5.
+
1f
F(t  fJ) cos fJ dfJ·
Show that the equation of Exercise 3 can be put in the form el F(t) = t ef
16.
F(I  fJ) cos fJ dfJ.
+
1f
e!3 F(fJ) dfJ.
(A)
Different iate each member of (A) with respect to r and thus replace the integral equation with a differential equation. Note that F(O) = O. Find F (t) by this method. Solve the equation
1f
F(t  fJ )e !3 dfJ = I
by two methods; use the convolution theorem and the bas ic idea introd uced in Exercise 15. Note that no differential equation need be solved in this instance.
J5.7
Transform Methods and the Vibration of Springs
303
115.71 Transform Methods and the Vibration of Springs All of the application studied in Chapter 10 gave rise to linear differential equation s with initial conditions. Those initial value problems were solved in that context by using the theory of linear differential equations developed in the earlier chapters. T he same initial va lue problems may of course be solved by using Laplace transformations. We ill ustrate these techniques by reexamining ome of the problems considered before.
EXAMPLE 15.17 Solve the spring problem of Example 10.1 of Section 10.2 with no damping but with F(t) = A sin wt. As before, the problem to be olved i X"(t)
+ rPx(t)
= sin wt,
( I)
x/(O) = Va.
(2)
w ith initial conditions x(O) = Xu,
Let L(x(t) }
= u(s). Then ( I) and (2) yield s 2ues)  SXa  Va
+ {J2u(s)
=
?
S
Aw
2'
+w
or u (s) =
+ Vo Aw + :;;::=:;::;:;:S2 + {J 2 (s2 + f32)(s 2 + ( 2 )
sxo
(3)
The la t term in (3) will lead to difi'erent inverse tran sforms according to whether w = {J or w =P f3. The case w = {J leads to re 'onance, which will be discussed in Example 15.20. If w =P f3, equation (3) yield s u(05)
=
SXo + Vo 05 2 + f32
+ w 2Aw  f32
(I+ S2
f3 2
I)
. 05 2 + w 2
(4)
From (4) it follows at once that x(t) = xocosf3t
+ vo{J  I sinf3t +
Aw ? ? sinf3t f3(w  f3)
A ? s in wt. w  f3
?
(5)
That the x of (5) is a solution of problem (1) and (2) is eai Iy verified. A tudyof (5) is simple and leads at once to conclusions sllch as that x Ct) is bounded, and so on. Thenrst two terms on the right in (5) yield the natural harmonic component of the motion; the last two terms form the forced component.
304
Chapter 15
Inverse Transforms
This is the same solution that was found for the sa me problem in Section 10.2 using a very different approac h.
•
EXAMPLE 15. 18 Solve the spring problem of Example 10.2 of Seclion 10.2 using Lap lace transformations . The initial value problem is x//(t)
+ 64x(t)
= 0;
xeO)
= ~,
x/CO)
= 2 .
(6)
We let L{x(t») = u(s) and conclude at once that s2 u(s)  ~s
+ 2 + 64u(s) = 0,
from which .!.s  2
£1(.1') =
3 S2
+ 64
.
Then
xU) = ~ cos 81  ~ sin 8f.
(7)
• EXAMPLE 15.19 A spring, with spring constant 0.75 Ib/ft, lies on a long, smooth (friction less) table. A 6lb weight is attached to the pring and is at rest (velocity zero) at the eq uilibrium position. A 1.51b force is app li ed to the support along the line of action of the spring for 4 sec and is then removed. Discuss the motion. We must solve the problem 12x//(t)
+ ~x(t) =
H(t) ;
x(O) = 0, x/CO) = 0,
(8)
in which
H (t)
= 1.5 , =0,
0 < t < 4, t > 4.
Now H(t) = 1.5[1  aCt  4)] in terms of the a function of Section 15.4. Therefore, we rewrite our problem (8) in the form x " (t)
+ 4x(t)
= 8[ 1  aCt  4)];
x(O)
= 0, x /CO) = O.
(9)
15.7
Transform Methods and the Vibration o./,Springs
305
Let L{x(t)} = u (s). Then (9) yields s2u(s)
+ 4u(s)
=
~(1
,
 e 4.,),
.I'
or 8(1  e 4 \' ) u(s) = ,.1'(.1'2 + 4)
= 2
(~ s
_ _ .)'2
.I' _ )
+4
([ _
e 41).
The des ired so lution is x(t) = 2(1  cos 2t)  2 LI  cos 2(t  4)]aCt  4) .
(10)
Of course, the sol ution (10) can be b roke n down into the two re lations : fo r
o :s t :s 4,
x (t) = 2 ( I  cos 2t) ,
for
t > 4,
X(/)
= 2 [cos 2(t 
4)  co 2,],
( I [)
( 12)
if those fo rm s seem si mpl er to use. Verification of the solu tion ( 10), or ( I I ) and ( 12) , i direct. The stude nt should show that lim XCI) = li m xC!) = 2( 1  cos 8) = 2.29
( ..... 4 
( ..... 4+
and lim x'(t) = Jim x l (t) = 4si n 8 = 3.96.
1 ..... 4
1 ..... 4+
From ( 10) or ( [ [) we see that in the range 0 < t < 4 , the maximu m deviation of the weight from the starting po int is x = 4 ft a nd occurs at f = ~JT = 1.57 sec. At t = 4, x = 2 .29 fl, as hown above . For f > 4 , eq uat ion ( 12) takes over and thereafter the motion is simp le harmonic with a maximum x of 3.03 Ft. Indeed, for t > 4, maxlx(t)1 = 2)(1  cos 8)2 + sin 2 8 = 2..)2.J J  cos 8 = 2.J2.29 10 = 3.03.
•
Examp[e [5.19 is one type of prob lem fo r wh ic h the Lap lace tra nsform techniq ue is particularly useful. Such proble ms can be so lved by the older c las ica l methods, but with much less simp li c ity and dispatch.
306
Chapter 15
In verse TrCIIHjorms
EXAMPLE 15.20 So lve the problem of undamped vibrati on of a spring of Example 15. 17 in the case w = fJ. Our prob lem is to solve X"(t)
+ (3 2x (t)
= A sin (3t;
x(O) = Xo, X'(O) = Vo ,
(1 3)
with the aid of SXo u(s) =
2
+ Va AfJ + (32 + (S2 + (3 2)2
(1 4)
We already know, fro m eq uation (8), Section 14.8 that L
1
1 f (3rco {(s +I(3 ) }=3(sin(3 2fJ 2
(3t).
2?
The refore, (14) leads us to the sol utio n x(t) = xa cos (3t
A
Va
+ 7i si n (3f + 213 2 (sin (3t 
(3 t cos (3t).
(15)
Again thi s soluti on is the same a the solution obtained in equ ati on (5) of Sectio n 10.3, and we have reso nance occurring.
•
EXAMPLE 15.21 Solve the problem of the example of Section 10.4,
Hx"(J)
+ 0.6x' (t) + 24x(t)
= 0;
x(O) =
J.. ~
x/CO) = 2.
( 16)
Pu t Llx(t)} = u(s) . T hen ( 16) yields (s2
+ 1.6s + 64)u(s) =
~(s  4.4),
from which we obta in
x t  1 L _ ( ) 
I {
3
1
S 
(s
4.4
+ 0.8) 2 + 63.36 _I {
S 
= :3 exp ( 0.8 r) L ?
.
}
5.2
s + 63.36
}
.
The refore, the desi red so luti on is
x(t ) = ex p ( 0.8 t )(0.33 cos 8.0t  0.22 sin 8. 0t) , a portion of its graph be in g shown in Figure 10.3 .
(17)
•
15.8 The Deflection of Beams
307
Exercises Each of the exerci es of Chapter 10 is an appropriate exercise here. It would be instructi ve to solve a prob lem both with and without the Laplace transform and to compare the two methods.
115.81 The Deflection of Beams As a further example of an application in which transform methods are useful, we consider a beam of length 2c, as shown in Figure 15.7 . Denote distance from one end of the beam by x, the deflection of the beam by y. If the beam is subjected to a vertical load W (x) , the deflection y must satisfy the equation d4y EI dx 4 = W(x)
for 0 < x < 2c,
(1)
in which E, the modulus of elasticity, and J, a moment of inertia, are known constants associated with the particular beam. The slope of the curve of deflection is y' ex), the bending moment is E f y'f (x), and the shearing force is E I y'" (x) . Common boundary conditions are of the following types:
= 0 and y' = 0 at the point. y = 0 and y" = 0 at a point.
(a)
Beam embedded in a support: y
(b)
Beam simply supported:
(c)
Beam free : y" = 0 and yll! = 0 at the point.
Problems in the transverse displacement of a beam take the form of the differential equation (1) with boundary conditions at each end of the beam. Such problems can be solved by integration with the use of a little algebra. There are, however, two reasons for employin g our transform method in sllch problems. Frequently, the load function, or its derivative, is discontinuous. Beam problems also give llS a chance to examine a useful device in wbic b a problem over a finite range is solved with the aid of an as ociated problem over an infinite range.
, ,
    +  
c
x
Figure 15.7
2c
308
Chapter 15 In verse Transforllls
EXAMPLE 15.22 Find the displacement y throughout the beam of Figure l5 .7, in which the load is assumed to decrease uniformly from Wo at x = 0 to zero at x = e and to remain zero from x = e to x = 2e . The weight ofthe beam is to be negli gible. T he beam is embedded at x = 0 and free at x = 2e . We are to solve the proble m d4 y Wo E I  ·4 =  [e x+(x e)a(xe) ] dx e
yeO) = 0, yl/ (2c)
= 0,
fo rO < x 2; and x(O) = x /CO) = yeO) = leO) = o.
9.
Write the olution of the sys tem of Example 15.24 in Section 15.9 in terms of convoluti on integrals .
10.
Use the results of Exercise 9 to obtain the sol ution of Example 15 .25 in Section 15 .9.
I; x(O)
= 3, x' CO) = 0, yeO)
In Exercises II and 12, write the the sol ution in tenns of convol utio n integrals.
11.
x'(t)  2y(t) = F(t), let) + 2x(t) = G(t); x(O) = 1, yeO) = O.
12 . 2x'(t) + 3y(t) = F(t), y'(t) + 2x(t) = G(t); x(O) = 2, y eO) = 1.
=  3.
316
Chapter 15
Inverse Transforms
13.
Consider the initial value problem
+ by + J(t), cx + dy + get);
x'(t) = ax y'(t) =
x(O) = c], yeO) =
C2,
where a , b, c, d, and c ], C2 are constants. Use an argument similar to that of Example 15.24 of Section 15.9 to show that the solution, if it exists, sho uld have the form
+ xp(t), Ye(t) + Yp(t),
x(t) = xc(t) yet) =
where xc(t) and Ye(t) depend on c ] and C2 whereas xp(t) and yp(t) depend on J(t) and get) . 14 .
Consider the initial value problem
+ 2y(t) = 0 x " (t) + 2/ (t) + 2y(t) x'(t)
= 2e ' ; x(O) = 1, x '(O) = 0, yeO) = O.
(a) Show that the Laplace transform method produces
x = 3  2e' ,
y = er .
(b) Verify that these function s satify the differential equation but do not satisfy the initial conditions. (c) By elementary el im ination, show that a solution of the system of differential equations has the form x =
c] 
2e r ,
y = / ,
and thus the initial conditions given are not compatible with the system of differential equations. 15.
For the network in Exercise 15 of Section 12.4, use the Laplace transform to show that the nature of the solutions depends on the zeros of the polynomial C 3 L 2 (R]
16.
+ R3 )m 2 + [C 3 (R ] R2 + R2R3 + R3 R I ) + L 2]m + R] + R 2 .
For the network in Exercise 16 of Section 12.4, use the Laplace transform to discuss the character of II (t) without explicitly finding the function .
[ 15.10 [ Computer Supplement A Computer Algebra System such as Maple is especially well suited for the Laplace transform techniques described in the preceding two chapters. We can
15.10
CompUle r SuppleJJlent
317
ask the machine to do the "dirty work" by having it find the necessary Laplace transforms and inverse. Maple finds L{si n kt} from Example 14.2 of Section 14.3 by >laplace(sin(k*t) ,t,s);
k S2
+k 2
We can also go in the inverse direction. Example 15.1 of Section 15.1 asks for L 
I
{? 15 }. s + 4s + ] 3
Maple uses the procedure invlaplace to solve this
problem : >invlaplace(15/(sA2+4*s+13) ,s,t);
5 e 2 ( sin(3 t) We could then combine these procedures to so lve a given different ial equation "by hand." Alternatively, we ca n have the macbine solve the problem di rectly using the usual commands. On the other hand, for problems inVOlving the step f un ction Ct, we need the Laplace option. Instead of the name Ct, Maple uses the name Heav iside. This is the method we would use for a problem such a Example 15 . 12 of Section 15.4.
+ 4x(t)
X"(t)
= 1/J'(t);
x(o) = 1, x'(O) = 0,
(I)
in which 1/I(t) is defined by 1/I(t) = 4t ,
=4, After converting the f unction
1/1
0:::/::: 1,
(2)
I Eqn2 : =D (D (x ) ) (t) +4 *x (t) =4 *t4 * (t1) *Heaviside (t1) : dsolve ({Eqn2, x (0) =1, D (x) (0) =O} , x (t) ,laplace) ;
x(t) = cos(2 t)
•
+t

(1
sin(2t) Sin(2t2») 4 Heavisid e(t  1)   1/4    2 4 8
 
Exercises
1.
Use a comp uter to so lve a variety of problems involving Laplace transforms.
2.
Use a computer to so lve a variety of p roblems involving inverse Lap lace transform .
3.
In Example 15. 12 of Section 15.4 as described above, the solution involves the Heaviside function. This does not, however, mean that the so lution is not continuous. Use a computer first to find the solution as above, and then have it plot the solution. Does it look continuous?
318
Chapter 15
In.verse Tran.4orms
TABLE OF TRANSFORMS Whenever It is used, it denotes a nonnegative integer. The range of validity may be determined from the appropriate text material. Many other transforms will be found in the examples and exercises.
f(s) = L{F(t)}
F(t)
f(sa)
ea'F(t)
f(as + b)
~exp(_ bt)F(!.)
1 _ e cs , c > 0
a(t  c) = 0, 0::::: t < c,
s
a
a
= I,
e f(s), c > 0 CS
f l (s)h(s)
a
c :::
t
F(t  c)a(t  c)
10
1
FI «(3)F2 (t  (3) df3
s til S" + I
1 sx+I '
x >  1
S  I/2
nl tX l(x + 1) (nt)  1/2
e at s+ a
tile  a, (s
+ a)II+1 k
S2 + k 2
s S2
+ k2 k
S2 
k2
s S2 
k2
2k 3 (S2 + k 2)2
n! sinkt cos kt sinh kt coshkt sin kt  kt cos kt
15. /0
ComputerSupplemen.t
TABLE OF TRANSFORMS (con ti nued) J(s)
=
L(F(t)}
F(t)
2ks (s2 + k 2)2
In (1 +
t sin kt
~)
J  e t
2sinhkt
s+k In  s k
In (1 
~~ )
2 (1  coshkt) t
In (1 +
~~)
(1  coskt) t
k
arctan s
2
sin kt
319
Nonlinear Equations
1
t 6. t
1
16
Preliminary Remarks T he existence and uniqueness theorem of Chapter 13 made no distinction between li near and nonlinear differential equations. We know from our study in the earlier chapters of thi book, however, that the methods we have found for actually determining solutions of a given equatio n often depend on the equation being linear. For example, in Chapter 2 we found that certain particul ar kinds of firs torder nonlinear equat ions ca n be solved, that is, if the equation is exact, separable, homogeneous, and so on. On the other hand, if a firstorder equation is linear, we have a melhod that can produce all possible solutions of the differen ti al equation. The fact is that there is no general method for solving firstorder nonlinear differential equatio ns even if the existence of such solutions can be shown by the theore ms of Chapter 13. Indeed, the determ in ation of sli ch solutions is often difficu lt, if not impossible. In th is chapter we discuss briefly a few of the special difficu lties that ar ise with non li near equations and a few techniq ues that will fi nd solutions for certain particular types of equations.
1t 6.21 Factoring the Left Member To ill ustrate the kind of complexity that may ari se in nonlinear situations, we cons ider fi rst a relatively simple complication. For an eq uat ion of the form f(x, y, y') = 0,
(1)
it may be possible to factor the left member. T he problem of solving (1) is then replaced by two or more problems of simp ler type. The latter may be capable of so lution by the methods of Chapters 2 and 5. Since y' will be raised to power in the example and exercises, let us simplify the printing and writing by a common dev ice, us ing p for y': ely  elx'
p
320
/6.2
Factoring lhe Left Member
321
EXAMPLE 16.1 Solve the differential equation xyp2
+ (x + y)p + 1 =
(2)
O.
The left member of equation (2) is readily factored . Thus (2) leads to
+ l)(yp + 1) = 0,
(xp fro m which it follows that either
+ 1=
0
(3)
xp+ 1= O.
(4)
yp or From equation (3) in the form
ydy+dx =O it follows that (5) Equ ation (4) may be written
x dy
+ dx
= 0,
from wh ich, for x =1= 0,
dy
dx
+
x
= 0,
so
(6)
•
We say, and it is very rough language, that the olutions of (2) are (5) and (6). Particular so lutions may be made up from these sol utions; they may be drawn fro m (5) alone, from (6) alone, or co nceivably pieced together by using (5) in some intervals and (6) in other. At a point w here a so lution fro m (5) is to be joined with a so lution from (6), the slope must remain continuous (see Exerc ise 2 1 be low) , so the piecing together m ust take place along the line y = x. Note (see Exercise 24) that the seco nd derivative, which does not enter the differential equation, need not be co ntinuous. The ex istence of these three sets of particular solutio ns of (2), that is, solutions from (5), from (6), or from (5) and (6), leads to an interesting phenomenon in initial va lue prob lems. Cons ider the problem of finding a so lution of (2) such that the solution passes through the point (2) . If the result is to be valid for the interval  1 < x <  ~ , there are two answers , which will be found in Exercise 25. If the result is to be valid for  1 < x < there is o nl y one an wer (Exercise 26), one of the two answers to Exercise 25. If the result is to be vali d in I < x < 2, there is only one answer (Exercise 27).
4,
4,
322
Chapter 16
Nonlin ear Equations
• Exercises In Exercises 1 through 18, find the solu tions in the sense of (5) and (6).
1.
6. p2  (x 2y + 3)p + 3x 2y = O. 7. xp2  (l + xy)p + y = O. 8. p2  x 2 y2 = O.
x 2 p 2  y2 = O.
2. xp2  (2x + 3y)p + 6y = O. 3. x 2 p2  5xyp + 6y2 = O. 4. x 2p2 + x p  y2  Y = O. 5. xp2 + (l  x 2y)p  x y = O. 11.
12. 13.
9.
+ y)2p2 = y2 . yp 2 + (x i)p (x
10. 3 p2  xy(x + y)p +x y3 = O. (4x  y)p2 + 6(x  y) p +2x  5y = O. (x  y)2 p 2 = y 2.
xy = O.
14 . xyp2 + (xy2  1)p  y = O. 15. (x 2 + i f p2 = 4x 2y2. 16. (y + x)2 p2 + (2 y2 + xy  x 2)p + y(y  x) = O. 17. xy(x 2 + y2)(p2  1) = p(x4 + x 2y2 + y4). 18. x p 3  (x 2 +x + y)p2 + (x 2 +xy + y) p xy = O. Exercises 19 through 27 refer to the examp le of this section. There the differential equ ati on xyp z + (x
wa
+ y)p + I = 0
(2)
how n to have the solutions
(5) and
y =  In Iczxi.
(6)
19.
Show that of the family (5), the only curve that passes through the point (1, 1) isy =(3  2x)J /2 andthatth issolutioni valid for x dsol v e( Eqn l ,y( x )) ;
2 v (x) = :;;:;~~;:;r======;;: .' (_ I +x)3/2 (x + 1)3/2 J  l + x 2 x3
x
+ ( I + x )3/ 2 (x + 1)3/ 2 J  I + x 2 A min or modi ficatio n LO the dsoJve comma nd will prod uce as ma ny term s of the series so lutio n as we choose, in th is case fi ve, >Orde r : =5 : >Sol l : =dsol v e(Eqnl , y(x) ,series); y (x) = (2 +x +4x 2
+ ~x3 3
+ 6 x4
+
0 (x 5 )
(2)
These are the fi rsl five le rm s of the power series solution and a n error term of the ord er of x 5 , Although this is an easy technique, for some ap pl icat io ns what we really want is the rec urre nce re lati on fo r the coeffi cients, Maple can fi nd th is as we ll , albe it with a littl e more work. We fi rst change the form at of the inpu t for the equation: >Eqn2 : =(1  x " 2)*diff{y (x) , x,x)  6*x *di f f( y {x) , x ) 4*y{x ) =O ; Nex t, we create the representati ve te rm s Cl k_2Xk  2 + soluti on: >SeriesSol : =sum(a [n J* x " n , n=k2 . . k+2 );
' , , + Clk+ 2X k+2
in a seri es
17.6
Computer Supplement
357
We then substitute this so lutio n into the di ffere nti al equation and solve th e resul ting equation for Ok, sim plifying as we go . >simp1ify(simp 1 ify(subs(y(x)=8eries801,Eqn2))) ; simp1ify(solve(coef f (lhs( " ) ,x"(k2) l,a[k))); (k
+ 2) Ok  2 k
T his re ult agrees with eq uation ( 16) of Sectio n 17. 5. To check that this rec urre nce re latio n agrees with our first res ult, we speci fy values for Go and 0 1 based on the initial co nditio ns, fi nd the coefficie nts a2 . . . as from the rec urrence relatio n, and fo rm the polynomial with these coefficients. >a ( 0 ) : =2: >a[ l ) : =1: >for k from 2 to 5 do ark ) :=(k+2)*a [k  2)/k
od : >80 1 2: =sum(a [ j ) *x" j , j =O . . 5 ) ;
2+x +4X2
5x 3
+ 3
7xS
+6x 4 + 
3
wh ich agrees with the fi rst fi ve terms of dl e o lution th at was fo und in equat io n (2) of this ectio n .
• Exercises 1.
Use a compu te r to solve a variety o f problems fro m the c hapter.
2.
For the equati o n given in Examp le 17.1 in Section 17.5, add the initial conditions yeO) = 1, y' (0) = 2, and fi nd tb e fir t In terms of the power . e ri es so lu tion for III = I . . · 5 .
3.
Have the computer plot the five functio ns from Exercise 2 on th e same axes alo ng w ith the actual solutio n of the initial value problem.
Solutions Near Regular Singular Points
18
11 8.1 1 Regular Singular Points Suppose that the poi nt x = Xo is a singular point of the equation boex )y"
+ b l (xV + b2 (x)y =
(1)
0
with polynomial coefficients. Then bo(xo) = 0, a boCx) has a facto r (x  xo) to so me power. Let us put eq uation (1) into the form y"
+ p(x)y' + q(x)y
=
o.
(2)
Because x = Xo is a singu lar pOint and because pCx) and qCx) are rational functions of x, at least one (maybe both) of p(x) and q (x) ha a denom inator that contains the factor Cx  xo) . In what follows , we assume that both p(x) and q (x) have been reduced so that in each ca e the numerator and denominator contain no common factors. If x = Xo is a singular poi nt of equation (2), if the denominator of p(x) does not contai n the factor (x  xo) to a higher power than one, and if the denominator of q (x) does not co nta in the factor (x  xo) to a power hi gher than two, then x = Xo is called a regular singular point CR.S .P.) of equatio n (2) . If x = xo is a singular point but is not a regular singular point, it is call ed an irregular singular point (LS.P.).
EXAMPLE 18.1 Classify the singuI ar points, in the fi nite plane, of the eq uation x(x  I )2(x
+ 2)/, + x 2y' 
(x 3
+ 2x 
For this eq uation x p(x) = (x _ 1)2Cx
+ 2)
and  (x 3
+ 2x 
q(x) = x(x _ 1)2(x
358
1)
+ 2)'
L) y = O.
(3)
18. 1 Regular Singular Poil1ts
359
The sing ular points in the finite plane are x = 0, 1,  2. Con ider x = 0. The factor x is absent from the denominator of p(x) a nd it appea rs to the first power in the den ominator of q(x). Hence x = 0 is a reg ul a r sin gular point of equati on (3). Now consider x = I. The facto r (x  1) appears to the second power in the denom inator of p(x). That is a hig her power than is permitted in the definition of a regular si ngu lar point. Hence it does not matter how (x  1) appears in q (x); the point x = I is an irregular sing ular point. T he factor (x + 2) ap pears to the fir tpowe r in the denominator of p(x), just as high as is permitted, and to the first power also in the denominator of q(x); therefore, x = 2 is a regular singu lar point. In summary, equatio n (3) has in the fi n ite plane the following singular point: regul ar si ngular points at x = 0, x =  2; in egul ar si ngu lar po int at x = I . The methods of Section 18. 10 w ill show that (3) also has an irregular sing ular point "at infinity."
•
EXAMPLE 18.2 Class ify the s ingular points in the finite p lane fo r the equatio n x 4 (x 2 +1)(xI)2y l/+4x 3 (x I )y'+Cx + l )y= O.
Here pCx) =
4
4 x(x 2
+ I )(x 
1)
=      x(x  i )(x
+ i )(x 
I)
and q(x )
x +1
= 4       x Cx
+ i)(x 
i)(x  1)2
Therefore, the desired classification is
R.S .P at x
= i,
 1, 1;
I.S.P. at x = 0.
•
Singular po ints of a li near eq uation of higher order are classi fied in m uch the sa me way. For instance, the s ingul ar point x = Xo of the eq uation y'l/
+ PI (x )y" + P2(X)Y' + P3(X) Y =
0
is called regu lar if the facto r (x  xo) does not appear in th e denom in ator of P I (x) to a power higher th an one, of P2(X) to a power higher than two, of P3(X) to a power h ighe r than three. If it is not regu lar, a ingul a r po int is irregu lar. Th is c hapter is devoted to the sol utio n of li near equations nea r regular singular points . Solutions near irregul ar sin gular points present a great deal more difficulty and are not studied in th is book.
360
Chapfer 18
Solutions Near Regular Sil/gular Poil1ls
• Exercises For each equation, locate and classify all its sing ular points in the fini te plane. (See Section 18.10 for the con cept of a singular point " at infinity.")
x 3 (x  l ) y" + (x  l ) y ' + 4x y = O. 2. x 2 (x 2  4) y" + 2x 3 y' + 3y = O. I.
y" + xy = O. 4. X 2 y" + y = O. 5. X 4 y" + y = O. 6. (x 2 + I)(x  4)3 y" + (x  4)2y' + y = O. 7 . x 2(x  2)y" + 3(x  2) / + y = O. 3.
8. x 2 (x  4 fy" + 3xy'  (x  4)y = O. 9. x 2(x + 2) y" + (x + 2) / + 4y = O.
14.
+ 3) y" + y'  V = O. + 4y = O. (x  l )(x + 2) y" + (x + 2)y' + x 2 y = O. (1 + 4x 2)y" + 6x y'  9y = O. (l + 4x 2 )2y" + 6x ( 1 + 4x 2 )y'  9y = O.
15.
(I
16.
+ 4)2y" + (x + 4) y' + 7y = (2x + J)4y" + (2x + I) y'  8y = O. x4 y" + 2x 3y' + 4y = O.
O.
Exercise 12, Section 17.3.
10. x (x
11. x 3 y" 12.
13.
17. 18.
+ 4x 2)2y" + 6x y' 
(x 
9y
= O.
1)2(x
19.
Exercise I, Section J7.3 .
26.
20.
Exercise 2, Section 17.3.
21.
Exercise 3, Section 17.3 .
27 . Exercise 13, Section 17.3. 28 . Exercise 14, Section 17.3.
22.
Exercise 4, Section 17.3.
29 .
Exe rcise 15, Section 17.3.
23 .
Exerc ise 7, Section 17.3.
30.
Exe rcise 16, Section 17.3.
24. Exerc ise 8, Section 17.3.
3 1.
Exercise 4, Section 17.5.
25 . Exerc ise 9, Section 17.3.
32.
Exercise 7, Section 17.5.
118.21 The Indicial Equation As in Chapter 17 , whenever we wi h to obtain oluti ons about a point other than
x = 0, we fir t translate the or igin to that point and then proceed with the ll sual technique. Hence we co ncentrate our attention on sol utions valid abo ut x = O. We shall restrict o ur study to the interval x > 0, and if we then wish to find
/ 8.2
361
The Indicia! Equation
solutions of the same differential eq uation vali d for x < 0, we can do so simply by substituting x =  u and tudying the resu lting equation on the interval u > O. Let x = 0 be a regular singular point of the equation
+ p(x)y' + q(x)y =
y"
(1)
0,
where p and q are rational functions of x. Then p (x) cannot have in its denomi nator the factor x to a power higher than one. Therefore, rex) p(x) =  ,
x
where rex) is a rational function of x and r ex) exists at x = O. We know that suc h a rational function, this rex) , has a powe r series expansion about x = O. Then there exis Ls the expans ion Po
P (x) = x
+ PI+
P2X
+ P3x 2 + ... ,
(2)
valid in some interval 0 < x < a. Bya imilar argument we find that there ex ist· an expan ion qo
q (x) = ? x
ql
?
+  + q2 + q3x + q4[ + . . .
(3)
x
valid in some interval 0 < x < b. We shall see in a formal manner that it is reasonab le to expect equation (I) to have a solution of the form 00
y = La" x"+ c = aoxc +a l x l +c +a2x2+c
+ ... ,
(4)
11=0
valid in an interval 0 < x < /1 , where h is less than both a and b. If we put the series for y, p(x) , and q(x ) into equation (1) and co nsider only the first few terms, we get eee  1)aox C  2 + (l
+ e)ealx c I + (2 + c)(l + e)a2xC + ...
+ [':0+ PI+ P2X + ... ] [caox c  I +
(I
+ c)alx + (2 + c)a2xl +c + ... ] C
+ [ qo2 + q I + q2 + .. . ] raox e + alx i+c + a2x 2+c + ... .] = 0. x
x
Performing the indicated multipli cations, we find that we have c(c  1)aox C  2 + ( I +
+ c)calxc I + (2 + c)( 1 + e)a2xC + .. . pocaoxc2 + [po CI + c)al + Pl cao]x c  I + .. . + qoaox c 2 + [qoa l + q IClO]X c 1 + .. . =
O.
(5)
362
Chapter 18
Solutions Near Reglliar Singular Points
From the fact that the coeffici e nt of x c [C(C  1)
2
must va nish , we obtain
+ PaC + qo]ao =
(6)
O.
We may insist that ao =j:. 0 because ao is the coefficient of the lowest power of x appearing in the solution (4), no matter what that lowest power is. So from (6) it follows that
c2
+ (Po 
l) c + qo = 0,
(7)
which is call ed th e indicial equal ion (at x = 0). The Po and qo are known con tants; equatio n (7) is a quadratic equation giving us two roots, c = C I and
c
= C2.
To distingui sh between the roo ts of the indici al equation, we shall denote by C I the root whose real part is not smaller than the real part of the other root. Thus, if the roots are real, CI ::: C2; if the root are im agin ary, m(Cl ) ::: ~l(C2 ). For brev ity we call CI the " larger" root. Superficially, it appear that there should be two solutions of the form (4), one from each of these values of c. In each olution the ao sho uld be arbitrary and the succeeding a' should be determined by equating to zero the coeffici ents of the higher powers of x (x c I, x c, X l+c , and so on) in identity (5). Thi s superficial conclusion is correct if the differe nce of the roots CI and C2 is not integral. If that difference is integral, however, a logarithmic term may enter the solut ion . The reaso ns fo r thi s strange behavior will be made c lear when we develop a method for obtaining the olutions.
11 8 .31 Form and Validity of the Solutions Near a Regular Singular Point Let x = 0 be a regul ar singu lar po int of the eq ua tion
y"
+ p(x)y' + q(x)y = o.
(L )
Then tbe functi ons xp(x) and x 2 q(x) have Maclaurin series expansions that are va lid in some common interval 0 < x < b. It can be proved that equation ( I ) always has a general solution either of the form
y= A
00
00
11=0
,,=0
L a"x" +C I + B L b"x,,+c2
(2)
or of the form 00
y = (A
00
+ B lu x) LO"X"+
o.
(2)
11=0
D irect substituti on of thi s y into (1) yields 00
L
2(n
+ c)(n + c 
l)ol/xl/+ c  I
00
+ L(n + c)allxl/+ c I
II ~
II~
00
+L
(n
+ c)allxl/ +c 
2
1/ = 0
L (/IiXI/+
C
= 0,
1/=0
or 00
L(n +
00
c)(2n
+ 2c 
l)allxl/+c 
I
+ L(n + c 
1/=0
2)a"x"+c =
o.
(3)
1/ = 0
Having co ll ected like terms , we next shift the index to bring all the exponents of x down to the sma ll est one present. Th is choice is used to get a recurrence relation for all rather than one for Cl II + 1 or some other Cl . In equation (3) we replace the index n in the second ummation by (11  I), thus gett ing 00
L(n + c)(2n + 2c 
00
I )ClIIX,,+c1
,, = 0
+ L(n + c 
3)al/ _ lx,,+c 1 = O.
(4)
11= 1
Once more we reason th at th e total coeffic ient of each power of x in the left member of (4) must van i h. The second summation does not start its contribution untiln = J. Hence the equations for the determination of c and the a's are 11
= 0:
c(2c 
n
~
(n
I:
I)ao = 0,
+ c)(2n + 2c 
J)Cl II
+ (n + c 
Si nce we may without loss of generality assume that eq uatio n, that which determines c, is c(2c 
J) = O.
The indicia I equation always cOllies ii"OIn the being used in this book is employed.
11
3)a,, _1 = O.
ao =f. 0, the indicial (5)
= 0 term when the technique
364
Chapter 18
Solut ions Near Regular Singular Points
!
F ro m (5) we see th at CI = and C2 = O. The difference of the roots is = C I  C2 = ~, whic h is no nintegral. W hen s is not an integer, the method we are usin g always gives two li nearl y independe nt so lutions of th e fo rm (2), one with each choi ce of c. Let us retu rn to the recu rrence relation usi ng the value C = C I = We have
S
!.
(n
+ !)(2n + 1 
I)a"
+ (n + ~ 
3)a,, _1 = 0,
(2n  5)a,, _1
a" = 
211 (211
+ I)
.
As us ual we use a vertical an ay a nd then fo rm the prod uct to get a formula fo r
a" . We have ao arb itrary (  3)ao a1 =    
2·3
(  I )a l a2 =    
4·5
a3
( I )a2
=  6 ·7
all = 
(211  5)a,, _1 211(2n
+ I)
,
so the produ ct y ie lds, fo r n ::: 1, ( 1)" [(3)(  1)(1)·· · (211  5) ]ao
a" = r2 . 4 . 6 ... (2n)] [3 . 5 . 7 . . . (2n
(6)
+ I)] .
The fo rmul a (6) may be simplifi ed to fo rm
a" =
( I )" · 3ao
2"n l (2n  3)(211  1)(211
(7)
+ 1) .
Us in g an = I, the (/" from (7), and tbe pertine nt va lue of c , C I = ~, we may now write a partic ular solution. It is
+L
( 1)"3x,,+1 /2
00
YI
=
X I /2
11 = 1
2"11 ! (217  3)(2n  I)(2n
+ 1)
.
(8)
The notatio n Y I is to e mphas ize that thi s particular soluti on corres pond s to the root C I of the ind icial equatio n. Our next tas k will be to get a partic ular solution Y2 correspo ndin g to the small er root C2 · T hen the general solution, if it is des ired, may be writte n at once as
Y = AYI wi th A and B arbitrary constants.
+ BY2
18.4
Indicial Equatio/1 with Differel1ce of Roots NOl1il1tegrai
365
In returning to the recurrence relation just above the indicial equation (5) with the intention of using c = C2 = 0, it is evident that the a's w ill be different from those with C = C I. Hence it is wise to change notation. Let us u 'e b's instead of a's. With C = 0, the recurrence relation becomes 11.(217  l)b ll
+ (11. 
3)bll _ 1 = O.
The correspo nding vert ical array is
b o arb itrary
( 2)bo bl =1. I (I)b l b2 =    
2·3
h
= _ (O)b2 3·5
b" = 
(/1  3)b ll _ 11.(211  I )
1
.
Then bll = 0 for n ::: 3 and, using bo = I, hi and b2 may be computed and found to have the va lues hi = 2 and b 2 = ihl = Therefore, a seco nd solution is
t.
Y2
I 2 = I + 2x +}x .
(9)
Since the differe ntia l equation has no sing ular po int, other than x = 0, in the finite plane, we concl ude that the linearly independent solutions YI of (8) and Y2 of (9) are valid at least for x > O. The vali dity of (9) is ev ident in this pa rticul ar examp le because the erie terminates. The student should assoc iate w ith each solution the region of validity guaranteed by th e general theorem quoted in Section 18.3, although from now on the printed answers to the exercises wi ll omit tIle statement of the region of validity.
• Exercises In Exercises I through 17, obta in two linearly independelll solu tions valid near the origi n for x > O. Always state the reg ion of val idi ty of each solution that you obta in.
1.
2x(x 2
+ l) y" + 3(x + 1)y' 
+ 4xy' + (4x Y" + 4xy'  (4x 2 +
2.
4x y"
3.
2
4x
2
y = O.
l)y = O. l )y = O.
+ 3y' + 3y = O. 2x2(l  x)y"  xO + 7x) y' + y 2xy" + S(l  2x)y'  5y = O.
4 . 4xy" 5. 6.
= O.
366
Chapter / 8
SO/lltiol1s Neal' Regular SiT/gular Poillls
7.
8X 2 y fl
+ 10xy' 
(I + x)y = O.
8. 3x yfl+ (2x)y'2 y =0. 9. 2x(x + 3)y"  3(x + I)y' + 2y = O. 10. 2x/, + (I  2x 2 )y'  4xy = O. 11.
x(4  X)yfl
+ (2 
x)y'
+ 4,11 =
O.
2
12.3x y" +xy'  ( I +x)y=O. 14 . 15.
+ (I +2x)y' +4y = O. 2xy" + (I + 2x)y'  5y = O. 2x2y"  3x(1  x) y' + 2y = O.
16.
2x2y"
17.
2xy "  (J
18.
T he equation of Exerci se 17 has a particu lar solution Y2 = ex p (!x 2 ) obtained by the series method. Make a change of dependent variabl e in the d i Fferential eq uation, u. ing y = v exp (~x 2 ) (the device of Section 9 .2), a nd thus obtain the genera l solut ion in "closed fo rm."
13. 2xy"
+ x(4x
 l )y'
+ 2x 2)y' 
+ 2(3x
 l )y = O.
xy = O.
In Exerciscs 19 through 22, use the power se ries method to fin d solutions valid for x > O. What is causing the recurrcnce relation s to degenerate into one term relati ons?
19. 20.
2X 2 y" + xy'  Y = O. 2x 2y"  3xy' + 2y = O.
21. 9X 2 y" 22 . 2x 2/'
+ 2y
= O.
+ 5xy' 
2y = O.
23.
Obtain dy/dx and d 2 y/dx 2 in terms of derivatives of y with respect to a new independent va ri abl e t related to x by I = In x for x > O.
24.
Use the resu lt of Exerc ise 23 to how that the cha nge of i nde pe nclent variable from x to I , where t = In x, transforms the eq uation I ?
d 2Y
ClX  dx 2
dy + bx~ + cy dx
= 0,
a, b , c con sta nts, into a linea r eq uat ion with constan t coefficients. Solve Exerci ses 25 throu gh 34 by the method implied by Exerc i. e 24, that is, by changing the independent variabl e to t
25.
Exercise 19.
= In x
for x > O.
26.
Exerc i e 20.
I An eq uati on slIch as the one or th is exerci c, whi ch contains only terms of the kind cx k O k y w ith
c con slam and k = O. I, 2, 3. .. . , i. ca lled an equation or Callchy type, or of Euler type .
18.5
27.
Exe rcise 2 l.
28.
Exercise 22.
29.
x 2 y"
33. 34.
2
Differentiation of a Product of Functions
367
30. X 2 y" + xy'  9y = O. 3 I. x 2 y"  3xy' + 4y = O. 32 . X 2 y"  Sxy' + 9y = O.
+ 2xy'  12y = O. y" + 5x y' + 5y = O.
x (x 3 D3 + 4x 2 D2  8x D + 8)y = O. You w ill need to extend the result of Exercise 23 to the third derivative, obta ining 3 3d y d3Y d2Y dy x = 3 2+ 2  . 3 3 dx dt dt dt
118.51 Differentiation of a Product of Functions It wi ll soon prove necessary for us to different iate efficiently a product of a number of function s. Suppose that (1) each of the u 's being a funct ion of th e parameter c. Let differe ntiation with respect to c be indicated by primes. Then fro m
In u
= In u , + In U2 + In U 3 + .. . + In Un
it follows th at
u'
u~
u; u;
u;,
= u, ++ +"'+ U . u Ll2 11 3 II
Hence
(2) Thus to differentiate a product, we may multiply the original product by a conversion factor (w hi ch converts the produ ct into its derivative) cons istin g of th e sum of the derivatives o f the logarithms of the separate factors. When the factors involved are th emselves powers of polynomials, there is a co nveni e nt way of forming the co nversion facto r men tally. That factor is the sum of the co nvers ion factors for the individ ual parts . The way most of us learned to di fferentiate a power of a quantity is to multiply the expone nt, the deri vative of the origi nal quantity, and the quantity with its exponent lowered by one. Th us. if
y = (ac+b)k , then
dy dc = y the divi sion by (ae
{ ka }
ac + b '
+ b) converting (ae + bl
into (ae
+ bl'.
368
Chaprer 18
Solwiol1s Near Regular Singular Poillls
EXAMPLE 18.3 If u = (4e _ 1)3(7e
+ 2)6'
then
du de =
{2 U
~
1
+e+I
12
42 }
 4c  I  7c
+2
.
Note that the denominator fac tors in the function u are thought of as numerator factors wi th negative exponents. '
•
EXAMPLE 18.4
If c+n y =         e(e
+ l )(e + 2) ··· (e + n 
1)'
then
d y= y 
de
{ I l  l  1  .. . e +n
c
c+J
c+2
I}
c+n  1 .
•
EXAMPLE 18.5 If W =
2"c 3 [(c + 2)(c + 3) . .. (c + n
~
+ 1)]2 ,
then
118 .61 Indicial Equation
• with Equal Roots
When the indicial eq uation has equal roots, the method of Sectio n 18.4 cannot yield two lin early independent so lutions. The work w ith one value of c would be a pu re repet ition of that w ith the other val ue of c. A new attack is needed.
18.6
linliciaf Equation
Willi
Equal
ROOIS
369
Consider the problem of sol ving the equation
x 2y"
+ 3xy' + (I
 2x)y = 0
( 1)
for x > O. The root of the indicial equat ion are equal, a fact that can be determined by setti ng up the in dicia l equation as developed in the theory, Section 18.2. Here 1  2x
3 x
p(x) = ,
q(x) =   7  '
x
= 3 and qo = I. The indicial equation i (.2 + 2c + I = 0, with roots CI = C2 = I .
so Po
Any atte mpllo obtain solutions by putting 00
c
(2)
Y = LClI/XI/+ 1/==0
into equ ation (1) is certain to force us to c hoo e C =  1 and thus get o nly one solution. We know that we must not choose c yet if we are to get two soluti o ns. Hence let us put the y of equation (2) into th e left member of equation ( I ) and try to come as close a we can to making that left member zero without chao ing c . It is co nvenient to have a notation for the left me mber of equation (I); let us use
L(y) = x 2 y"
+ 3xy' + (I
 2x )y.
(3)
For the y of equation (2) we fi nd that 00
00
L( y) = L(n
+ c)(n + C 
I)al/ xl/+c
1/=0
+L
+ c)al/xl/+c
3(n
1/=0 00
+L
al/x Jl + 0,
11= 1
in which
1'1 ::: 1 :
all(e)
2"
= [(c + 2)(c +3)·· · (c + n +
1)]2
.
(7)
T he y of equatio n (6) has been 0 determined th al for that y, the right member of equ atio n (4) mu st red uce to a single term , the 17 = 0 term. T hat is, fo r the y(x , c) of eq uation (6) , we have
(8) A solution of the original diffe re ntial equ ation is a f un c ti on y for whic h L (y) = O. Now wesee why the choice c =  1 yields asolut ion; it makes th e right me mber of eq uatio n (8) zero.
Indicia I Equation with Equal Roots
18.6
371
But the factor (e+ 1) occurs squared in equation (8), an automatic consequence of the equa lity of the roots of the indicial equation . We know from elementary calculus that if a function contai ns a power of a certai n factor dependent upon e, then the derivative with respect to e of that function contains the same facto r to a power one lower than in the original. For equation (8), in particular, differentiation of each member with respect to e yields
a
 L[y(x, e) ] = L
[a y
ac
(x, e) ] = 2(e ac
+ l) x + (e + 1) 2 x C
C
ln x,
(9)
the right member contain ing the factor (c + 1) to the first power, as we knew it must fro m the theorem quoted above. In (9), the order of differentiations with respect to x and c wa interchanged. It is best to avoid the need for justifying such steps by verifying the solutions, (13) and (14) below, directly. The verificatio n in thi s in stance is straightforward but a bit lengthy and is omitted here. From eq uations (8) and (9) it can be seen that two solutions of the equation L(y) = 0 are YI
= [ y(x, c)L=_ 1 = y(x, 
1)
and
Y2 = [ay(x,
ac
C)] c=I
because c = 1 makes the right member of each of equations (8) and (9) van ish. That YI and Y2 are linearly independent will be evident later. We have 00
y(x, c) = X C
+ LCln(C)X
Il
+c
(6)
11=1 and we need ay(x, c)/ac. From (6) it follows th at ay(x, c)
ac
=
.
XC
Inx
~ + L....Ci,,(c)x'
I+C
Inx
~ + L....a,,(c)x" I
,,=1
+
c,
11= 1
which simplifies at o nce to the form ay(x,c)
ac
= y(x, c) Inx
~
I
+ L....al/(c)x" 1/ =1
+ c.
(10)
372
Chapler 18
Solutions Near Regular Singular Points
The olutions)l l and Y2 will be obtained by putting c = 1 in equations (6) and (10), that is , 00
)' 1
+ L a ( l )x" I ,
= x I
(11)
ll
11=1 00
Y2
= )l lln x
+ La:,(I)x" I •
(12)
11 = 1
Therefore, we need to evaluate all (c) and a;, (c) at c =  1. We know that all(e) =
2" [(c
+ 2)(e + 3)· . . (e + n + 1)F
,
from wh ich, by the method of Section 18.5, we obtai n immediately
{1
I + I  + ... + } . '
a (e) = 2a//(c)  II c+2 I
We now use c
=
c+3
c+I1+1
1 to obtain 2// a (1)  //

( n.1)2
and
a;, (1) =
 2 2"(n !)2
{I
} + 2I +I 3 +I ... + It .
A frequently used notation for a partial sum of the harmonic series is useful here. It is HII = 1
We can now write
I
1
1
2
3
n
+  +  + ... + 
a;, ( 1) more
1 L . k 11
=
k=l
impl y as
I
a,,( l )=
2,,+ IHII
(n l)
2
'
Finally, the desired solutio ns can be written in the form 00
)' 1 = x  I
+L
2"x"  1 (11 !) 2
(13)
11 = 1
and
(14)
18.6 Indicia I Equation with Equal Roots
373
The general solution, valid for x > 0, is
with A and B arbitrary co nstants . The li near independence of YI and Y2 should be evident because of the presence of In x in Y2 . In detail, x YI has a power series expansion about x = 0 but XY2 does not, so that one cannot be a constant times the other. Examination of the procedure used in solving this differential equation shows that the method is in no way dependent upon the specific coefficients, except that the indicial equation has equal roots. That is, the success of th e method is due to the fact that the n = 0 term in L (y) contains a square factor.
• Exercises Obtai n two linearl y independent solutions va lid for x > 0 un les otherw i e instructed.
+ Y = o.
1.
x 2 y"  x (1 +x)y'
2.
4x 2 y" + (l  2x)y = o. x 2y" + x(x  3)y' + 4 y = O. x 2y" + 3xy' + (l + 4x 2)y = O.
3. 4.
8.
+ x)y" + (l + 5x)y' + 3y = O. + 3x)y' + (l  6x)y = O. 2 x y" + x(x  l )y' + ( 1  x)y = o. x(x  2)y" + 2(x  1)y'  2y = o.
9.
Solve the eq uation of Exercise 8 about the point x = 2.
5. 6.
7.
x(I
x 2 y"  x (l
+ (x
+ xy =
10.
Solve about x = 4: 4(x  4) 2y"
11.
xy" + y' + xy = O. T hi s is known as Bessel's equation of index zero. It is widely encountered in both pure and applied mathematics. (See also Sections 19.5 and 19.6.)
12.
xy"
13.
Show that
+ (1 
 4) (x  8)y'
O.
x 2 )y'  xy = O.
and apply the result to simplification of the formula fo r Y2 in the answer to Exercise 12. 14.
x 2 y" + x (3 + 2x)y' + (1 + 3x)y = O. In simplifying given in Exercise 13.
15.
4x 2 y"
16 .
x
2
+ 8x(x + I)Y' + y = O. y" + 3x(1 + x)y' + (l  3x)y =
O.
Y2,
use the form ul a
374
Chapter 18
Solulions Neal' Regular Singular Points
17 . xy"+(l  x)y' y=O.
J 8.
Refer to Exercise 17. There one solution was fo und to be YI = eX. Use the change of dependent variab le y = vex, to obtain the general solution of the differential equation in the form
118 .71 Indicial Equation with Equal Roots : An Alternative In Section 18.6 we saw that when the indicial equation has equal roots, two li nearly independent solutions always appear of the form
C2
=
C I,
00
YI
=
X CI
+L
a ll x ll +CI ,
(1)
11 = 1 00
Y2 = YI In x
+L
bIl XIl+CI,
(2)
,,=1
where CI, all, bll are dependent upon the coefficients in the particular equation being solved. It i possible to avoid some of the computational difficulties encountered in Section 18.6 in computin g the bll of (2) by first determining CI and YI, then substituting the Y2 of (2) directly into the differential equation and finding a recurrence rel ation that must be satisfied by b". The resulting recurrence relation may well be difficu lt to so lve in closed form, but at least we can successively produce as many of the bll as we choose.
EXAMPLE 18.6 For the differential eq uation of Section 18.6, L(y)
= X 2 y" + 3xy' + (1 
2x)y = 0,
(3)
we saw that the roots of the indicial equation were both  1 and that a nonlogarithmic solution was
X
 I
2"x ll (n !)2 . 1
00
YI =
+ "~
(4)
11=1
We know that a logarithmic solution of the form 00
Y2
b II I = YI 1n x + " ~ "x 11 = ]
(5)
J8.7
Indicial Equation with. Equa l Roots: An Alternative
375
exists and we shall determ ine the bll by fo rc.ing this Y2 to be a sol ution of (3) . We have 00
YzI = YII In x.
+ x. 1YI + "L...., ( n 
l)bIl X. ,,  2 ,
,,=1 Y~ = y;' ln x
+ 2x  1Y; 
X
2YI
00
+ L (n  1)(n 
2)b"x ,,  3,
,,=1
so th at 00
L(Y2) = L(YI) ln x
+ 2YI + 2x y ; +
L (n  1)(/1 
2)b" x,,  1
,,= 1 00
00
+L
3(n  l )h"x 'l  1 +
11 = 1
00
L b" x ,, l  2 L b"x ,, = 1
l1
,
,, = 1
fro m which 00
L(Y2 ) = 2YI
+ 2xy ; + h i + L (n 2 h" 
2h l1 _l) x ,, I .
,,=2
T he logarithmi.c term vani he because L(YI) = 0. Substituting fro m (4) for YI yields L(Y2 ) = 2x  1 + 2
211xll  1 L 11= 1 (n!) 00
 2

2x  J
+2L 00
11 = 1
2"(n  l )x "I 2
(n!) 00
+ b l + L (n 2 b" 
2bll _ l )x,,  I ,
,,=2 or L(Y2) = b l
+4 +
f:
11
[ n2hl1  2bl1 _ 1+ n2 +1] 11=2 (n l )
]
X,,  I .
If Y2 is to be a so lution of eq uation (3) , then b l =  4 and hl1 must satisfy th e recurrence relation ? n211 + 1 nb"  2b,,  J +  2 = 0, (n !)
n ::: 2.
(6)
A simple calculation yields b2 = 3 and b3 = 22/27, but a closed form for bl1 is difficult to obtain from (6) . In Section 18.6 we fo und the val ues of bl1 to be I b  2"+ H I I n (n!) 2
(7)
376
Chapler 18
Solutions Near Regular Singular PoinlS
It is not diffic ult to show that th is expression ati sfi es the recurrence relation (6), but it is difficult to obtain the form (7) from (6). Even so, for computational purposes the alternative for m for Y2 given by the series (5) and the recurrence relation (6) is useful.
•
EXAMPLE 18.7 Solve the differential equation
°
+Y =
x 2 y"  x(1 +x)y'
(8)
of Exerc ise 1 of Section 18.6. The two roots of the ind icial eq uation are foun d to be nonlogarithmic solutio n is
CI
= C2 =
I and the
X"+ I
00
L n! '
YI =
(9)
1/=0
We seek a second solution of the form 00
Y2 = yl ln x
+ Lbl/xl/ + I , 1/ = 1
so that 00
y~ = y~ In x +X1YI + L(n
+ l )bl/x",
n= 1
y{ = y ;' ln x + 2x  1Y;
00

X
2
YI
+
L n(n + l )b"x,,  I. ,, = 1
Substitution of these expressions into (8) gives 00
00
2xy;  2YI  XY I + L
1
2
n bl/x"+  L(n
1/ = 1
+ 1)b"xl/+2 =
0,
,,=1
or 00
2x)l;  (2 + X)Y I + b l x 2
+ L(n2bl1 
nbl/ _ l )x" +1 = 0.
11=2
Using the series (9) for 00
2x+ ~ ~
11= 1
2(
n
+ 1)
n!
gives
)II
00
11 + 1
x"+ 1 _2x2~ ~x2 ~ n! 1/= 1 00
 L n= 1
11 + 2
_x_
n!
00
+ b l x 2 + L(n 2b 11=2
ll

nbn_dx"+1 = 0,
J8.8
IndicialEquation with Difference o.lRoots a Positive Integer: Non logarithmic Case
377
which may be written
It follows that b 1 =  I and 2
n b ll

nb ll _
1
+
1
= 0,
(n  1)1
(1 0)
n ::: 2.
If thi problem i solved by the method of Section 18.6, we obtai n  HII
bll =   · 171
It is easy to show th at this expression satisfies the recurre nce relati on (1 0) .
•
Exercises
•
For Exercises 2 through 8 of Section 18.6, fin d the logarithmic solution by fi ndi ng a recurrence relation for the b" of equation (5) of this section.
118.81 Indicial Equation with Difference of Roots a Positive Integer: Nonlogarithmic Case Co nsider the equatio n
xy"  (4 + x)y'
+ 2y =
(1)
O.
As usual, let L(y) stand for the left member of (I ) and put 00
allx"+c .
y= L
(2)
,, = 0
At once we fi nd that fo r the y of equation (2), the left membe r of equation (1) takes the form 00
00
L(y) = L [(n
+ c)(n + c 
1)  4(n
+ c )]a"x"+ c  I
11=0

L (n
+c 
2)a l x"+c ,
11=0
or 00
L(y) =
I)n + c)(n + c  5 )a x ll
11 = 0
00 ll
+c 
1 
LCn 11 = 1
+ c
3)a ll _ l x ll +c 
l
.
378
Chapter 18
So/wiolls Near Regular Singu lar Points
The indicial eq uation is c(c  5) = 0, so CI
= 5,
We reason that we may hope for two power series solutions, one startin g with an xo term and the other with an x S term. I f we use the large root c = 5 and try a eries 00
'"' a x,,+5, ~II ,, = 0
it is ev ident that we can get at most one solution; the xo term would never enter. On the other hand, if we use the smaller root c = 0, then a trial solution of the form 00
o La"x"+ 11 = 0
has a chance of picking up both solutions because the n = 5 (n = s) term does contai n xS. If s is a posi tive integer, we try a seri es of the form (2) Ll sing the smaller root C2. If ao and as both turn out to be arbitrary, we obtain the general olution by th is method. Otherwise, the relation that should determine a.l · will be impossi ble (with our usual assumption that ao =f. 0) and the general solution will involve a logarithm as it did in the case of equal roots. That logarithmic case will be treated in the next secti on. Let us return to the numerical problem. Using the smaller root c = 0, we now know that for 00
(3)
y = L all x" 11=0
we get 00
L(y) =
L n(n. 
5)al
x"
1 
11 = 0
L Cn  3)a,,_lx" 1. 11= 1
Therefore, to make L(y) = 0, we must have
n= 0:
o . ao =
n:::l:
n(n  5)all
0 (ao arbitrary), 
(n  3)a n _1 = 0.
Since division by (n  5) cannot be accomplished until n > 5, it is best to write out the separate relations through the critical one for as. We thus obtain
18.8
Indicial EquCllioll wilh Differellce of Roots a Posi/ive lJUeger: Non/ogarithmic Case
n. = 1 : n= 2: n = 3: n.
=4 :
1'1
= 5:
379
+ 2aO = 0, 602 + a l = 0, 603 + 0 · 02 = 0,
 4a I 
= 0,
 404  a 3
°.
05 
= 0, 3)all _1
2a4
( 1'1 
n(n  5) It follows fro m these re lations that _ 01 
I 2: 00,
a2
I I = O.
•
• Exercises
In Exercises I through 6, the si ngular poi nts in the fi nite plane have al ready been located and class ified . For each equati on, determine whether the point at infini ty is an ordi nary point (O.P.), a regular si ngu lar po int po int (R.S.P.), or an irregular singu lar point (I.S.P.) . Do not solve the problems.
1. 2.
2
2
x (x 
+ xy
3.
y"
4.
x 2 y"
5. 6.
+ (x  1)/ + 4xy = O. (Exercise 1, Section 18.1.) 4) y" + 2x 3 / + 3 y = O. (Exercise 2, Sectio n 18. 1.)
x 3 (x  1)/'
X X
= O. (Exercise 3, Section 18.1.)
+y =
O. (Exercise 4, Section 18. 1. )
4
y"
+y
4
y"
+ 2x 3 / + 4y
= O. (Exercise 5, Section 18 .1.) = O. (Exerci e 18, Section 18.1.)
In Exercises 7 through 19, fi nd solutions val id fo r large positive x unless otherwise instructed .
7.
X
4
+ x(1 + 2x2) y' + 5y = O. y"  x(2  5x)/ + y = O.
y" 3
8.
2x
9.
x(l  x) y"  3/
10 .
X
3
y"
+ x(2 
+ 2y =
0, the equation of Exercise 8, Section 18.8.
3x)/  (5  4x)y = O. See Exercise 13, Sec tion 18.6.
2
+ (x + l)y
I t.
2x (x  I )y" +x(5x  3)y'
12.
Solve the equation of Exerc ise 11 abo ut the poi nt x = O.
13.
2x 2 ( 1  x)y"  5x( 1 + x)y'
14.
Solve the equation of Exerc ise 13 abo ut the po in t x = O.
15.
x(l + x)y" + (l + 5x) /
16.
17 .
+ (5 
= O.
x)y = O.
+ 3y = 0, the equation of Exerc i e 5, Section 18.6. x (4 + X )y" + 2x(4 + x 2 ) / + y = O. x( 1  X)y" + (l  4x) y'  2y = 0, the eq uati on of Exercise 18 in th e 2
2
M iscell aneous Exercises at the end of this chapter. 18. x(l + 4x)y" + (1 + 8x)y' + y = 0, the eq uation of Exercise 49 in the M iscell aneous Exercises at the end of this chapter. 19. The eq uation of Exe rci e 6.
118.1 1 I ManyTerm Recurrence Relations In so lvin g an equation near a reg ul ar singular poi nt, it will sometimes happen th at a manytellll rec urrence relation is encou ntered . In nonl ogarithmic cases, the
18. 11
MallyTerm Recurrence Relations
389
methods developed in Chapter 18 are easily applied and no complications result except that u uall y no explicit formula will be obtained for the coefficients. In logarithmic cases, th e methods introduced in Chapter 18, co nstmcting y(x , c) and ay(x, c)/ac, can become awkward w hen a manyterm recurrence relation i present. There is anothe r attack which bas its good points. Consider the problem of solving the equation L(y) =
X
2
y"
+ x(3 + x)y' + ( 1 + x + x 2 )y =
0
(1)
for x > O. From co
CIlx"+c
y = L
(2)
11=0
it is eas ily shown th at 00
00
L(y) = L
(n
00
+ c + 1)2a"x"+ c + L(n + C)ClII _ IX"+c + L
,, =0
11=1
Cl" _2x"+c .
(3)
11=2
Therefore, the indicia] eq uation is (c + 1)2 = O. Since the roots of the indicial equ ati on are equal, c =  J,  1, it follows that there ex i t the solutions (4)
YI = "LCI"x III , ,, = 0 00
Y2 = YI Inx
+ Lb"x,,  I ,
(5)
,, = 1
val id for x > O. The region of validity is obtained from the differential eq uatio n; the form of the solutions can be seen by the reaso ning in Section 18.6 . We shall deternune the CI", It > 0, by requ iring that L(YI ) = O. Then the b" , n ~ 1, will be determined in term of the a" by requiring th at L(Y2) = O. From L(YI ) = 0 it follow s that 00
00
00
n2a"x"  1 + L(n  '1 )a,,_ l x,,  1 + La,, 2 x"  1 = O.
L ,,=0
1/ = 1
,, =2
Let us choose ao = 1. Then the rest of the a's are determined by
n = 1: n
~
al
2:
n
2
+ O· ao = 0, a" + (n  1)a,,  1 + a,,2 =
O.
Therefore, one solut ion of the d ifferential equation (1) is
"a 00
)1
l  L 11=0
II
X,,I
t
(6)
390
Chapter 18
Solutions Near Regular Singular Points
in w hich ao = 1, a l
= 0, (n  l)al/  I + al/  2 aI/ =   ::  
n ?: 2 :
/1
2
Next we wish to req uire that L(Y2) = O. Fro m 00
+ "'b ~ IIX 11 
Y2 = YI In x
1
(5)
,, = 1
it follows that
Y~ = Y; ln x +X I YI
+ L(n 1)bl/x,,  2 ,, = 1
and
Y~ = y ;' l nx
+ 2x  1Y;
 X 2 YI
+ L(n 
1)(n  2)b ll x,,  3.
11 = 1
Now direct co mpu tation of L(Y2) yields 00
L(Y2) =L(YI)l n x+2xy; Y I +XY I +3Y I
+ Ln 2 bl/x"  ' 1/ =1
00
00
'(n +" ~
l. )b ,, _ IX /I 
1/=2
I
" bIl  2X 11 + '~
1.
11= 3
Since L(YI) = 0, the requirement L(Y2) = 0 leads to the eq uati on 00
00
L n 2b"xl/ 1/=1
1
00
+ LCn l)bll _ lx,,1 + Lb/l _2 x"  1 11=2
11 =3
00
00
" 2l1al/x ,,  1  '~ " al/ _I x II  I , =  '~ 11 = 1
(7)
11 = 1
in which the righ t mem ber has been simplified by using equation (6). Fro m the identity (7), relations for the determination of the bl/ from the all follow. T hey are :
n = 1:
b l =  2a l
n = 2:
4b 2
n ?: 3 :

ao ,
+ b I =  4a2  0 I , 2 n b + (n  l )b,, _ 1 + b ll
ll 
2
= 2no"  a,,  I .
/ B. 1 J
ManyTerm Recurrellce Relations
391
Therefore, the original differential equation has the two linearly independent solutions given by the YI of equation (6) and by 00
Y2 = yl ln x
+L
I
(8)
b"x"  ,
11= 1
in whichb L =  1, b2 = n ::=: 3 :
!, bll = 
(n  l)b,,  1 + b,,  2
n2
2a" a,, _ 1    .
n2
n
If the indi cial eq uation has roots that differ by a positive integer and if a logari thmic solution exists, then the two olutions will have the form co '"' ) 1I ~
aII X" +C1 ,
,, = 0 co
Y2 = YI In x
+L 11 =0
b X"+ C2 , "
where C I is the larger and C2 the smaller root of the indicial equation. The all and b ll can still be determi ned by the procedure used in this section .
• Exercises So lve each equation for x > 0 unless otherwise instructed.
I. 2. 3. 4. 5.
X2y" + 3xy' + (1 + x + x 3 )y = O. 2x(1  x)Y" + (1  2x)y' + (2 + x)y = O. xy" + y' + x(l + x)y = O. x 2y" + x(i + x)y'  (1  3x + 6X 2)y = O. Show that the series in the answer to Exercise 4 start out as follows: ) 1 
I +2+ 4x 2  ~x3 ao(x. 2
4 + .!lx 5
~ x5 ~
+ ... )
3 + a2(x  i3 x 2 + !2 12 x
6.
X/I + xy' + (1 + x4)y
_
4 z'x 6
+ ~x5 + .. .). 72
= O. Here the indic ial equatio n has roots c = 0, c = 1, and an attempt to get a complete solution without In x fail s. Then we put
'"'a 00
Y] ~
11
X" + I )
11 = 0 00
Y2 = y llnx
+ Lb ,,=0
ll
x
l1 •
392
Chapter J8
Solutions Near Regular Singular Points
The coefficient h i (s = I) turns out to be arbitrary and we choose it to be zero. Show that the indicated )I I and)l2 are solutions if aD
=
1, a I
=
=
 1, a2
n > 5 : ali = 
(n.
+
~, a3
=
k, a4 = ~,
1)a,, 1 + a,,  5 n(n
+ 1)
,
and if the h's are given by
bo
=
n
1 , b l
~
= 0 (so chosen),
h2 = 1, h3 = ~, b4
nb ll _ + h,, 5 5: h" =      n(n  1)
n(n  1)
7.
For the)l l in Exercise 6, prove that the the terms of )I I out to the x 7 term .
8.
x(x  2)2 y"
9.
Solve the equation of Exercise 8 for x > 2.
+ (1
 x )y'  (I
+ 2y =
+ x)y =
tk,
(2n  l)a,, _ 1 + al1  2
1
 2(x  2)y'
=
all
alternate in sign. Also compute
O.
O.
10.
2xy"
11.
Show that the answer to Exercise 10 are also given by
10[ fi exp (
Y2 = eX, YI = eX
~j32) d,8.
118.12 1 Summary Confronted by a linear equation
L(y) = 0,
(1)
we first determine the location and natu re of the singular points of the equation. In practice, the use to which the results will be put will dictate that solutions are des ired near a certain point or points. In seeking solutions valid about the point x = xu, always first translate the origin, putting x  Xo = v. Solutions valid near an ordinary point x = 0 of equation (1) take the form 00
y =
Lall x"
(2)
,,=0
with ao and al arbitrary, if equation (1) is of econd order. If x = 0 is a regular singular point of equation (1) and we wish to get sol utions valid for x > 0, we first put 00
y
=L
,,=0
c
a l1 x "+ .
(3)
/S. /2
Summary
393
For the y of (3), we obtai n the series for L(y) by substitution. Fro m the n = 0 term of that seri es, the indicial equation may be wri tten . When the difference of the roots of the indic ial equation is not an integer, or if the roots are equal, the techn iq ue is stra ightforward fo ll ow in g the method of Section 18.4 or 18.6. W hen the roots differ by a no nzero integer, the sol ution may, or may not, involve In x. T he recurrence relatio n for n = s, where s i the difference of the roots , is the critical one. We mu st then determine whether the re latio ns for n = 1, 2, . . . , s leave ao and as both arbitrary. If they do , two power e ri es of the form (3) wi ll be solutions of the differential equation. If ao and as are not both arbitrary, the case is logarithmic. Then the device of Section 18.9 may be used. The techn ique ca n be varied, if desired , by always choosing the all in term of c so that the power series for L(y) reduce to a sin g le term. T hus a series of the fo rm
y(x, c) = ao [xc
+
f
f;,(C)X II +c ]
(4)
11=)
will be determined for whi ch (5)
where k = 0 or I for the equations being treated here and c ) and C2 are the roots of the indicial eq uati on. Then it can be determ i ned from the actual coeffic ients in (4) whether the use of c = c) and c = C2 w ill res ult in two sol utions of the di fferen ti al eq uatio n. If C I = C2, the results would be identical and the u e of 3y(x, c)/3c is ind icated. The other logarithmi c case will be identified by the fact that some o ne or more of the coefficients .(,,(c) will not exist when C = C2, the sma ll er root. Then again the differentiation process is needed , after the introduction of ao = c  C2 . The method sketched above has a di . advantage in that it seems to tempt th e user into automatic app licati on of rul es, always a dangerous procedure in mathe matics. W hen a student thoroughly under tands what is happen ing in each of the four possibl e cases, th i method may safe ly be lI sed and saves some labor. Extension of the methods of this and the preceding chapter to li near eq uations of higher order is direct. As an example, a four th orde r eq uati o n whose indicia l equation has roots c = 2, 2, 2, ~ would be treated as foll ows. A seri es wo ul d be determi ned for y(x, c), y(x, c) = ao
[xc+ f
f;,(C)X'I+c]
11=1
for whi ch the left member of the orig inal equatio n reduces to o ne term , such a L[y(x, c) ] = (c  2) 3(2c  l )aoxc.
394
Chapter 18
Solutions Near Regular Singular Points
Then four linearly independent solution s could be obtained:
Y2= [ BY (X,e )] ; Be c=2
Yl = y(x, 2);
B2 y (X ,
Y3
= [ Be2
e)]
. c=2
'
• Miscel laneous Exercises In each exercise, obtai n sol utions va lid for x > O.
1. 2. 3. 4. 5. 6.
xy"  (2 + x)y'  y = O. xy"  (2 + x)y'  2y = O. x 2y" + 2x2y'  2y = O. 2x2y"  x(2x + 7)y' + 2(x + 5)y = O. x 2(l + xl)y" + 2x (3 + x 2)y' + 6y = O. (1  X 2)y"  10xy'  18y = O.
7. 2xy" + (I + 2x)y'  3y = O. 8. y" + 2xy'  8y = O. 9. x (l  x 2 )y"  (7 + x 2 )y' + 4xy = O. 2x 2 y"  x(l + 2x) y' + (1 + 4x )y = O. 11. 4x 2 y"  2x(2 + x)y' + (3 + x)y = O. 12. X 2 y"  x(l + x 2 )y' + (l  x 2 )y = O. 13. 2x y" + y' + y = O. 14. x 2 y" + x(x 2  3)y' + 4y = O. 15. 4x 2y"  x 2 y' + y = O. 16. (1 + x2)y"  2y = O. 17. 2x 2y"  x (I + 2x)y' + (1 + 3x)y = O.
10.
18. 19 . 20. 21. 22. 23. 24. 25. 26.
y'" +xy
= O.
4X y" + 3x 2 y' + (1 + 3x)y = o. xy" + (1  x 2 )y' + 2xy = O. 4X 2y" + 2x2y'  (x + 3)y = O. xC I  X2 )y" + 5(1  x 2 )y'  4xy = O. 2 X y" + x(3 + x)y' + (1 + 2x)y = O. 2 X y" + xy'  (Xl + 4)y = O. x( 1  2x)y"  2(2 + x)y' + 18y = O. xy" + (2  x)y'  y = O. 2
Y4 = y(x, ~ ).
J8. J2
+ 4(1 + x)y = O. + 6y = O. 4x 2 y" + 2x(x  4)y' + (5  3x)y = O. x 2 y"  xC3 + 2x)y' + (3  x)y = O. x(l  x)y"  (4 + x)y' + 4y = O. 4x 2 y" + 2x(x + 2)y' + (5x  l)y = O.
27.
x 2 y"  3xy'
28 .
y"  2xy'
29. 30.
3l. 32.
Summary
33 . Solve the equation of Exercise 31 about the point x = 1. 34. 4x 2 y" + (3x + l)y = O. 35. x(l  x)y" + (1  4x)y'  2y = O. 36. Show that the solutions of Exercise 35 may be written in the form Y2 = (1  x)  2(lnx  x).
37. 38. 39. 40. 4 l. 42. 43 . 44. 45 . 46. 47 . 48. 49. 50. 51. 52.
+ (l  x)y' + 3y = O. + x(3x  l)y' + (3x + 1)y = O. 2x(1  x)y" + (1  2x)y' + 8y = O. x 2 y" + xC4x  3)y' + (8x + 3)y = O. x 2 y"  3x(1 + x)y' + 4(1  x)y = O. 2(1 + x 2)y" + 7xy' + 2y = O. 2xy" + (3  x)y'  3y = O. xy"  (l + 3x)y'  4y = O. xy" + 3y'  y = O. x 2 y"  x(3 + 2x)y' + (4  x)y = O. 3xy" + 2(1  x)y'  2y = O. 2x(l  x)y" + y' + 4y = O. xy" + (3  2x)y' + 4y = O. x(l + 4x)y" + (l + 8x)y' + y = O. 2x2y" + 3xy'  (1 + x)y = O. x 2y" + x(2x  3)y' + (4x + 3)y = O.
xy"
x 2 y"
395
Equations of Hypergeometric Type
19
119.1 1 Equations to Be Treated in This Chapter W ith the methods stud ied in C hapters 17 and 18, we are able to so lve ma ny equations that ap pear frequently in phys ics and engineering as well a in pu re mathe matics. We shall consider briefl y the hypergeometric equation, Bessel' s eq uation, and the eq uat ions that lead to the study of Laguerre, Legendre, and Hermite po lynom ials. There are, in mathematical li terature, tho usands of researc h pape rs devoted entirely or in part to the study of the fun ctio ns th at are solutions of the equation to be studied in this chapter. Here we do no more than call to the attention of the stu de nt the existence of th ese spec ial f unctions, which are of such great va lue to theoretical phys icists, engineers, and many math emat ic ians. An introd uction to th e properti es of these and oth er spec ial fu nctions can be fo und in Rai nvi ll e. 1
119 .21 The Factorial Function It w ill be convenient for us to empl oy a notation that is widely e ncountered in adva nced mathematics. We defi ne the fac torial functio n (a)1l for n eq ual to zero or a pos itive integer by (a)1l = a(a + l )(a + 2) . . . (a + n  1) fo r n :::: 1; (1) (a)o = 1 fo r a :j:. O. Thus the symbo l (a)1I denotes a product of n factors starting with the factor a, each facto r bein g one larger than the fac tor before it. Fo r instance, (7)4 = 7 . 8 . 9 . to, (  5)) = ( 5)(4)(3),
(D3= (  ~) ( ~) ( ~ ). T he fac torial fun ction is a generalizati on of the ordinary factorial. I ndeed, (l)Il=I·2 · 3· · ·n=nl . E. D. Rainville, Special Func tiol1 ,l' (New York: Macm illan Publishi ng Company, 1960),
396
(2)
19.3
The Hypergeomefric Function
397
In our study of the gamma f unction in Section 14.9, we derived the functional relation rex
+ 1) =
for x > O.
x r(x)
(3)
By repeated use of the relation (3), we find that if n is an in teger, rea
+ n) = =
+ 11 (a + 11 
+ n  I) l)(a + n  2)f(a + n 
l)r(a
(a
= (a
+n 
l)(a
+n 
2)
2)· . . (a)f'(a)
= (a)"r(o). Therefore, the factorial function and the gamma function are related by
Co)" =
rCa + n) r(a)
n integer, n > 0, and a > O.
'
(4)
Actually, (4) can be shown to be valid for any complex a except zero or a negative integer.
119 . 31 The Hypergeometric Function Let us now co nsider any secondo rder li near differential equation that has only three singular points (one could be at infinity). Suppose that each of these si ngularities is regu lar. It can be shown 2 that such an equation ca n be transformed by change of variables i.nto the hypergeometric equation x(l  x)y lJ
+ [c 
(a
+ b +I)x ]y' 
aby = 0,
(1)
in which a , b, c are fixed parameters. Let us solve equation (1) about the regular singular point x 0. For the moment let c not be an integer. For (1), the ind icial equation has roots zero and (l  c) . We put
in eq uation (1) and thus arrive , after the usual simplificatio ns, at 00
Ln(n
00
+ c l)e"x
ll 
I
,,=0
 L(n
+ a)(n + b)e'lx"
= 0.
(2)
l)e,, I x,,  l = O.
(3)
,, =0
Shift the index in (2) to get 00
Ln(n IJ= O
+ c 1)e"x n  1 
00
LCn
+a 
1)(n
+b 
,,=1
2 See, for exam ple, E. D . Rai nville, Jl1l ennediale Differential Equations, 2nd ed. (New York: Macmill an Publ ishing Company, 1964), Chapter 6.
398
Chapter /9
Equa/ions of Hypelgeomelric 7)'p e
We thus find that eo is arbi trary and for n ::: I, ell =
(n + a I ) (n + b  I ) n(n +e l)
(4)
ell _ I ·
The rec urrence relat io n (4) may be solved by our custo mary device. T he result is, for n ::: 1, a(a + I)(a + 2) ··· (0 + n  I) . b(b + I )(b + 2) ··· (b + Il  J)eo ell
=
n ! e(c+ 1)(c+2) · ·· (e+I1  1)
(5)
But (5) is greatly simplified by use of the factorial function. We rew rite (5) as ell = (0)11 (b )11 eo .
(6)
n ! (e)1I Let us choose eo = 1 and write our first solution of the hypergeometric equation as ~ (a)"(b),,x" YI = 1 + 0 ,,=1 (e)II n.!
(7)
The particular solution YI in (7) i called the hypergeometric f un ction and a common sy mbol for it is F(a, b; e; x). That is , F(a, b; e; x) = 1 +
(a) (b) x" L 11= 1 (e)lIl!! (X)
II
II
,
and YI = F(a, b; e; x) is a solutiol1 of equation (I). The other root of the indicial equation is (1  e) . We may pu t
Y=
L f"X,, + Ic
,, =0 into equation (l), determine 1;, in the usua l manner, and arrive at a second ol ution (8)
In the hypergeometri c notation this seco nd solution (8) may be written Y2 =x lc'F(a+ l c, b +lc; 2c; x), which means exactl y the same as (8). The solutions (7) and (8) are valid in o < x < 1, a region extending to the nearest other singular point of the differential equation (1 ). If e is an integer, one of the solutions (7) or (8) is correct, but the other involves a zero denominator. For example, if c = 5, then in (8) , (2  e)" = (3)" and as soon as n ::: 4, ( 3)1) = 0, for (3)4 = (3)(2)( 1)(0) = O.
19.4
Laguerre Polynomials
399
If c is an integer but a and bare nonintegral, one of the solutions about x = 0 of the hypergeometric equation is of logarithmic type. If c and one or both of a and b are integers, the solution mayor may not involve a logarithm. To save space, we omit the logarithmic solutions of the hypergeometric equation.
119.41 Laguerre Polynomials The equation
+ (1
xy"
 x)y' +ny = 0
(1)
is called Laguerre's equation. If n is a nonnegative integer, one soluti on of equation (l) is a polynomial. Consider the solution of (1) about the regular singular pointx = O. T he indicial equation has equal roots c = 0, O. Hence one solution will involve a logarithm . We seek the nonlogarithmic solution. Let us put
into (1) and obtai n, in the usual way, 00
L
00
k2akxk  1  L(k  I  n)ak_ lx k  1 = O.
k=O
(2)
k= 1
From (2) we find that
k
~
(k 1  n)ak  I
I :
k2 (n)(  n
+ 1) .. . (n + k 
l )ao
(k !)2
( nh = (k!) 2 ao·
If n is a nonnegative integer, (nh = 0 for k > equal to unity, one solution of equation (1) is 1/
}
_
) I 
f:o
'""'
(
)
kX (k!)2
 /1
11..
Therefore, with ao chosen
k
(3)
The right member of (3) is called the Laguerre polynomial and is usually denoted by L I1 (x): 11
LI/(X)={;
(n)kxk 1/ (lln!xk (k!)2 ={;(k!)2 (nk)!'
(4)
400
Chapter 19
Equations of Hypergeometric Type
The student sho uld prove the equivalence of the two summation in (4) by showi ng that ( I )kn ! (n)k =   (n  k)! One solution of (1) is YI = Ln(x). The associated logarithmic solution may, after co nsiderable simplification, be put in the form
Y2 = L,,(x) In x
~ (n)k(HII k  HII  2Hk )X k
+~
I ?
(k .)
k=1
b 00
+
( l)"n!(k _ l)! xk+" [(k+n)!] 2
(5)
The solution (3) is valid for all finite x; the solution (5) is va lid for x > 0.
119.51 Bessel's Equation with Index Not an Integer The equation (1) is call ed Bessel's equation of index n. Equation (l) has a regular singul ar point at x = 0 , but no other singular points in the finite plane. At x = 0, the roots of the indicia l equation are CI = n, C2 =  no In this section we assume that n is not an integer. It is a simple exercise in the methods of Chapter 18 to show that if n is not an integer, two linearly independent solutions of (1) are 00 (_ 1)k x 2k+" Yl = { ; 22kk! (l
+ n)k'
(2)
(3)
valid for x > O. The f un ction 1 )13
= 2"r(l
+ n)
00
YI
= {;
( lix 2k+" 2 2k+"k!
i(k
+n+
1) '
also a so lution of equatio n (1), is called ill (x), the Bessel function of the fi rst kind and of index n . Thus 00 (llx2k +" Y3 = i,,(x) = {; 22k +ll k ! r(k + n + 1) (4) is a solution of (1) and the general so lution of (1) may be wlitten
y=A i,,(x )+BL,,(x),
ni= an integer.
(5)
19.6
Bessel 's Equation. willi Index Clnlnteger
401
That Ln(x) is a solution of the differential equation (1) should be evident from the fact that the parameter n enter ( I) only in the term /12 . It is also true that L
2lIro _ n)
LII(x) =
Y2·
119.61 Bessel's Equation with Index an Integer In Bessel's equation
x 2 y"
+ x y' + (x 2 
n 2 )y = 0,
(I)
let n be zero or a positive integer. Then I
(_1)k x 2k+1i
00
YI = JII(x) = { ; 22k+llk! r(k
(2)
+ n + .J)
is one solution of equation (1 ). Any solution linearly independent of (2) must contain In x . We have already o lved (1) for n = 0 in Exercise 11 of Section 18.6, and for n = 1 in Exercise lOaf Section 18.9. For n an integer :::: 2, put 00
y =
LGjX
HC ,
j =O
a ac y(x,
proceed with the techniq ue of Section 18.9, determine y(x, c) and and then use c =  n to obtain two solutions: ( _ 1)k x 2k
00
y?  "
c),
1I
  ~22k I (ln)II _ I(k  n)!k!
(3)
and II  I (lf x 2k 1I
)1

3 
yin x 2 .
+ x  II + "~ :::,22k (1 _ ) kI n
k= 1
+L 00
k
(

k=1I
A shift of index in (3) from k to (k
.
l)k+ 1(H H H ) . 2k 1I k II + k 11  1 x 2k I 2  (1  n)II _ I (k  n)!k!
(4)
+ n) yields (l) k+1I x 2k+1I
00
Y2 = { ; 22k+211  1(1 _ n)II _ l k! (k + n)!'
But for n :::: 2, (1 
/1)11 _1
= ( 1),,  I(n  I)!, so
I Y2 = 211  1(11  I)! J,,(x ).
We can therefore replace solution (3) with Yl
=
JII(x).
(5)
402
Chapter 19 Equations of Hypergeometric Type
By imilar manipulation, we replace solution (4) with
( l l+ I (n  1)! X 2k 1/ 22k+ 1 l/ kl (1 _ ) k= O . n k
11  1
Y4 = JI/(x) ln x
+L
+I
00
'\"'
( l )k+ J(H + H k
2~
k+1l
)x 2k +1I
(6)
22k+llk!(k+n)!
For n an integer> 1, equations (5) and (6) can be used as the fundamental pair of linearly independent solutions of Bessel 's equation (I) for x > O.
119.71 Hermite Polynomials The equation y"  2xy'
+ 2fty =
0
(1)
is called Hermite's equation . Since equation (1) has no singularities in the finite plane, x = 0 i an ord inary point of the equation. We put 00
Y = L ajx J j=O
and employ the methods of Chapter 17 to obtain the general so luti on Y 
.
ao. 1 +
[
+ al
2k ( n)(  n
Loo
+ 6
2)X 2k ]
(2k)!
k= 1
~ 2k(l
[x
+ 2)· · · (n + 2k 
n )( 1  n
+ 2)··· (1 (2k + 1) !
n
+ 2k 
2)X 2k + l ] '
(2)
valid for all finite x and with aa and a I arbitrary. Interest in equation (1) is greatest when n is a posi tive integer or zero. If n is an even integer, the coefficient of aa in (2) terminates, each term for k ~ ~ Cn + 2) being zero. If n is an odd integer, the coefficient of aJ in (2) terminates, each term for k ~ (n + 1) being zero. Thus Hermite's equation always has a polynomial solution, of degree n, for n zero or a positive integer. It is e lementary but tedious to obtain from (2) a single expression for this polynomia l solution. The res ul t is
t
[~I/] H ,,(x) = { ;
(l/n! (2x)Il  2k
k! (n  2k)!
'
in which [~n] stands for the greatest integer :::: ~n. The polynomial HI/ (x) of (3) is the Hermite polynomial; y sol ution of equation (1).
(3)
HI/(x) is a
19.8
Legendre Polynomials
403
119.81 Legendre Polynomials The equation (l  x
2
y"  2xy'
)
+ n(n + 1) y
=
°
(1)
is called Legendre's equation. Let us solve (1) about the regular singu lar point x = 1. We put xI = v and obtain the transfor med equation v(v
el 2 y
ely
+ 2) elv 2 + 2(v + 1) dv
 n(/1
+ l)y
= 0.
(2)
At v = 0, eq uation (2) has, as roots of its ind icia1 equation , C = 0, 0. Hence one so lution is logarithmic. We are interested here only in the nonlogarithmic solution. Following the methods of Chapter 18, we put 00
Y=
Laki k=O
into equation (2) and th us arrive at the results: ao is arbitrary and
k
:::: 1:
Cik
=
 (k  n  l)(k+n)ak_1
(3)
Solve the recurrence relation (3) and th us obtain ( L)k ( n)k (I
ak =
+ nhao
2k(k!)2
with the factorial notation of Section 19.2. We may now write one sol ution of equation (1) in the form YI = 1 +
~ ( l/(  nh(n
~
+ l)k(X  II
2k(kl)2
(4)
k= 1
Since k! = (l)k, we may put (4) in the form
~ (nh(n + l)k
YI = 1+ ~
(l)kk!
k =1
(1  x)k 
(5)
2
The ri ght member of equation (5) is an example of the hypergeometric function that we met in Section 19.3. In fact,
Yl = F
2. (n, n + 1; 1; IX)
(6)
If n is a positive integer or zero, the series in (4), (5), or (6) terminates. It is then called the Legendre polynomial and designated PI/ (x) . We write our nonlogarithmic so lution of Legendre's equation as
. (n, n + 1; .1; 2 I X) .
YI = PI/(x) = F
(7)
Partial Differential Equations
20
120.1 1 Remarks on Partial Differential Equations A partial differential equat ion is one that contains o ne or more partial derivatives. S uch equations occur frequen Uy in appli cations of mathematics. T he subject of partial differential equations offers sufficient rami hcations and difficulties to be of interest for its own sake. In this book we devote the space allotted to partia l differentia l equations almost entire ly to a kind of boundary val ue problem that enters applied mathematics at every turn. Partial differential equations can have sol utions invo lving arbitrary fu nctions and so lutions involving an unlim ited number of arbitrary constants. The general sol ution of a linear partial differential equation of orde r n may involve n arbitrary functions. The general solution of a parti al d iffere nti al eq uation i a lmost never (the wave equation is one of the few exceptions) of any practical use in solving boundary value problems assoc iated with that equation .
120.21 Some Partial Differential Equations of Applied Mathematics Certain partial differential equations enter applied mathematics so freq ue ntly and in so many connect ions that their study is remarkably remu nerative. A suffic iently thorough study of these equat io ns would lead the studen t eventu ally into every phase of clas ieal mathematics and, in particu lar, would lead alm o t at once to contact w ith special functions that are widely used in quantum theory and e lsew he re in theoretica l physics and engineering. The derivation of the differentia l equations to be listed here is beyond the scope of this book. Some of the ways in which these equation are useful will appear in the detailed applications in C hap ters 23 and 25. Let x , y , z , be rectangular coordinates in ordinary space. Then the equati on
a2 v a2 v a2 v +  +  2 =0 2
ax
404
ay2
az
(1)
20.2
SOllie Pm'rial DifFerel1tial Eqllatiol1S
405
is ca ll ed Laplace's eq uation. It enters problems in steadystate temperature, e lectrostatic potential, fluid flow of the steadystate variety, and so on. If a problem involving equation ( I ) is such that a physical object in the problem is a circula r cylinder, then it is possible that cylindrical coordinates will fac ilitate sol uti on of the problem. We shall encounter such a problem later. It is possible to change equation (I) into an equation in which the independent variables are cy li ndrical coordinates r, e, Z, related to the x , y, Z of equation (I) by the equations
x = r cose ,
y = r sin e,
z·
Z=
The resu lting equation, Laplace's equation in cylindrical coordin ates , is
a2v 1 av 1 a2v a2v  +++  = 0. ar2 r ar r2 ae2 az2
(2)
Note that the use of Z in both coordinate systems above is safe in making the change of var iables only because Z is not involved in the equations with other variables. That is, in a change of independent variables such as
x=
XI
+ Y I + ZI,
Y=
XI 
Y I,
=
Z
ZI,
or its equivalent XI
=
~(x
+Y

Z),
YI
=
~ (x  y  z) ,
ZI
= z,
incorrect conclusions wou ld often result from any attempt to drop the subscript on the ZI , even though Z = ZI. For instance, from the change of variables uncler di scussion , it follow s that
av lay Jav av  =  +. az 2 aX I 2 aYJ aZI Hence
av
av aZI
=j=
az
even though Z = ZI. Let us return to Laplace's equation. In spherical coordinates p, to x, Y, Z by the equations
Y = p sin ¢ sin e,
x = psin¢cose,
Z
=
e, ¢. related
p cos (P ,
Laplace's equation is
a2 V ap2
2
+
av
J
a2 v
pap + p2 a¢ 2 +
cot ¢ p2
av csc 2 ¢ a2 v a¢ + ~ ae 2 = O.
(3)
With an additional independent vatiable t representing time, and with a constant denoted by ct , we can write the wave equation in rectangular coord inate, 2 2 a2 v =a  (a2 V V) +av+a

at 2
?
ax2
ay2
az2 .
(4)
406
Chapter 20
PClrlial Differelltial Equations
Equation (4) occurs in prob lems involvi ng wave motions. We shall meet it later in the problem of the vibrating string. Whenever the physica l problem suggests a different choice for a coordinate system, usually by the shape of the objects involved, the pertinent partial differential equation can be transformed into one with the desired new independent variables. Suppose that for some solid under consideration, u represents the temperature at a po int with rectangular coordinates x , y, z and at time t. The origin of coordinates and the initial time t = 0 may be assigned at our convenie nce. If there are no heat sources present, the temperature u must satisfy the heat equation
au at
=h
2
(a2 u a2u a2 u) , + + ax 2 ay2 az2
(5)
in which 112 is a physical constant called thermal diffusivity. Equatio n (5) is derived under the assumption that the density, specific heat, and thermal co nductivity are con tant for the solid being studied. More comment on the question of validity of equation (5) will be made in Section 23.2 . Equation (5) is the eq uation that pertains in many types of diffusion, not only when heat is being diffused. It is often called the equation of diffusion. In the subject of elasticity, certain problems in plane stress can be solved with the aid of Airy's tres function ¢, which must satisfy the partial d ifferential equation
a4 ¢ ax4
+2
a4 ¢ + a4 ¢ =0. ax 2ay2 ay4
(6)
Numerous other partial differential equations occur in applications, though not w ith the dominating insi stency of equations (1), (4) , and (5). In tbis book two methods for solving boundary value problems in partial differential equations will be examined. The Laplace transform, which was tudied in Chapters 14 and 15, i a useful tool for certai n kinds of boundary va lue problems. T he transform technique will be developed further in Chapter 24 and Llsed in Chapter 25 . A second method, the classical one of separation of variables, wi ll be discussed in the remainder of this chapter. We shall find that two other topics, orthogonal sets and Fourier series, need to be treated before we can proceed, in Chapter 23 , to use separation of variables efficiently to solve prob lems involving partial differential equations.
120.31 Method of Separation of Variables Before attacking an actual boundary value problem in partial differentia l eq uations, it is wise to become somewhat proficient in getting solutions of the differential equat ion s. When we have acquired some facility in obtai ning sol utions, we
20.3
Method of Separation of Variables
407
can then tackle the tougher problem of fitting them together to satisfy stipulated boundary conditions. The device to be exhibited here is particularly useful in connection with linear equations, although it does not always apply to such equations. Consider the equation
au
2
cPu
 = h  2 at ax
(1)
with h constant. A solution of equation (1) will in general be a function of the two independent variables t and x and of the parameter h . Let us seek a solution that is a product of a function of t alone by a function of x alone. We put u = f(t)v(x)
in equation (1) and arrive at 1'Ct)v(x) = h 2 f(t)v"(x),
(2)
where primes denote derivatives with respect to the indicated argument. Dividing each member of (2) by the product f(t)v(x), we get
l' (t) f(t)
(3)
Now equation (3) is said to have its variables (the independent variables) separated; that is, the left member of equation (3) is a function of t alone and the right member of eq uation (3) is a function of x alone. Since x and t are independent variables, the only way in which a function of x alone can equal a function of t alone is for each function to be a constant. Thus from (3) it follows at once that (4) h 2u"(X) , = k, vex)
(5)
in which k is arbitrary. Another way of obtaining equations (4) and (5) is this: Differentiate each member of equation (3) with respect to t (either independent variable could be used) and thus get
d 1'(t) =0, elt f(t) since the right member of equation (3) is independen t of t. First we obtain equation (4) by integration and then (5) follows from (4) and (3).
408
Chapter 20
Partial Differential Equatiolls
Equation (4) may be rewritten
dj = kj, dt from which its general soluti on
fo llows immediately. Before going further into the sol ution of our problem, we attem pt to choose a convenie nt form for the arbitrary con ta nt introduced in equations (4) and (5) . Equation (5) suggests that k be taken as a multiple of h 2 . Let us then return to equations (4) and (5) and put k = h2fJ 2, so we have f'(t) = h 2fJ2 f(t)
(6)
h,2V" (X) ? ? =lrfJ·
(7)
and vex)
Using real fJ and the choice k = h 2 fJ 2, we are imp lyin g that the constant k is positive. Later we shall obtain solu tion corresponding to the choice of a negative constant. From eq uations (6) and (7) we find at once that j(r) =
CI
exp (h, 2fJ 2t)
(8)
and vex)
= C2 cosh fJ x + C3 sinh fJx .
(9)
Since u = f(t)v(x), we are led to the re ult that the partial differential equation
au at

2 ? au =h
(1)
ax2
has o luti ons
u = exp (11, 2 fJ 2r)[a cosh fJx
+ b sinh fJx] ,
(10)
in wh ich fJ , a, and b are arbitrary constants. The a and b of eq uation (10) are respective ly given by a = CI C2 and b = C I C3 in terms of the co nstants of equations (8) and (9). If we return to equations (4) and (5) with the choice k = _h 2 a 2 , so k is taken to be a negative constant, we find that the partia l differenti a l equation ( 1) has so lutions Ll
= exp (h 2 a 2 t) [A cos a x
in wh ich a, A , and B are arbitrary constants.
+ B si n ax ] ,
(11)
20.3
Me /hod of Separatioll of Va riables
409
F inall y, let the co nstant k be ze ro. It is straig htforwa rd to determin e that the corresponding 01utions of the differen tial equati on ( I ) are
u = C,
+ C2x,
(12)
in which C, and C2 are constants. Di rect verifi catio n that equations (10), (1 1), and (12) are actu ally solutions of ( I ) is sim ple. Sin ce th e partial differential equ ation (1) is li near, we may construct solu tions by form ing linear combin ation s of so luti o ns. Thus from (10), (11 ), and (1 2)wi th varyi ngcho icesofa, /3, A, B, a, b, C" C2 , wecan constructasmany sol utions of ( 1) as we wis h. The disti nctio n between equ ations (10) and ( 11 ) is depende nt upo n the parameters and variables remaini ng real. Our aim is to develop tools for solving physical problems; hence we do intend to keep thin gs real.
• Exercises Except where other inst ructions are given, use the method of separat ion of va riables to obtain sol utions in real form for each differential equation.
I.
2.
a2 u
a2 u .
_ = a2 _
at 2 ax 2 2 2 av av +=0. ax 2
ay2
a u + 2bau = alat 2
3.
7
au
2 (12_ , ax 2
Show that th is equation has the sol utions
u = g(t)(B , cos kx
+ Bz sin kx),
where g(t) can assume anyone of th e forms ebt(A ,eYI+A2e YI),
y2=b 2 _a 2k 2
or
or
and the sol utions
Find also so lutions con tai ning e h and e  b · . 4.
aw
aw
ay
ax
=Y·
ifb 2
 a 2k2>0,
410
Chapter 20
Parlial Differential Equations
aw aw ax ay 6. Subject the partial differential equation of Exercise 5 to the change of dependent variable w = uI y and show that the resultant equation for v is
5.
x  =w+y.
av
au
x  = y. ax ay 7.
Show that the method of eparation of variables does not succeed, without modifications, for the equation
a2u

ox2
8.
a2u
a2u
oxot
ot
+ 4 +52 = O.
For the equation of Exercise 7, seek a solution of the form u = e kl I(x)
and thus obtain the solutions
u
= exp [kef
 2x )[A I cos kx
+ A2 sin kx],
where k, A I, A2 are arbitrary. Show also that the equation of Exercise 7 has the solutions It = Ax + Band u = Ct + D, with A, B, C, D arbitrary. 9.
For the equation
a2 u
au
a2 u
 2 + 4x  + =0 ax ax oy2 ' put u = I(x )g (y ) and thus obtain solutions 2ky u = [A l e
in which II (x ) and equation
+ A2e 2kY][BI fl (x) + B2h(x)],
h ex) are any two linearly independent solutions of the
r + 4xF + 4k2I = O. Obtain also the solutions that cone pond to k = 0 in the above . For k =f. 0, the functions I I and 12 should be obtained by the method of Chapter 17. They may be found in the form .
./1 (x ) =
1+
00
L
(4Y'1 (k 2)(k 2 + 2)(k 2 + 4) .. . (k 2 + 2m  2)X 2117 (2m) ! '
111 = 1 00
.h(x ) = x
+
L
( _ 4)111 (k 2
+ 1)(k 2 + 3)(k 2 + 5) ... (k 2 + 2m (2m
+ I)!
m= 1
Find similar solutions involving co 2ky and sin 2ky.
_ 1)x 2111 + I
A Problem on the Conduction of Heat in a Slab
20.4
10.
411
Use the change of variable u = ef3 1 g(x) to find solutions of the equation
a2u a2u a2u + 2  +  2 =0. 2 ax
11.
axBt
at
Show by direct computation that if fl (y) and 12 (y) are any functions with continuous second derivatives, 1(' (y) and I~' (y), then
hex + at)
u = II (x  at) +
satisfies the simple wave equation (Exercise I)
a2u at 2
?
a2 u . ax 2
=a 
12.
Show that v = I(xy) is a solution of the equation of Exercise 6.
] 3.
Show that w =
L4.
Show that u = II (t  x) Exercise 10.
I
(2x
+ y2) is a solution of the equation of Exercise 4. + xhCt 
x) is a sol ution of the equation of
120.41 A Prob1em on the Conduction of Heat in a Slab Among the equations or applied mathematics already stated is the heat equation in rectangular coordinates,
au at
=h
2
(a2u a2u a2u) + + ax 2 ay2 az,2 '
(1)
in wh ich
z=
x , y,
rectangular space coordinates,
t = time coordi nate , 2
= thermal diffu
II
= temperaLure.
h
iviLy,
The constant h 2 and the variables x, y , z, t , u , may be in any consistent et of units. For instance, we may measure x, y , Z in feet, t in hours, u in degrees Fahren heit, and h 2 in square feet per hour. The thermal diffusivity (assumed to be con tant in our work) can be defined by ?
K
Ir= 
uo'
in terms of quantities of elementary physics, K = thermal conductivity, u
= specific heat,
0= den ity, all pertainin g to the material composing the solid whose temperature we seek.
412
Chapler 20
Parlial Dijferellliai Equatiolls
For our first boundary value problem in partia l di fferential equations it seems wise to set up as simple a problem as possible. We now construct a temperature problem that is set in such a way that the temperature is independent of two space variables, say y and z. For such a prob lem, u wi ll be a function of only two independent variables (x and t), which is the smallest number of independent variables possible in a partial differential equation. Consider a huge slab of concrete or some other material reasonably near homogeneity in texture. Let the thickness of the slab be c units of length. Choose the orig in of coordinates on a face of the slab as indicated in Figure 20. 1 and assume that the slab extends very far in the y and z directions. Let the initial (t = 0) temperature of the slab be f (x), a functio n of x alone, and let the surfaces x = 0, x = c be kept at zero temperature for all t > O. If the slab is considered infinite in the y and z directions, or more specifically, if we treat only cross sections nearby (far from the distant surface of the slab), then the temperature u at any time ( and position x i determined by the boundary value problem
as t
?
for 0 < t , 0 < x < c;
(2)
for 0 < x < c;
(3)
0+,
1I ?
f
(x)
as x
?
0+,
II ?
0
forO < f ;
(4)
as x
?
c ,
U ?
0
for 0
k Ib w(x)x kf,,(x) dx = 0 III
w(x)flll (x)fll (x) dx =
{/
(/
k~
by ( I), si nce eac h k involved is less than n. If In > 11 , interchange the roles of m. and 11, and repeat the argument. If In = n in (2), then all =F 0, and we have
1 0
w(x)f}(x) dx =
I>k Ib w(x)x kj,,(x) dx II
k=O
1/
= all
I
/;
a
w(x)xll.f,,(x)dx
=F O.
a
T hu s we see that the co nd ition (l) is sufficie nt fo r the o rth ogonali ty o f th e set
{.t;, (x) } .
I
Next suppose that the {f" (x ) satisfy the co ndition for orthogonality as laid dow n in Secti on 21. 1. Since { j;,(x) } is a simple set, we can write k
Xk =
L blll j;l/(x).
(3)
111=0
If k
0 over a < x < b, then the zeros of j;.(x) are distinct and all lie in the open interval a < x < b . Proof. For n ~ 1, each of the polynomial f,,(x) does change sign in the interval a < x < b because, by Theorem 2 1.1 (with k = 0),
lb
w(x)f,,(x)dx=O
and w(x) does not change sign in a < x < b. Suppose 1" ex) changes sign in a < x < b at precisely the di st inct points x = ai, a 2, . .. , as. The a's are preci ely the zero of odd multiplicity of j;/(x) in the interval. Since f,,(x) is of degree n , it has n zero , mul tiplicity counted. Thus we know that s .::: n. FotTI1 the fu nction (1)
Then, in a < x < b, 1/t(x) changes sign at x = a i, a 2, . . . , a s and nowhere else. If s < n , 1/f(x) is of degree less than n, so b
r w(x)f,,(x)1/f(x) d x = 0 le,
(2)
by the applicati on of Theorem 21.. 1 to each term in the expanded form of 1/f(x) . But the integrand in (2) does not change sign anyw here in the interval of integration, because w(x) > 0 and the function s 1" (x) and 1/f (x) change sign at precisely the sa me points. Therefore, the integral in (2) canno t vani sh and the assumption s < n has led us to a contradiction. Thus we have s = n. That is , among the n zeros of /" (x) there are precisely n of odd multiplicity in the open interval a < x < b. Therefore, each zero is of multiplicity one ; the zeros are distinct. The proof of Theorem 21.2 is complete.
•
422
Chapter 2 1 Ortlwgonal Sets of Functiolls
121 .51 Orthogonality of Legendre Polynomials The Legendre polynom ials
l ~X)
PIl(X) = F(  n,n+ I ; I ;
(1)
of Section 19.8 were obtained by 01ving the differential equation
(1  x2)y"  2xy'
+ /l(n + l) y =
(2)
O.
T he PI/(x) form a simple set of polynomials for which we now obtain an orthogona li ty property. From (2) we have 2 ( I  x )P,:'(x)  2xP,:Cx) 2
D [ O  x )P,:(x)]
+ n(n +
+ n(1l + l )PII(x)
I )PII(x) =
= 0,
d D  dx'
0;
(3)
For index m we have 2 D [ (1  x )P,:,(x) ]
+ m(m + l)P,"(X)
=
o.
(4)
We are interested in finding the product PII,(x)P,,(x) . Hence we multip ly (3) thro ughout by Pili (x), (4) throughout by PI/(x), and subtract to obta in 2 2 PIlI (x)D [ (1  x ) P,: (x)]  PII (x)D[(lx ) P,:,(x)]
+ [n(n + I) 
m(m
+ 1)]Pili (x)PIl(x)
= O.
T he equatio n above may be rewritten (n 2

m2
+n 
m) PII/(x) PIl(x) 2
= PII(x)D [ Cl  x )P,:, (x)]  PIlI(x)D[Cl  x
2
)
P,: (x)].
(5)
Now, by the formula for differentiatin g a product, we get 2 2 D [ (1  x )PIl (x) P,:, (x)] = PIl (x)D[(1 x ) P,:,(x)]
+ (1
2  x ) P,:(x) P,:, (x)
and
Hence,
D[(J  x 2 )l PII (x) P,:) (x)  P,:(x) P", (x) }] = PIl(x)D[(l x 2 )p):,(x)]  PI/l(x) D[(I  x2) p,;(x) ].
2 1. 5
Furthermore, n 2 (5) as (n  m)(n

m2
+n 
m
= (n 
+ In + l)Pm(x)P,,(x)
Orthogonality of Legendre Polynomials
m)(n
423
+ m + J). Therefore, we can write
= D[(lx 2 ){p,,(x)P,:,(x )  P,:(x) Pm(x) lJ.
(6) We have expressed the product of any two Legendre polynomial as a derivative. Derivatives are easy to integrate. Equation (6) yields (n  m)(n
+ Tn + 1)
Ib
PIl/(x)PI/(x) dx
= [(1  x 2 )/ PI/(x) P,:,(x)  P,: (x) PI/l (x ) }] ~ .
(7)
We may choose any a and b that we wish. Since (l  x 2 ) is zero at x =  1 and x = I, we concl ude th at (n  m)(11
+ Tn + 1) j
l p",ex) PI1 (x) dx = O.
(8)
 1
+ m. + 1 =1=
Since nand m are to be nonn egative integers, n. In =1= n , /'l  m =1= 0 and (8) yields
t
O. It follows that if
Pm(x) PI/(x) dx = O.
11
(9)
The Legendre polynomials are real , so J~I p} ex ) =1= O. We have shown that the Legendre po lynomials / P" (x)} form an orthogonal set witb re pect to the weight function w(x) = lover the interval  1 < x < 1. Since / P,,(x) } is a imple set of real polynomials, the lheorems of Sections 21.3 and 21.4 appl y to it. Further study of P" (x) would occupy more space than seems appropriate in an elementary differenti al eq uations text. We now list a few of the simplest from amo ng hundreds of known propertie of these interesting polynomials: 00
(1  2xt
+ t 2) 1/2
=
L
P,,(x)t l1 ,
(10)
,,=0
t
P}(x)dx = _2_ . ,
11
2n
+I
I ? d = 2"n! D"(x  1)'" D=' dx' xP,:(x) = nPn(x) + P,:_I (x) ,
P (x) n
(11 )
( 12) (13)
(x 2  I )P,:(x) = nx P,,(x)  nP,, _1(x) ,
(14)
nPn(x ) = (2/'l  l) xP,,_ I (x )  (n  1) P,, 2 (x).
(15)
424
Chapter 2 I
Orthogonal Sets of FUlictions
121 .61 Other Orthogonal Sets In Chapter 19 we solved several diffe ren ti al eq uations of hypergeometric type. In Section 19.4 we encountered the Laguerre polynomial n (nhxk n ( I/n!xk Ln(x) = { ; (k!)2 = { ; (k!)2(n _ k)!
(1 )
as a ol ution of the ditferential equation xL;;(x)
+ (1
 x)L;,(x)
+ nLn(x)
= O.
(2)
Equation (2) can be put in the form D[xe XL;,(x) ]
+ ne X Ln(x)
= 0,
(3)
from which the orthogo nali ty of the set of Laguerre polynomials foHows. (See Exerc ise 1.) The Herm ite polynomial of Secti on 19.7, [n /2] ( l)kn! (2x),, 2k H" (x)
= {;
k! (Il _ 2k)!
( 4)
'
satisfies the differential equatio n H,:' (x)  2x
H,;(x) + 2n H" (x) = o.
(5)
Eq uation (5) can be put in the form D[exp (_x 2 ) H;' (x)]
+ 2n exp (x 2) H,,(x)
= 0,
(6)
from wh ich the orthogonality of the set of Hermite polynomials follows. (See Exercise 3.) The Bessel function 1" (x) of Section 19.6 can be shown to have orthogonality properti es also, but they are beyond the scope of thi book. I
• Exercises 1.
Use equation (3) above and the method of Section 21.5 to show that the set of Lag uelTe polynomials is orthogonal with respect to the weight function e x over the interval 0 ::: x < 00 .
2.
Show, with the aid of Exerc ise 1, that the zeros of the Laguerre polynomial LIl(x) are distinct and positive.
3. Use equation (6) above and the method of Section 21.5 to show that the set of x2 Hermite polynomials is orthogonal with respect to the weight function eover the interval 00 < x < 00. 4 . Show, with the aid of Exercise 3, that the zeros of the Hermite polynomials HII (x) are real and distinct.
I See, for example, R. V Churchi ll and J. W. B rown, Fourier Series and BOllndary Valu e Problems, 5th ed . (New York: McG raw Hili Book Company, 1993).
22
Fourier Series
122. 11 Orthogonality of a Set of Sines and Cosines T he functio ns sin ax and cos ax occur in the forma l solution of certai n boundary value problems in part ial differential equations, as we indi cated in Chapter 20. We shall now obtai n an orthogonality property for a set of s uch fu nctions with a pecified. A n interval mLl st be involved; let the ori g in be chosen a Lthe center of the interval so the latter appears in the ymmetric form  c :s x :s c. We shall show that the set of functions
n = I,2,3 , . .. } , n = 0, 1,2, . . .
Sin(lln x/c), { cos (nnx/c),
(I)
or } sin (nx/c), sin (2nx/c), sin (3nx/c), ... , in (/1rrxjc), ... { I , cos (nxjc) , cos (2rrx/c), cos (3nx/c), ... , cos (nnx/c), . . . '
(1)
is orthogonal with respect to the weight function w(x) = l over the interval c :s x :s c. That is , we shall prove th a t the integral from x = c to x = +c of the product of a ny two different members of the set ( I ) i . zero. F irst cons ider the integral of the product of any of the s ine functions in ( I ) and any of the cosine functions in (1). The resu lt c
11 =
1
. nrr x kn x sin   cos   dx = 0
c
c
C
fo ll ows at once from the fact that the integrand is an odd function of x; in this instance the resu It does not depend upon the fact that k a nd n are integers. Next consider the integral of the product of two different si ne functions from the set ( I ), c
12 =
1 c
nnx kn x sin   sin   dx,
c
c
k
i= n.
Let us introduce a new vari able of integration for simp li city in writing; put nx =(3,
c
425
426
Chapter 22
Fourier Series
from which
c
= df3. rr
dx Then [ 2 can be written
12 = c rr
1'"
sin nfJ sin kf3 df3 .
rr
Now from trigonometry we get the formula sinnf3 sinkfJ = ~ [cos (11.  k)fJ  cos (n
+ k)f3],
w hi ch is useful in performing the desired integratio n. Thu .it follow s that the integral becomes
~ 2rr
12 =
1'"
_".
[cos (n  k)(3  cos (n
+ k)f3] df3
= ~ [Sin (11.  k)f3 _ sin (n + k)f3] '" , 2rr 11.  k 11. + k _".
since neither (11.  k) nor (11. + k) can be zero. Because nand k are positive integers, sin (n  k)fJ and sin (11. + k)f3 each vanis h at f3 = rr and f3 = rr ; then
12 = 0 for n, k = 1, 2, 3, .. . ; k =/::. n. Finally, consider the integral of the product of two different cosine functions from th e set (1 ), c
h=
1
nrrx krrx cos   cos   dx ,
c
c
c
where 11. , k = 0, 1, 2, 3, . .. ; k =/::. n . The method used in 12 works equally well here to yield
+ sin (11. + k)f3 ] '" n +k _".
[ Sin (n  k)f3 2rr n k
13 = 
C
=
o.
It is easy to see that the integral of the square of any f unction from the set (1 ) will not vanish; its integrand is positive except at an occasional point. The values of those integrals are readily obtai ned. The integral c
14
=
1
nrrx sin 2   dx
c
c
has an even integrand. Hence it can be written as c
[4 =
2
1 o
?
nrrx
sin   dx. C
22.2
Fourier Series: An Expansion Theorem
427
Elementary methods of integration yie ld
fo c(1 
14 =
= [X
cos 2n;x) dx
e 2n.n x ] C 2n:rr sin  e0 = e.

Therefore,
l
c
sin 2
nnx
 dx = e , for 11. = 1, 2, 3, e In the same way it fo llows that for n > 0, n integral, 
c
Is =
l
c
c
2 nn x cos   dx e
e 2nn x ] C = [ x + 2nn sin  e0 = e. For n = 0 the integral Is becomes
h
=
l
c
1 . dx = 2e.
c
Thus
l

c
I1.n x cos 2   dx = e, c e = 2e,
for n = 1, 2, 3, for n = O.
We have shown that the set sin (nn x/e), { cos (mnx/e),
11. = 1, 2, 3, ... } m = 0, 1, 2, .. .
( 1)
is orthogonal with respect to the we ig ht fu nction w(x) = over the interval e ~ x ~ e. We have also evaluated the integrals of the squares of the functions of the set (1 ) .
122.21 Fourier Series : An Expansion Theorem With the assumptio n that there exists a series expansion of the type 1 f(x)= ao + 2
f.. (allcos+b"sin nnx nnx)  ,
w 11=1
e
(1)
e
valid in the interval  e ~ x ~ e, it is a imple matter to determine the coefficients, and bll . 1 Indeed, di sregarding the question of validity of interchanging of order of summation and integration, we proceed as follows . all
1 A reason ror the apparen tly peculiar notation ~ ao fo r the constant term will be seen quite soon.
428
Chapler 22
FOicrier Series
Multiply each term of equation (1) by sin (brxjc) dx , where k is a positive integer, and then integrate each term from  c to +c, thus arriving at c
hex 1 f(x) sin   dx =  ao c 2
I
c
ICsin brx dx c
c
I
C nn:x kn:x ] / Ccos .nn:x kn:x +L [ a" sin   dx + b" sin   sin   dx 00
c
,, = 1
C
c
C
C
.
(2)
C
As seen earlier,
I
nn:x kn:x cos   'in   dx = 0
G
C
c
for all k and n ,
(3)
C
and
I
c
nn:x kn: x si n   sin   dx = 0
c
C
for k
"# n;
k, n
= I , 2,
3, ....
(4)
C
Therefore, each term on the righthand side of equation (2) is zero except for the term n = k. Thus equatio n (2) reduces to
I
c
kn: x f( x ) sin . dx = bk
Ie

C
c
C
 G
sin 2
kn:x
(5)
dx.
Since
I
c
si n2
kn:x 
we have J
bk = c
clx = c ,
C
 G
Ie
kn:x
.r(x) sin 
dx,
k = 1, 2, 3, .. . ,
C
 c
from which the coefficients b" in equation (1) follow by mere replaceme nt of k with n; that i , I b}/ = c
Ie  c
nn:x f(x) sin   dx , c
n = J, 2, 3, .. . .
(6)
Let us obtai n the a" similarly. Using the multiplier cos (kn: x j c) dx throughout the equation (1) and then integrati ng term by term from x = c to x = +c, we get
I
e
c
kn:x 1 f(x ) cos   dx =  ao c 2
~ + L..., ,,=1
I
C
kn:x cos   dx c c
[Ie nn:x kn:x Ie sin .nn:x kn:x ] a" cos   cos   clx + b" cos   dx . c
C
C
c
C
C
(7)
22.2
Fourier Series: All Expallsiol1 Theorem
T he coefficient of bll in (7) is zero fo r alln and k. If k c
1
nnx knx cos   cos   dx C
c
C
c
1
I
a" = 1 C
C
all,
#
k,
=c
fo r n = k,
Clk
C
fro m which Clk> and therefore Thus we get
0 , we know that
for n
1 c
knx f(x) cos   d x =
c
#
=0
and also the coefficient of ~ao is zero . Thus, for k
#
429
0 , equati on (7) reduces to
kn x cos 2   dx, C
c
can be fo und in the way bk was determi ned.
nnx fex) cos  dx,
_ ('
Il
(8)
= I , 2, 3, .. ..
C
Next let u determine ao . Suppose that k = 0 in equation (7) so we have the eq uation c
1
f(x) dx =
~ao 2
c
I
C dx
c
+
f I
C
[all
n=1
cos nnx dx
c
i:
+ b"
I
C
c
C
s in nnx dX ] . C
The terms involving n :::: 1 are each zero. Hence
f(x) dx =
~ao(2c),
fro m which we obtain
l/Cf(x) dx.
(9)
ao = 
C
 c
Equ ation (9) fits in with eq uatio n (8) as a special case n = O. Had the factor ~ not been inserted in equation (I), a separate formula would have been needed. As it is, we write the fo rm al expansion as follows :
1 f(x) = ao 2 with all
= 1 C
bll = 1 C
I I
C
c C
_ ('
~ ( an cos nn x nnx) + L.. + b" si n  n=1
C
nnx f(x) cos  dx,
(10)
C
n = 0, 1, 2,
(1 I)
n = 1, 2, 3,
( 12)
C
f(x)si n nnx  dx, C
Before proceeding to specific examp les and a pplication s, it behooves us to state co ndition s under which the eq uality in (10) makes sense. When all a nd bn are given by (1 1) and (12) above, then the eries on th e right hand si de of equation
430
Chapter 22
Fourier Series j( x )
c
0
.£
c
x
2 0
0
Figure 22.1
(10) is cal.l ed the Fourier series, over the interval c .:s x .:s c, for the function (x) . A state ment of conditions s uffi cient to ensu re that the Fourier series in (10) represents the f unction f(x) in a reasonably meaningful manner follows. Let f(x) be continuous and differentiable at every po int in the in terval c .:s x .:s c except fo r, at most, a fi ni te nu mber of points, and at th ose point let f(x) and /,(x) have right and lefthand lim its. Such a fu nction i exhib ited in F igure 22. 1.
f
Theorem 22.1 Under the stipulations of the preceding paragraph, the Fourier series for f (x), namely the se ries on the right in equation (10) with coefficients given by equations (11) and ( 12), converges to the value of f ex) at each point of continuity of f (x); at each point of discontinuity of f(x) the Fou rier series converges to the arithmetic mean of the values approached by f (x) from the right and the left. Since the Fo urier series for f (x) may not converge to the val ue f (x) everywhere (e.g., at d iscontinuities of the f unctio n), it is customary to replace the equals sig n in eq uation (10) by the sy mbol "', which may be read "has for its Fou ri er series." We write
2::: 00
1 + ( nnx f(x) '" .a all cos  o 2 c 1/= 1
nnx) + bl/ sin  , C
(1 3)
wi th a" and bl/ given by equation (1 1) and ( 12). An in teresting fac t and one often useful as a check in numeri cal problems is that ~ao is the average value of f(x) over the in terval c .:s x .:s c. The sine and cosi ne functions are peri odic with period 2TC, so tb e term s in the Fourier seri es (13) for f(x) ar e periodic with period 2c. Therefore, the series represents (co nverges to) a function that is described above for the interval c < x < c and repeats that struchl re over and over out ide that in terval. For
22.3
Numerical Examples of Fourier Series
431
I(x)
/
L 2c
c
3c 2 0_
c
•
0
/
./ /
c
c
•
"2
2
3c 2
2c
x
0_
Figure 22.2
the fu nctio n ex hib ited in Fig ure 22.1, the corresponding Fourier series would converge to the period ic function shown in Figure 22.2 . Note the convergence to the average va lue at the discontin uiti es, the periodicity, and the way in which the two together dete rmine the value Lo which the seri es converges at x = c and x = c. T hese taLements will be amply illu strated in the numerical exerc ises of the next ec ti o n.
122 .31 Numerical Examples of Fourier Series We shall now construct th e Fouri er eries for two specific f un ctiom
EXAMPLE 22.1 Co nstruct th e Fourier series over the interval
2.:::: x.:::: 2 fo r the fu ncti o n defi ned by f(x) = 2,
=x,
 2 < x.:::: 0, < x < 2,
°
(1)
and sketch the graph of the function LO whi ch the seri es converges. F irst we sketch f(x) it elf, the result being exhibited in Figure 22.3. Note that f(x) is undefi ned except fo r x between x =  2 and x = +2. For the fu nction described in (l),
I
f(x) '"  ao
2
in which
1 a = II 2
12 2
( It1TX +~ ~ 0" cos   + b
,, =1
n1TX
f(x) cos   d x ' 2 '
2
It
m TX )
ll
sin   ,
= 0, 1,
2
2, ... ,
(2)
432
Chapler 22
Fourier Series j(x)
 O. We hall determi ne the temperature in the sphere fo r positive f under the ass umption that the heat eq uation
au = at
( a2u+a2u+a2u) ax 2 ay2 az2
h2
_
(1)
is vali d. Si nce the o bj ect under stud y is a sphere, we choose the origin at the center of the sphere and introduce spheri ca l coord inate , related to x, y, z, by
z=
y = ps in cpsin e,
x=ps in cpcos8,
p coscp .
Then the heat equ ati on becomes 2 2 2 2 au = h 2( a u+ ~ au + ~ a u+ cotcp ~ + csc ¢ a u). at ap2 p ap p2 acp2 p2 a¢ p2 ae 2
(2)
For the problem we wi sh to solve, th e temperature is independent of th e coordin ates e and cp , so equ ati on (2) reduces to
(aap2u + ~p aapu) . 2
au = at
h2
Let R be the radiu s of the sphere and problem confronti ng us is
.f (p)
(a
au = h2 u + ~ au) at ap2 pap
(3)
be the in itial temperatu re. Then the
2
as p
~
R , u
as t ~ 0+ ,
U
0
~
~
for 0 < t , 0 < P < R;
f
for 0 < t ; fo r 0
(p )
~ p
(5)
< R.
The student can easily show that the chan ge of dependent variabl e, v U
(4)
= ,
p
(6)
(7)
transform the problem (4) through (6) in to the problem
av

at
2
a2 v
= h 
ap2
for 0 < t, 0 < P < R;
(8)
458
Chapter 23
Boundary Value Problellls
as p
~
R, v
~
0
v asp ~ 0+,  ~ a limit p
as f ~ 0+,
V ~
pf (p)
for 0 < t;
(9)
for 0 < f;
(10)
for 0 < p < R.
( 11)
The added condition (10) is a reflection of the fact that the temperature u is to exist at p = 0 in spite of relation (7). The new problem (8) through (11) is much like those treated at the beginning of this chapter. Its so lution is left as an exercise. The corresponding problem of finding the temperatures in a solid cylinder is less elementary and involves series of Bes el functions. It may be found worked out in many books.2
• Exercises 1.
Solve the problem (4) through (6) by the method outlined above.
2.
A sphere of radius R is initially at a constant temperature Uo throughout, then has its surface held at temperature U I for t > O. Find the temperature throughout the sphere for t > 0 and in particular the temperature U c at the center of the sphere.
123.51 The Simple Wave Equation If an elastic string held fixed at two points is taut and then is displaced from equilibrium position and released, the subsequent displacements from the position of equilibrium may be determined by olving a boundary value problem. Figure 23 .7 shows a representative displacement of the string, which is to be fixed at x = 0 and x = c . The displacement y for 0 < x < c and 0 < t is to be found from the known initial displacement I(x) , the initial velocity ¢(x) , and the fact that y must satisfy the onedimensional wave equation 2 a2 y 2 Cl y 2 a 2 Clt Clx '
(1)
in which the parameter a is a constant that depends upon the physical propelties of the string. The boundary value problem to be so lved is for 0 < x < c, 0 < f;
(2)
2 See, for example, E . D. Rainville, Imerm ediate Dil/erell/ia! Equatiolls. 2nd ed. (New York: Macmillan Publishin g Company, 1964), or R. V. Churchill and J. W. Brown, Fourier Series and BOlllldary Valu e Problems. 5th eel. (New York: McGrawHi]] Book Co mpany, 1993).
23.5
The Simple Wave Equation
459
f( x )
r~ x
o
c
Figure 23.7
as x + 0+, Y + 0
forO < t;
(3)
fo r 0 < t;
(4)
as t + 0+ , Y + f(x)
for 0 < x < c;
(5)
oy as t + 0+,  + 0, S > 0, we know how to evaluate L  I\e  h·,'j(s)} by Theorem 15.3 of Section L5.4. Indeed, L  I {e  h'\' f(s)} = F(t  h)a(t  h) ,
h > 0, s> O.
(12)
We therefore rewrite (11) as f (s )
2f(s)e  CS
sinh (cs)
I  e 2cs
(13)
because we can use the power serie (I) to expand (l  e 2cs ) 1 in a series of exponentials with negative arguments. From (1) we get 1
L exp (2ncs), 00
=? = 1  e_ CS
1/=0
so, by (13), /(s)
00
. =2L/(s)ex p(  2ncs cs). slll h (cs) 11 =0
(14)
24.1
Po wer Series Clnd Inverse Transforms
469
We now use ( 12) to obta in , for c > 0, s > 0, L
I {
. f(s) } = 2 L,OO F(t  2ne  c)cx(f  2ne  c) . smh (cs) n=O
(15)
It is important to realize that the series on the right in (15) is a finite series. No matter how large the value of t nor how sma ll the (positive) e, the argument of the a function wi 11 become negative for sufficiently large n and for all succeeding n values. Thus each term of the series will be zero for all n such that (2n + l )c > f .
•
The procedure used in this example is of value to us in app lications invo lving boundary value problem in partial differential equations, to be discussed in Chapter 25.
EXAMPLE 24.2 Evaluate L
{I I}  te
. By (2) we obtain
L 00
e I =
(_
1)11 til n!
1l= 0
= 1+
I)" til
L It!00
(_
11 = 1
Therefore, we may write
A shift of index from n to (n
+ I) yields
l  e
I
=
t
f
(  1)"f"
1l=0 (n
+ I )!
We know that L { t" } = II . Hence n.! S" + I  e I L { t
}
=
L
Il=o
(
(n
1)"
+ l)S,, +1'
so comparison with (9) above yields I
L { J t e
}
= In
(1 + ~) ,
s > O.
(16)
The restriction s > 0 may be obtained by examining the integral definition of the left member of (16) . Note al 0 the con nection with Exerci e 19 of Section 14.10 .
•
470
Chapter 24
Additional Properties of the Laplace Transform
EXAMPLE 24.3 Evaluate L I {In s + 1 } . From (10) we have s 1
s+1
In . sl
Now L
1+.!. = In _ _ = 2"
;7:0 (2n+l)s2"+1
1~
I {_21I} =~. s 11+
1
00
.1
(2n)!
.
Hence 211
00 t s + 1} L  1 {I n   = 2 " , s 1 (2n + I)!
;7:0
which, with the aid of (6), yields L  1 {lnS + S 
I} = ~sinht.
1
(17)
•
t
• Exercises 1.
sin kt } Evaluate L {  t .
2.
Evaluate L {
5.
Eval uate F(t) = L
I{ 3(1 1 e2.\). }and compute F(5).
6.
Evaluate F(t) = L
I {s3cos1h (2s) } and compute F(12).
7.
Let 0, s > 0, and let L  1 (f(s)} = F(t) . Prove that L 1
{
I  cos kt } t·
I {
.
{ Sinh (kt) } t .
4.
Evaluate L
{ 1  cosh (kt) } t .
3
}. Compute 4>(10).
} =2I:(lt F (t  2nC  C)Ct(t2nCC).
cosh (cs) 9.
Evaluate L
S
s smh (3s)
/(s)
3.
n=O
I
Let c > 0, s > 0, and let L I {J(s) = F(t). Prove that 00
L  I {J(s) tanh (cs)}
= F(t) +
2
L( 1)'1 F(t 
2nc)a(t  2nc).
11 = 1
10.
Let 0 < x < 1, where x does not depend on s. Find the inverse transform y(x, t) of S3 (e·\·
+ e
S)
and then compute y(!, 5), assuming continuity of y.
24.2 The Error Function
471
11 .
In Exercise 4 of Section 12.4, replace the alternatingcurrent eleme nt E sin wt by EQ(t , c) , in which Q is the squarewave function of Figure 14 .4 of Seclion 14.10.
12.
In Exercise 4 of Section 12.4, replace E sin wt by EF(t), in which F(t) is the halfwave rectification of sin wt as described in Exercise 17 of Section 14.10.
124 .21 'I 'he Error t:unction The error function, abbreviated "erf," which was mentioned briefly in Section 5.6, is defined by eli'x =
2 r ft 10
exp (_ f32) df3.
(1)
This function arises in many ways. It is sometimes studied in elementary courses. We also encounter erf x in evaluating inverse transforms of certain simple functions of s. We know that L  I {s  1/2 } = (TC t)  1/ 2 and therefore that
I} =.jiit' e{.JS+T I
L
1
Then the convolution theorem yields
LOn the right in (2) put
L
I {
I} = 10t 1. ~ df3 N .
s.JS+l
v1J =
I {
s
y. Then {3  1/2 df3 = 2dy and we obtain
~} s
(2)
+1
=
~
v TC
[0
10
exp (_y2) dy .
That is,
L
I {
~}
s s+1
= erf(.Jt).
(3)
A few basic properties of erf x are useful in our work and will now be obtained. Directly from the definition (1) it follows that the derivative of erf x is given by
d
2
 erfx = exp(  x 2 ). dx ft
(4)
From (1) and the power series for exp (_f32) we get 2
erf x =
00
ft ~
(IYx 211 + 1 (2n + l)n!
(5)
472
Chapter 24
Additional Properties of the Laplace Tran~form
In elementary calculus we found that
1
00
o
?
exp ( f3)
.jli
dfJ =  . 2
(6)
From (6) we get lim elfx = 1. ..\" 4
(7)
00
The values of erf x are easily computed for small x from (5) above and for larger x from the asymptotic expansion I exp (_x 2 ) oo (1)"[1·3·5··· (2n  1)] (8) erf x '" I . 'rr n=O 2"X 211 + 1 y;l
L
It is convenient in our work to use what i call ed the complementary error function , denoted by erfc x and defined by erfcx = 1  eifx,
(9)
which means also that
Joo exp(f32)df3.
2 x erfcx=,fir
(LO)
The properties of erf x are readily converted to properties of erfc x. It is important that for any fixed m, lim
XIII erfc
x
X """7 00
= 0,
(11 )
which the student can demonstrate by considering the indeterminate form eric x and using the derivative of erfc x as obtained from (4) above. See the exercises at the end of this section for other properties of erf x and erfc x. A transform that is impoltant in certain appl ications (Sections 25.3 through 25.6) is
in which k is to be independent of t and k > O. By definition of eric x we have erfc (
kr;) =
yt
~ yT!
[00
exp (_f32) df3.
(12)
k/ ,fi
I See, for example, E. D. Rainville, Special FUllctions (New York: Macmi ll an Pub lishing Company, 1960), pp. 3638. The function erf x is tabulated under the name "The Probability Inlegral," in B. O. Peirce and R. M. Foster, A Short Table of Integrals, 4th ed . (Lex ington, Mass.: Ginn and Company, 1956), pp. 128132.
24.2
The Error Function
473
t to v = o. Since df3 = _~ kV3/2 dv, we obtain (using the minus sign to reverse the order of integration) 2 erfc (~) = ~ v3/2 exp (_ k ) dv . (13)
In (12) put f3 = k/.JV so that the limits of integration become v =
t
~
,JJi Jo
v
The integral on the right in (13) is a convolution integral. Hence L {erfc
(~ ) } =
In
L{ I} . L
{t
3 2 /
exp ( _
~2 ) },
or (14)
Now let (IS)
Note that the functions till exp (_k2 / t) are of class A, Sectio n 14.6, fo r each m. From (15) it follows, by Theorem 14.8 of Section 14.8, that
dA ds
t
= L {  t _ 112 exp (k 2 ) }
(16)
2 d A  = L {1 t /2 exp (k 2)} . ds 2 t
(17)
and
But also, by Theorem 14.5 of Section 14.7, L
L~/12exp (  ~2) }
2
= sL {tl /2 exp (_ k
t
)} _
l~~+ [tl/2 ex p (_ ~2 ) ] ,
or L
{!t1/2 exp ( _ ~2 ) + k2t  3/2 exp ( _ ~2 )
}
= sL
{t 1/2 exp ( _ ~2 )
} o. (18)
Because of (15) , (16), and (17), eq uation (18) may be written 1 dA 2 d2 A . 2 ds ds 2 Therefore, the desired function A (s) is a solution of the differential equation
   +k A = s 
d2 A ds
s2
Id A +
2 ds
2
k A = O.
(19)
474
Chapter 24
Additional Properties of the Laplace Transform
We need two boundary conditions to go with equation (19). We know that as s 7 00, A 7 O. Now consider what happens as s 7 0+ . By (15)
5> 0+
.1>0+
=
2
1
00 e sl t 3 / 2 exp ( k )
lim A(s) = lim
t
0
dt
10[ 00 t~/2 exp (k2) '( dt. 0
Equation (13) yields (with y replacing t) 2
k ) 1[0y v 3/ 2 exp ( ;
dv
k ) . = k.fii erfc ( v'Y
(20)
Therefore, lim A(s) = .fii lim etfc k y>oo
s>O+
(~) v'Y
=
.fii erfcO = .fii . k
k
To get the general solution of the differential equation (19), we change independent variable 2 from s to z = /5. Now by the chain rule of elementary calculus d z dA 1 dA I dA dA  = ds 2z dz ds d z 2/5 d z and dA
=      
4s
d z2
4s/5 dz .
Thus d 2A
J d2AIdA 4 dz 4z dz
s=   2 2 ds
and equation (19) becomes 2
dd zA 4k2A= O. 2
(21)
The general solution of (21) is A = b l exp (2k z)
+h
exp (2kz),
so the general solution of (J 9) is A = b l exp (2k.jS)
+ b2 exp (2k.jS).
(22)
2 Such a cha nge of variable is dictated by the test on page 16 of E. D. Rainville, Intermediate Differential Equations, 2nd ed. (New York: Macmillan Publishing Company, 1964).
24.2
The Erro r FUI1C1ion
475
We must determine the constants hi and b2 from the conditions that A + 0 as + 00 and A + fi I k as s + 0+. As s + 00, A will not approach a lim it unless b2 = O. Then, letting s + 0+, we get
S
ft

k
=b l ·
Therefore, A(s) = L
{t3/2
ex p ( _
~2)}
=
~ exp(2k.Js).
We return to (14) to write the desired transform
L{erfc (~)} =~eXp(2kJS),
k > 0, s > O.
(23)
We shall use (23) in the form 1
L  {} eX P (2kJS } =
erfc(~) ,
k > 0, s> O.
(24)
In Chapter 25 it will be important to co mbine the use of equation (24) and the series methods of Section 24.1. Consider the problem of obtaining L
I
{ Sinh (xJ'S) } ssinh J'S '
0 < x < 1, s > O.
(25)
If x were greater than unity, the inverse in (25) would not exist because of the behavior of sinh (x J'S) I sinh J'S as s + 00 .
Because we know (24), it is wise to turn to exponentials . We wr ite sinh (xJ'S) exp (xJ'S)  exp (x J'S)  = = exp (J'S)  exp (J'S) . sinh J'S
(26)
As in Section 24.1 , we seek a series involv ing exponentials of negative argument. We therefore multiply numerator and denominator on the right in (26) by exp (J'S) and find that sinh (x .J s) sinh ,.fi
=
exp [
(1 
x)J'S]  exp [
(1 + x)J'S]
1  exp ( 2J'S)
(27)
Now I
L exp ( 2nJS). 00
 =Ie 1  exp(2ys)
=
(2 8)
11=0
Therefore, ~ J sinh (xJ'S) . J'S = ~ {exp s smh s 11 = 0 s
[ (1 
x
+ 2n)JS]
 exp [(I
+ x + 2n)JS]).
476
Chapter 24
Additional Properties of the Laplace Transform
For 0 < x < 1 the exponenti als have negative arguments and we may use (24) to concl ude that
L
 I
{ Sinh (xv's) S sinh v's
f
}=
[ erfc
11=0
(1 20 x 2n) _ (I +
erfc
+
x + 2n) ]. (29) 20
Exercises 1. S how that for all real x , Ierf x I < 1. 2. Show that erf x is an odd function of x. erf x 2 3. Show that lim   = ~. x + o
4.
y If
X
Use integration by parts to show that
r
Jo
1
erf y d y = xerfx  f i [1  eXp ( x
2
)].
5. Obtain the result of equ ation (11 ). 6. Start with the power series for erf x , equation (5), and show that L {t 
7.
I 2 /
erf (0)} =
2
]
'ylfS
yS
c;: arctan In'
S
> O.
Use the fact that 1
 == =
] + .J1+S
1 ,.J1+s 1 1+S =   + =  +== 1 (l +s) s s S s,.J1+s
1  ,.J1+s
and equation (3) to show th at L
8.
I
{
I} = 1 +
,.J1+s 1+ l+s
e r
erf (.Jt) +
c;
ylft
=
e I c; ylft
Use eq uation (3) to conclude that L
I
{
1 v's}=e'erfC.Jt) s
(s  ])
and therefore that L
I
{v's
~
s ( s + 1)
r
} = e erfc (.Jt).

erfc (.Jt).
24.2
10.
Eval uate L 
J 1.
Define the function 0, s > O. (ll)
s
25.1
Boundary Valu e Problems
483
Because of (9), the w of (11) is to approach a limit as x + 00. The first two terms on the right in (11 ) approach limits as x + 00, but the term with the positive exponent, exp (4sx), will not do so unl ess
C2(S) = 0.
(12)
That is, (9) forces (12) upon us. The w of (11) then becomes 1
w = 2 +c,(s)exp(4sx), x> 0, s > 0. s Application of cond iti on (8) to the w of ( 13) yields 2
1

S3
= c,(s)   ; S2
2
c,(s) =  3 S
(13)
1
+ 2 · s
Thus we find that w(x, s) =
~ 2+ s
(: + s.,
~) exp (4sx), s
x > 0,
S
>
o.
(14)
We already know that if
L  '! eC.rf(s» ) = F(t  c)Q".(t  c).
( l5)
Therefore, the applicatio n of the operator L  , throughout ( 14) gives us y(x, f ) =  t
+ [(t
 4x) 2 + (t  4x)]Q".(t  4x),
x > 0, t > O.
( 16)
It is OUf conten tion th at the y of (16) sati s fie s the boundary value problem ( I) through (5). Let us now verify the sol uti o n in detaiL From ( 16) it follows at o nce that
ay = at

 1 + [2(/  4x)
+
1]Q".(t  4x),
x > 0, t > 0,
f
t= 4x.
(17)
Note the discontinuity in the derivative for t = 4x. This is forcing us to the admi ssion that we obtain a solution of the problem only on each s ide of the line t = 4x in the first qu adrant of the xf plane. Our y will not satis fy the differential equation along that line because the second derivative cannot exist there. Thi s is a reflection of the fact that (1) is a "hyperbolic different ial eq uation." Whether the "solution" does or does not satisfy the differential eq uation alo ng what are call ed characteristic lines of the equat ion depends upon the specific boundary conditions. We shall treat each problem individually with no attempt to exami ne the ge neral si tu ation. From (17) we obtain
a2 y at 2 =
2Q".(t  4x),
x > 0, t > 0, t
t= 4x.
(18)
484
Chapter 25
Partial Differential Equations: Trallsform Methods
Equ ation ( 16) also yields ay ax = [8(t  4x)  4]aCt  4x) ,
x > 0, t > 0, t
i= 4x,
(19)
and
a2 y 
ax
2
x > 0, t > 0, t
= 32a(t  4x) ,
i= 4x .
(20)
Equations (18) and (20) combine to show that the y of (16) is a solution of the differential equation ( J) in the xl region desired, except along the li ne t = 4x, where the second derivatives do not ex ist. Next we verify that ou r y atisfies the boundary conditions. To ee whether y satisfies condition (2), we must hold x fixed, but positive, and then let t approach zero thro ugh pos ilive values. As t ~ 0+, Y + 0 + [( _4x)2 + (4x)]a( 4x) = 0
fo r x > O.
T hus (2) is satisfied. Note that a ( 4x) would not have been zero fo r negative x. Fro m ( 17) , with x fixed and positive, it follows that as t + 0+, ay
at
~
 1 + [2(  4x)
+ l ]a(4x)
=  1
for x>
o.
T hus (3) is satisfied. O nce more the fact that x is positive plays an impo rtant role in the veri fica tion. Consider condition (4). In it we must hold f fixed and pos itive. T hen, by (16), as
x + 0+ , Y +  { + (t 2 + t)a(t)
=  t + t 2 + t = t2
for t > O.
T he n (4) is satisfied . Finall y, the y of (16) satisfies condition (5), si nce
lim y(x , 1) =  l x ~oo
+0 =
t
for t > 0,
because for sufficiently large x and fixed t , (t  4x) is negative and therefore aCt  4x) = O. T his comp letes the verification of the so lution (16).
•
• Exercises [n each exerci. e, . olve the problem and verify your sol ution completely.
1.
ay ay  +4
=  8f ax at t ~ 0+, Y ~ 0
x ~ 0+, Y +
2{ 2
for 1 > 0, x > 0; for x > 0; for t > O.
25.2
2.
ay + 2 aY ax at
= 4t
The Wave Equation
485
for t > 0, x > 0; for x > 0; for t > O.
t +
0+, Y + 0 x + 0+, Y + 2t 3 3.
Solve Exercise 1 with the condition as t + 0+ replaced by t + 0+, Y + x.
4.
SolveExercise2withtheconditionas t + O+ replacedbyf + 0+, y + 2x.
5.

a2 y
ax
a2 y
= 16 
at 2 t + 0+ , y + 0 ay t + 0+,  + at 2
x + 0+, y + t
lim y(x, t) exists
x>oo
6.
a2 y
a2 y
for t > 0, x > 0; for x > 0; 2
for x > 0; for t > 0; for t > O.
= 4 2 2
for t > 0, x > 0;
t + 0+ , y + 0
for x > 0;
at
t
ax ay + 0+, at
+ 2
for x > 0;
x + 0+, Y + sin I for t > 0; lim y(x, t) exists for t > O. x*oo
125 .21 The Wave Equation The transverse displacement y of an elastic string satisfies the onedimensional wave equation
a2 y
2
a2 y
=a
at 2
ax 2
of Section 23.5, in which the positive constant a has the dimensions of a velocity, centimeters per second, and so on. Suppose that a long elastic string is initially taut and at rest so that we may take, at t = 0,
y= 0
and
for x
~
o.
We assume the string long enough that the assumption that it extends from x = 0 to 00 introduces no appreciable error over the time interval in which we are interested. Suppose also that the end of the string far distant from the y axis is held fixed, y + 0 as x + 00, but that at the yaxis end the string is moved up and down according to some prescribed law, y + F(t) as x + 0+, with F(t) known. Figure 25.1 shows the position of the string at some t > O.
486
Chapter 25
Partial Differential Equations: Transform Methods
+=~+ x
o
Figure 25.1
The problem of determining the transverse displacement y in terms of x and t is that of solving the boundary value problem: 2 a2 y 2a y = aat 2 ax 2 t + 0+ , Y + 0 ay t + 0+,  + 0 at x + 0+ , Y + F(t ) lim y(x, t) = 0

for t > 0, x > 0;
(1)
for x ::: 0;
(2)
for x > 0;
(3)
for t ::: 0;
(4)
for all t ::: O.
(5)
x~oo
The prescribed function F(t) must van ish at t = 0 to retain continuity of the string. This problem satisfies the criteria, Section 25 .1 , that suggest the use of the Laplace transform. Let L(y(x,
t) } =
u(x, s),
L( F(t)}
= f(s).
(6)
Note that F(t) must be continuous because of its physical meaning here. The operator L converts the problem (1) through (5) into the new problem
S2u
= X
d 2u for x > 0; dx 2 + 0+ , U + f (s); a2_
li m u(x, s) = O. X""" 00
(7) (8)
(9)
25.2
487
The Wave Equation
From (7) we wri te at once the general solution u(x, s ) =
CI
'; . ( SX) + C2(S) ex p (~X)
(1 0)
(S) ex p  ;
With s > 0, x > 0 , the condition (9) requires that ( I I)
Thu s (10) becomes u (X, s) =
CI
(s) exp (_
S;),
( 12)
and (8) requires th at f(s) =
CI
(s).
We therefore have u(x , .\')
SX) , = I(s) exp ( ;
> 0.
( 13)
0, t > 0,
(1 4)
x > 0,
S
Equation ( 13) yields the desired soluti on , y(x, t)
= F (t  ~D a (l  ~) ,
X>
in which we assume that F(t ) is defi ned in some manner fo r negative argument so th at T heo rem 15.3 of Sec ti o n 15.4, can be used. Verifica tion of th e oluti on ( 14) is a simp le matter. Note that
aroy =F' (t;;X) (t;;X) , Ct
oy
ox
= _ ~ F' a
(t  ~) (f  ~) a a Ct
and
aox'2y2 =
I
a 2 F"
(
I 
X) Ct (t  ;;X)
;;
.
We are forced to ass ume the ex istence of two derivatives of the prescribed f un ction F (t) . It is particul arly convenient to c hoose F(t) so that F' (0) and F " (0) vanish a long with F(O), so th at the co ntinuity of y and its deri vatives are not intelTupted al ong the li ne x = a f. Co mpl etion of the verifi cat io n of the sol utio n is left to the student. Tn Section 23 .5 we stud ied th e transverse disp lacement of a string of fin ite length held fi xed at both ends. Fo urier seri es methods seem superior to Lapl ace transform techniqu es fo r uc h problems. Try, for instance, transform methods on Exercise I of Sectio n 23 .5.
488
Partial Differential Equations: Tran.~form Methods
Chapter 25
• Exercises 1.
Interpret and solve the problem:
at2
=
for t > 0, 0 < x < 1;
ax 2
t + 0+ , Y + x  x 2
for 0 < x < ( ;
+ ay t+O , +0
for 0 < x < 1;
x + 0+, Y + 0
for t > 0;
x + 1 , y + 0
for t > O.
at
Verify your solution directly.
125.3 1 Diffusion in a SemiInfinite Solid Consider the solid defined by x ::: 0, occupying one half of threedimensional pace. If the initial temperature within the solid and the conditions at the surface x = 0 are independent of the coordinates y and z, the temperature u will be independent of y and z for all t > O. We may visualize, for example, a huge flat lab of concrete with an initial temperature distribution dependent only upon the distance from the plane sUlface of the slab. If the temperature at that surface is thereafter (t > 0) maintained at some specified function of t, or if the surface is insulated, the problem of finding the temperature for all positive x and t is one involving the simple heat equation (2) of Section 23 .1.
EXAMPLE 25.2 Consider a semiinfinite slab x ::: 0, initially at a fixed temperature u = A and thereafter subjected to a surface temperature (x + 0+) which is u = B for o < t < to and then u = 0 for t ::: to. Find the temperature within the sol id for x > 0, t > O. The boundary value problem to be solved is
au at
2
a2 u ax 2
= h t + 0+ ,
U
for x > 0, t > 0;
(1)
+ A
for x > 0;
(2)
(3)
x + 0+,
U
+ B
for 0 < t < to,
x + 0+ ,
U
+ 0
for t > to;
lim u (x , t) exists
x"'" 00
for each fixed t > O.
(4)
25.3
Diffusion in a SemiInfinite Solid
489
In this problem A, B, and h 2 are constants. We use the a function to reword the boundary condition (3) in the form
x + 0+,
for t > O.
u + B[l  a(t  to)]
(5)
Note also that the physical problem dictates that th e value of the limit in (4) is to be A. This furnishes us with an additional check on our work. The problem satisfies the criteria, Section 25.1, that uggest the use of the Laplace transform. Let x > 0, s > O.
L{u(x, t)} = w(x, s),
(6)
The equation (1) with condition (2) is transformed into x> 0, or
d2w dx 2
S A h2 W =  h 2 '

X
(7)
> 0.
Conditions (4) and (5) become > 0
(8)
w + [1  exp (loS)].
(9)
lim w(x, s) ex ists
for fixed
S
X"'" 00
and
B s
The differential equation (7) has the general solution
( xfi) +
w = c, exp  h
C2
exp
(xfi) h
A, + ~
x > 0, s > 0,
(0)
c,
in which and C2 may be functions of s, but not of x. As x + 00, the w of (10) will approach a limit if, and only if, C2 = O. Hence condition (8) yields the result (1)
and the w of (10) becomes w =
c, (_x;)+ ~. exp
(12)
By letting x + 0+ and using (9), we obtain B
[1  exp(tos)] =
s
A
c, +.
(13)
S
Therefore, the solution of the problem (7) through (9) is w(x, s) =
~ [1
exp (
x;)]+ ~
exp (
X;)[l_
exp (tos)]. (14)
490
Chapter 25
Partial Differential Equations: Transform Methods
We know that x >
o.
(15)
Hence we may write L
I {~exp( x .Js)exp(toS)} = erfc ( h
S
x
2hlt  to l
1/2 )a et  to),
(16)
where absolute value signs have been inserted to permit t to be used in the range in which range the a function will force the right member of (16) to be zero. We are now in a position to write the inverse transform of the w of equation (14). For x > 0 and t > 0,
o to to,
uex, t) = A
[1  elfc (2:0) J + B [erfc (
I.) 
x 2h..; t
erfc (
1/2) a(t  to)J,
x 2h It  to I
(17)
or u(x , t) = A erf
(2:0)
+B [erfc( 2h..;x r;)t  erfc( 2h lt x to l1/2 )a et 
)J.
to
(18)
The u of (17), or of (18), is the desired solution. It is a matter of direct substitution to show that each term of (18) is a solution of the onedimensional heat equation. That the conditions (2), (3) , and (4) are also satisfied follows rapidly from the properties lim erf z = 0,
z+ o
lim erf z = 1
z+oo
and the corresponding properties of the erfc function. Indeed, for the u of (18), as x + 0+,
U
+ A· 0 + B[l  a (t  to)] = B[1  a(t  to)]
as t + 0+ , u + A . 1 + B(O  0) = A as x + as x +
00,
u + A . I + B . 0 = A
00,
U
+ A . 1 + B(O  0) = A
for t > 0; for x > 0; for 0 < t < to; for t > to.
25.4
Canonical Variables
491
125 .41 Canonical Variables As we attack problems of increasing complexity, it becomes important that we simplify our work by the introduction of what are called canonical variables. These variables are dimensionless combinations of the physical variables and parameters of the original problem. We now illustrate a method for selecting such variables. In Section 25.5 we shall solve a diffusion problem that can be expressed in the following way: 2 au 2a u = 112 at ax
for t > 0, 0 < x < c;
(1)
t + 0+ ,
U
for 0 < x < c;
(2)
x
u + O u+O
for t > 0;
(3)
for t > O.
(4)

+ 0+,
x+c  ,
+ A
A consistent set of units for the measure of the various constants (parameters) and variables in this problem is u = temperature (OP), t =
x
time (hr),
= space coordinate (ft),
112 = thermal diffusivity (ft2/hr), c = length (ft), A = initial temperature (OP). We seek dimensionless new variables t; , variables x , t , u. Por the moment let x =
f3t;,
T,
1/1, proportional to the physical (5)
u = 81/1,
t = yT,
in which f3, y, 8 are positive constants to be so determined that the new variables will each be of dimension zero. The changes of variable (5) transform (1) through (4) into
o a1/l y
11 2 8 a2 1/1

ar
f32 at;2
T
+ 0+,
t;
+ 0+ ,
f3 t;
+
C ,
for r > 0,
81/1 + A 1/1 + 0 1/1 + 0
°
0;
for
T
> O.
492
Chapter 25
Partial Differential Equations: Trallsform Methods
Because of (7), we choose 8 = A and f3
= c. Because of (6), we choose
h2
y
f32'
from which
Y
=
c2 h2 '
We thus find that the introduction of the new variables
u 0/ =, A
x
{= ,
c
(10)
transforms the problem (1) through (4) into the canonical form
d2 0/ do/ d{ 2 dr r + 0+, 0/ + 1 { + 0+, 0/ + 0 s + 1, 0/ + 0
for r > 0, 0 < { < 1;
(11)
for 0 < { < 1;
(12)
for r > 0;
(13)
for r + O.
(14)
s
Note that the canonical variables in (l0) are of dimension zero; has dimension feet over feet, and so on. The sol ution of (11) through (14) is independent of the parameters h 2 , c, and A of the original problem, a fact of great importance in applications. The solution of the original problem (1) through (4) is a function of two variables and three parameters, u = f(x, t, c, h, A).
(15)
The olution of (11) through (14), for which see Section 25.5, is a function of two variables
0/ = F(s, r) ,
(16)
so (15) actually takes the form U
= AF
X
(
h2t) .
,c c2
(17)
The function F , of two variables, can be computed and it thus yields the solution of the original problem no matter what the values of c, A, and h 2 . There are problems, such as in the study of temperatures in a concrete dam, in which it is important to know the mean value with respect to x of the temperature u of (15) over a range 0 < x < c. That mean value may be computed by using
25.5
Diffusion in a Slab of Finite Width
493
(16), and the result is a function of the one vruiable r. Thus a single curve can be drawn in the 1/Ir plane to give the pertinent mean temperature for all problems (1) through (4).
1'25.51 Diffusion in a Slab of Finite Width We shall now solve by transform methods the slab problem of Section 20A for the special case f (x) = A . Let the thickness of the slab be c units of length. Let the coordinate x denote distance from one face of tbe slab and assu me that the slab extends very far in the y and z directions. Assume that the initial temperature of the slab is a constant A and that the slllfaces x = 0, x = c are maintained at zero temperature for all t > O. If the slab is co nsidered infinite in the y and z directions or, more specially, if we treat onl y cross sections nearby (far fro m the distant surfaces of the slab), then the temperature u at any time t and position x is determined by the boundary value problem:
au at
?
a2 u ax 2
=11  
for t > 0, 0 < x < c;
(1)
A
for 0 < x < c;
(2)
t
7
0+,
x
7
0+,
U 7
0
for t > 0;
(3)
X 7
c,
U 7
0
for t > O.
(4)
U 7
We shall solve the corresponding problem in canonical variables. That is, in
0) through (4) we pu t ~
In the new variables
~,
x =  , C
a2 1/1
ar
a~2
7
~
7
~ 7
h2 t
(5)
 2 '
c
r, 1/1, the problem to be solved is
a1/1 r
r =
1/1 0+ , 1/1 1, 1/1
0+,
for r > 0, 0
O.
Section 14.10 7.
9.
2k(3s 2  k 2 ) ?
(s
3
+ k) = s I (J
L(F'(t»)
?
 e 2.\·), s > O.
13.
I
cs
 tanh. s2 2 k .I'  2 ·cothTr . s2 + k 2k
Answers to Oddnumbered Exercises
15.
 1
_. s I
1  exp c(l s) ,s> 1. 1  ex p (  cs)
17.
513
W
s2+ w 2·1  exp(sn/w)·
Chapter 15 Section 15 . 1 I  I . ,., 1. je SIO .) /. 3.
e 21 (3 cos 3t  2 sin 3t) .
9.
e4r(2t~t2) .
Section J 5.2 J L  (I  e a/ ). a
2+e cos at  cos bt 9. b2 _ a 2 Secti on 15 .3 I . y = e r + I. _ 1er 3 . Y  le2/ 3 3· 5. y=cosat. 7. y = ~el  e 21 + ~e3r.
e 2r .
l
3.
y = e
21
+ 2t
9. 21. 27.
x(t) = 2t 2
43.
x(t)
y

5.
te 2/ •
7.
e 2/ (2cos2t
5. 7.
3e 21
11. 13. 15. 17. 19.
 L 6t + 7  8e  1 + e 2/ •

+ ~ sin 2t) .
3  21.
t  2 + el
+ e 2/ .
= e2/ (2t 2  2t  I ). yet) = 2 cosh t + sinh t  2 cos t. x(t) = 1 + ~t  i sin 2t. x(1) = si n t  cos t + 2e y(x) = 4 e·\· + cos 3x  2 sin 3x .
x(t)
l
= e2x (cl cosx + C2 sin x) + JOx  8 + !e3x .
= ~(l  t)2e 2/ .
Section 15.4
9. 11 .
4 ;
+ e  2,
( 2
1)
[3 .
s2  ; .
2. + e' (~ _ 2. _ 2.) _ 3e 2s . s3
s
s2
s3
17.
~(t4)2exp [  2(t4) ] Ct(t4) . F(l)
23.
x(t) = 3  2cos t
25 .
x(t)
27 .
= 1, F(3) =  1, F(5) = 4. + 2[1  4  sin (t 
3(1: e T.\·) S +9
4) ]a(t  4).
= HI  aCt  2n)] (2sint  s in 2t). x( l ) = 2 + e I , x(4) = 1 + 3e 1 + 4e 4 •
Section 15.5 3 I. S2(S2
3.
1
15.
S
19 .
l  exp(2s2) s+1
+ 9)·
6 S4(S 
5.
~(J  e 2/ ).
7.
~(sin t  t cos t).
1) .
111
9.
y(t)
=
1 [.
X(I)
= e  31 [ A
k 0
H (t  f3)si nh kj3dj3.
+
(8
+ 3A)t ] +
1/
j3e  3/J F(t  j3) d{3.
.
514
Answers to Oddnumbered Exercises
Section 15 .6 1.
F(t) = 1 +2t.
3.
F(t)
5.
F(t) = t 3
13.
= t + ~t2
F(t) =
=
11 .
llWQc2x2  l..wQcx 3 +
1.
Ely(x)
3.
Ely(x) = ~woc2x2 
2
9.
+ tots. 4 + ~t2 + ftt 4 .
Section 15 .8
t *t
F(t) = H (t) = Se 2,
7.
480
40
g(x)
~[5cx4 120c
iig wacx3 + ftwo [x4 
4
.
+ 4e
1

6te  l •
= e x (l  x)2.
 x S + (x  c)s a(x  c) ]. (x  c)4a(x  c)].
Section 15.9 1. x(t) = 2  ~el  ~e31  e 21 , yet) = 7t +5  e' + ~e21. 3.
+ 2t)e' + 2e 31 , y(i) = (1  t)e'  e31 . =  t  ~s int + ~ sin 21, y(t) = \ + ~ cost = 1 + e' , yet) =  Ie', z(t) = 1  e t  tel.
x (t) = (I
5.
x(t)
7. 9.
xU)
~ cos2t.
x(t) = CI cosh t + C2 sinh t + J~ [cosh,8 F(r  ,8) + sinh ,8G(t  ,8)] d,8,
y(t) = CI sinh t + C2 co h t + J~ [cosh,8G (t  ,8) + sinh ,8 F(t  ,8)] d,8.
11.
x(t)
= cos 2t + J~ [cos 2,8F(t  ,8) + si n 2,BG(t  ,8)] d,8, + J~ [cos 2,8G(t  ,8)  si n 2,B F(t  ,8)] d,8.
yet) =  sin 21
Chapter 16 Section 16.2 J. Y=CIX,XY=C2. 3. y = CIX2 , Y = C2 x3 5. Y = CI exp (~x2), Y = In IC2x l· 7. y = In IC lxl, x = In IC2y l.
9. 11. 13. 15.
x = y In IeIY!, y(2x + y) = C2 . y(x2 + CI) 2, x 3 3 1n IC2y l .
=
17 . 25.
y2(y2 + 2x 2 ) = CI, y2 = 2x21n I C2xi. y = (3  2X)I /2; and y = 2  ln2  In (x) .
27 .
y
= (3 
Section 16.5 9.
(a) x 2(\ 2
2x) 1/2 for x ~ 1; y
+ 12x2y) = 0;
(b) x
=1= 0,
=
In x for 1 ~ x.
12x 2y
=
x =  y In IeIY!' y(2x  y) = y2  x 2 CIY, y(3x 2 + yZ)
=
I.
2
4y=0; ( b)x 4y=0.
11.
(a)x
13. 15.
(a) (y2  x)(y2 + x) = 0; (b) same as (a). (a) y8(yZ  2x)(l + 2xy2 + 4x 2) = 0; (b) Y
= 0, yZ = 2x.
Section 16.7 3. c2 + cx 2 = 2y; sing. sol. , 8y = _x4. 5. 2c 3 x 3 = I  6c 2 y; sing. sol.,2y = x 2 7. y = cx + kc 2 ;sing. sol., x 2 = 4ky. 9. x 2 (l + cy) = e2 . I \, xy = c(3cx  1); sing. sol., \ 2x2 Y =  1. 13. c(xc  Y + k) + a = 0; sing. soL, (y  k)2 = 4ax. 15 . xy = c(c 2x  1); sing. sol. , 27x 3y2 = 4.
C2.
= C2.
AllSwers to Oddnumbered Exercises
17. 19.
xc 3  ye 2 + 1 = 0; sing. sol.,4y 3 = 27x 2. 3x = 2p + Cp  I/2 and 3y = p2  ep l/2.
21.
x = 4p In Ipcl and y = p2[l + 2 1nlpcl]' 2x = 3p2 + Cp  4 and 3y = 9p  1 + 2cp3. p3(x + p)2 = c and 2y = 6xp + 5p2. p3(x + 2p)2 = c and y = 3xp + 5p2 x2 = Cp4/3  2p 1 and y = xp + x 3p2.
23. 25. 27. 29.
Section 16.9 1. x = el sin (y + C2) ' 3. x=c ly ]n lc2yl · 5. (y  C2)2 = 4a(x  CI). 7. y3=3(C2XCIY) .
515
y = X I + X + CIX 2 + C2 . CI)' + C2 =  In ICle x  I I. e>'cos(x+CI)=C2. x=c2+c ly(l+cf) ln ly+c ll. 9. x = CI + y In IC2yl . 3y=c2+ ln lx3+c ll . 21. Gen. 01.: 2y = CIX 2  2cfx + C2; fami ly of sing. sols.: 12y = x 3 + k. 23. 27cI (y + CI)2 = 8(x + C2)3. 33. y = In (4  x). 25 . y=~x2+3x3+9 In (3x). 35 . x= ln(cscycot y). 27. 24y = x 6 + 9x 2 + 2. 37 . y = 1 + In (secx + tan x). 29. Y = CI cos fJx + C2 si n fJx. 39. x = Cl Inlc2yl  cos y. 12+x 41. 2y l =(x  2?+8 In(x+2) . 31. y=I+2 In   . 2x 43 . 2y = I +x 2 +2s inx. I 1. 13 . 15. 17. 19.
Miscel laneous Exercises 1. exy + 4x + e 2 = 0; sing. sol., xy2 = 16. 3. ey3(x  c) = I ; sing. sol. , x 2i = 4. 5. x 2 (y  c 2) = e; sing. sol. ; 4x4 y = 1. 7. x = Cp  3/2  P and 2y = 6Cp I/2 _ p2. 9. x 3 (2ey  1) = e 2; sing. sol. x 3 y2 = I. I I.
5x = 4 p 3 + epl /2 and 15y = 9p4 + ep3/2 . x 2c3  2xyc 2 + lc + 1 = 0; sing. sol. 27x
= 4i. x 2 = 2(y  CI), y = In IC2XI · 8x = 3p2 + Cp  2/3 and 4y = p3  Cp l/3 . 19. x 2c 2  (2xy + I)c + l + 1 = 0; sing. sol., 4x 2  4xy  1 = O. 21. x(x2y)=cl,y=xlnlc2x l. 23. y = ex + c3  c 2 ; sing. sol., x = 2ex  3ex 2 and y = ex 2  2ex 3. 25. py = eexp (p I) and px = y(p2  P + I) ; sing. sol., y = x. 27 . x 2 = 4c(y  8c 2); sing. so l. , 8y3 = 27x 4 13. 15. 17.
Chapter 17 Section 17.3 1. x = 2i, 2i. 3. None. 5. x = o.
7. x = i, i. 9. x = 0, l. 11 . x=O.
516
Answers to Oddnumbered Exercises
x = 0, 3,  3. 15 . x = 3.
13 .
 i.
17 .
x = 0, i,  i .
19.
x =
b
(_ 3)kx2k+ 1 ] 3·5·7 ... (2k + I) ;
i.
Section 17.5 I . y = aocosx+alsin x.
b 00
3.
y = C/o
[
1+
(_3)k X2k ] 2kk!
[
+ al
00
x
+
val id for all finite x . 22kx2k+ 1
00
+ al L
5.
y = ao(l  4x 2 )
7.
y = ~ao L(  l)k(k
2 ; valid for k=O 4k  1
00
+
1)(2k
+
1) (2k
Ix l
k > O.
5.
F(t)
7.
344.
II.
I(t)
1.
sk
00
= ! L (t  2n?aCt = E
R
2n) ; F(5)
= 17 .5.
t) + R L(~ J) exp (t RC
exp (  RC
2E
"
11= 1
nc) aCt  nc).
a
526
Answers to Oddnumbered Exercises
Section 24.2
9.
1 1: '\I nt
e' erfc (.Jr) .
Chapter 25 Section 25. 1 1. y(x. / ) = _1 2 + 3(t  4x)2a (t  4x). 3. y(x, t) = x t 2 + [3(t  4x)2 + ~(t  4x) ]a(t  4x) . 5. y = 3(t  4x)aCt  4x)  2t.
it 
Section 25.2 00
1.
y = x_x 2 _ t 2+ L(1 )" [(tnx)2 a (t  n  x)+(tnl +x)2a (tn  l+ x)]. 11=0
Section 25.5 1.
u = 1
~ ;S)(
l )1I
[erfc (2n2.Jr + x) + erfc (211 +2.ji 2  X)] .
Index
Abel's formula , 163 Alpha function, 286 Analytic functions, 342 Beams bending moment, 307 boundary conditions, 307 deflection of, 307310 sheari ng force , 307 Bernoulli 's Equation, 8689 Bessel's equation of index one, 385 of index zero, 373 with index an integer, 401402 with index not an integer, 400401 cdiscriminant equation, 325326 Canonical variables, 491493 Catenary, 338 341 Cauchy type equation, 366 Change of variable equation of order one, 26, 8486 in Cauchy type equation, 366 linear equation of order two, 152156 Characteri tic equation, 198 Characteristic polynomial, 198 Chemical conversion, 6569 Clairau t's equation, 330334 Class A functions, 261263 Complementary function, 108 Compound illtere t, 69 Concrete dams, temperature of, 453454 Constalll coefficients, linear equations with, 117 Continuing method s, 5861 Convergence, improvement in rapidity of, 444445
Convolution theorem , 294297 Cooling, Newton 's law of, 64 Critical damping, 173 Damped vibrations, 172177 Damping factor, 172
Damping, critical, 173 Dampi ng, overdamped, 173 Degree of homogeneous function, 24 Dependence, linear, 102 Dependent variable defined, 2 missing in equation, 334335 Difference equation, 35 j Differen tial operator , 109111 exponential shift, j 14 laws of operation, 1 111 13 products of, 109 Differentiation of a product, 367368 Diffusivity, thermal , 4 11 Dimensionless variables, 49 1 Drugs, dissipation of, 72 Eccentricity of a conic, 182 Eccentrici ty of an orbit, 183 Eigenvalues of a matrix , 196 Eigenvectors of a matrix, 196 Electric circuits, 232235 Electric networks, 235 241 Eliminating the dependent variable, 328 330 Envelope, 325 Epidemics, 72 Error function application, 488490 complementary, 472 defined,471 Escape, velocity of, 6264 Euler type equation, 366 Euler' s method, 4548 modified, 4849 Eu ler's theorem on homogeneous functions, 25 Exact equations, 2934 Existence of solutions, 1415, 243246 Exponential function with imaginary argument, 123 125 Exponential order, funct ions of, 258261 Exponential shift, 1 14
527
528
Index Factori al function, 396397 Families of curves as solutions, 5 10 geometric interpretation, 10 13 F lux of heat, 448 Fourier se ries at discontinuities, 430 convergence theorem, 430 cosine series, 441443 defined,430 improvement in rapid ity of convergence, 444445 numerical analysis, 443444 numerical examp!cs, 431438 periodicity, 431 ine series, 438441 Gamma function, 267269 General ol ution homogeneous linear equation, 106107 non homogeneous linear equation , 107109 Gravitation, Newton's law of, 178 Halfwave rectificat ion, 270 Harmonic series, 372 Heat conduction quarter infin ite so li d, 496499 semiinfi ni te solid, 488490 slab, 41 14 17, 448, 455 slab of fin ite width, 493496 sphere, 457458 Heat. eq uaLion cartesian coordi nates, 4 1 I spherical coord inates, 457458 validity of, 453454 Hermite polynom ials defin ed,402 orthogo nality of, 424 Homogeneous coefficients, equations with , 2529,83 Homogeneous functions, 2425 Euler's theorem on, 25 Homogeneous li near equation with constant coefficient, 117 with variable coefficients, 106 107 Hooke's law. 165 Hyperbolic functions. 127 133 Hypergeometric func tion, 397 399 Impedance, steadystate, 239 Im pressed force, 166 Inclined plane, motion on a, 68 Independent variable, 2 Independent variab le missing, 335338 Indi cial equaLion defined,362 difference of roots integral logarithm ic case, 38 1385 non logarith mic case, 377 381
difference of roots noni ntegral, 363367 eq ual roots, 368374 alternate method, 374377 Infinity poin t at. 386 solution s near, 386 Initial value problem , 245 Insulation, 448 In tegral equations, speci al, 298303 Integral transforms. 252 Integrating factor for equations with homogeneous coefficients, 83 found by inspection, 75 79 linear first order equation, 35 liTegu lar singular point. 358 Isocl ine. 13 J 5 Kepler's first law, 180182 Keple r's second law, 179, 180 Kepler's third law, 182 184 Kernel of an integral tran. form, 252 Kirchhoff's Jaws, 233 Laguerre poly nom ials defi ned, 399400 orthogonali ty of, 424 Laplace operator, 253 Lap lace transform a convolution theorem, 294297 defi ned,253 deflection of beams, 3073 10 derivatives of, 266267 in verse of, 274 277 linearity of, 253, 275 obtained by power series, 46747 1 of a step fu nction, 286293 of derivatives, 263266 of ele men tary functi ons, 253257 of functions of Class A, 26 1 263 of func ti ons of exponentia l order, 258 26 1 of period ic function s, 269273 of sectiona ll y con tinuous function s, 257 258 table of, 3 18 vibration of springs, 303 307 Laplace' equation in three dimensions cartesian coordinates, 405 in two di mensions, 46 1463 Legendre polynomials defined,403 orthogonality of, 422423 Linear coefficients. equat ions with, 89 94 Linear combination of function, 99 100 Li near dependence, 102 Li near equation change of variable in, 8994 defined,4 homogeneou with constant coefficients, 11 7
In dex homogeneous with variable coefficients, 106107 irregul ar singular point of, 358 nonhomogeneous with constant coefficients, 134 nonhomogeneous with variable coefficients, 107 109 order one, 3543 ordinary point of, 345 solutions near, 347 Regular singular point of, 346 regular ingul ar poi nt of, 358 systems of, J 86 l87 undetermi ned coefficients, 139144 variation of parameters, l56 l6 I Linear independence of a set of fu nctions, 102 of a set of vector functi ons, 200 of a set of vectors, 200 Lineari ty of differential operators, 112 l J 3 of inverse Laplace transform, 275 of Laplace transform , 253 Lipschi tz condition, 246 Logarthmic soluti ons, 381 385 Logistic growth, 6973 Manyterm recurrence relation, 388392 Matrix algebra characte risti c equation, 198 characteristic polynomial, 198 eigenvalues, 198 eigenvectors, 198 reviewed, 189197 Mi xture problems, 65 69 Networks, electric, 235 241 Newton's constant of gravitation, l78 Newton's law of cooJi ng, 6465 Newton's law of gravitation, l78 Newton's second law of motion, 178 Nonelementary in tegrals, 9498 Normal linear equation, 99 Nu ll fu nction, 274 Nu merical methods continuing methods, 58 6 l Euler's method, 4548 Euler's method, modi fied, 4849 RungeKulla, 5458 successive approximati on, 49 5l Tay lor's theorem, 5254 Operator Laplace, 253 Order of a di fferential equati on, 2 Ordinary differential eq uation, 3 Ordi nary point defi ned,345 solutions near, 347
529
vali di ty of, 347 Orthogonali ty of Hermite polynomials, 424 of Laguerre polynomials, 424 defi ned,4 18 of Legendre polynom ials, 422423 of polynomials, 4 19420 zeros of, 42 1 Overda mped motion, 173 pdiscrimi nant equation, 326328 Parameters, varimion of, 156 161 Partia l differential eq uations change of variables , 405 defined,404 of applied mathematics, 404406 Pendul um, si mple, 177 178 Periodic extension, 43 1 Phase plane complex eigenva lues, 222 226 defi ned, 220 phase portrait, 220 real distinct eigenvalues, 22 J stable node, 22 1 trajectory, 220 unstable node, 221 unstab le sadd le, 22 J Phase portrait, 220 Pl anetary motion, J78 179 Polynomials Herm ite, 402 Laguerre, 399400 Legendre. 403 orthogonal ity of, 419420 simple sets of, 419 Power erie convergence of, 343345 table of. 468 Recu rrence relation defined,35 1 manyterm, 388392 Reducti on of order linear equati on, 152 156 non linear equation, 334335 Regu lar in gul ar poin t, 358 manyterm recurrence relation, 388392 solutions near, 362 Regular ingu lar point of linear equation, 346 Resonance, 169172 Retarding force, 165 RLC circu it, 233 RungeKUlta , 5458 Sectionall y contin uous functions, 257258 Separation of variables ordi nary differential equati ons, 1823 partial di fe rential equation ,40641 1 Series, computati on wi th , 443444
530
Index Shearing force, beams, 307 Simple sets of polynomials, 4 19 Singul ar point defined , 346 irregul ar, 358 regul ar, 358 olutions near, 362 Singular so lutions, 323325 Solution curve, I I Soluti on 01' a differential equati on, 3 Spring, vibration of, 165 constant, 165 critical dampi ng, 173 dam ped, 172177 forced, 166 overdamped , 173 resonance in , 169 172 transform methods, 303307 undamped, 167 169 Squarewave fun ction, 271 Steadystate temperature, 461 String, elastic, 45846 1, 485488 Successive approximation, 49 51 Superposition of solutions, 147 Systems of equations Laplace transform method, 31 03 16 matrix methods complex eigenval ues, 206212
real distinct eigenval ues, 197206 repeated eigenva lues, 212220 Taylor's theore m, 52 54 Tractri x, 68 Trajectory, 220 Triangu larwave function, 272 Und amped vibrati ons, 167 169 Undetermi ned coeffici e nts, 139144 Variable dependent, 2 independent, 2 Variation of parameters, 156 1. 61 Velocity of escape, 62 Voltage law, Kirchhoff's, 233 Wave equation one dimen ion, 458461 , 4 85488 Wei ght functi on, 4 18 Wronskian Abel's Formula for, 163 of olutions of a system, 200 of sol ution s of an equation , 103106 Zeros of orthogonal polynomi als, 421
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