Elementary Differential Equations [8 ed.] 0-13-508011-8

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Eighth Edition

EE ea Cne nest a Differential

- Equations wn Fg

a 4

F

™ _

+ by (x, a

d*w ay

dw ae

+ b3(x, yo = oe bale

ge

+ Bote, yyw = R(x, y)

is the general second-order linear partial differential equation with two indepen-

dent variables. Exercises In Exercises | through 16, state whether the equation is ordinary or partial, linear or nonlinear, and give its order.

dx

2

ag tHe =0.

a? w

oe

2 aw

age

1.3

Families of Solutions

di

3.

(x? + y*) dx + 2xydy =0.

10.

LT + Ri=E.

4.

ae rey = Q(x).

Hl.

(+ y)dx + Bx? — dy =0.

12.

ay"?

=

&. 7 8.

9.

/

=

2

F

oe

ee: OU tu —+—54+— =0. Ox? ay? Az? d4 =

-

= w(x).

Pp

xP[“y yoo x =a.

+ (y’y* —y=

0.

14,

Bw \ dw\* +yw=0. -—2(—] ax* dx ay -=1-xy+y’.

15.

y”+2y’ —8y = x* + cosx.

16.

= ada+bdb=0.

13.

au

3

(—~])

dx

17.

Verify that sin kt is a solution of the equation in Exercise |.

18. 19. 20.

Verify that e~* is a solution of the equation in Exercise 5. Verify that 3e~?* + 4e* is a solution of the equation in Exercise 5. Bessel’s function of index zero is defined by the power series oO

ny2n

Jo(x) = u alco

Verify that Jo(x) is a solution of the differential equation

xy" +y+xy 21.

Verify that for x > 0, (2//3)x3/*

=0.

is a solution of the equation of Exer-

cise 6.

Families of Solutions Every student of calculus has spent a significant amount of time in finding the solutions of first-order differential equations of the form dy

dx

_

tM)

(1)

This antiderivative problem is often written y=

f fade

+e

(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 function 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 one-parameter

families of solutions, the parameter being the so-called arbitrary constant c of equation (2). If one considers the general first-order differential equation

dy

;. = f(x, y),

(3)

the problem of finding solutions, 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

one-parameter families of solutions. In Chapter 2 we shall study a number of methods for finding families of solutions for some particular types of first-order equations, but in general there is no method of attack that will solve every such equation. We content ourselves for the moment by illustrating what happens in a few simple examples. EXAMPLE

1.1

The differential equation

“*d = 8sin4x

(4)

y = —2cos4x +c,

(5)

dx

has the family of solutions

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 = Q, we are forced to choose c = 8. We then say that y=

—2cos4x +8

is the solution of the initial value problem

d oY =8sindx,

y=6, whenx =O.

dx

EXAMPLE 1.2 From calculus we

learn that the derivative of the function

f(x)

=

ce?*

is

f(x) = 2ce**. Phrased in the language of differential equations, we say that the differential equation dy

In

_

2y

(6)

1.3

Families of Solutions

7

has the family of solutions y=ce™,

(7)

If we seek a solution of equation (6) that satisfies

dy

oY =2y,

y=4, whenx =0,

dx

(8)

then from equation (7) we see that c = 4 and the solution of (8) is

y =4e™.

EXAMPLE

1.3

Consider the second-order equation

y= Be.

(9)

Integration of both sides of this equation with respect to x yields

y =4x7 +e).

(10)

y=x' texte.

(11)

A second integration produces

In this example there are two arbitrary constants, so we have a two-parameter family of solutions. This means that to single out one member of this family we

need to provide two pieces of information. These are usually given by specifying the values of both y and y’ for the same value of x. For example, suppose we want the solution to (9) that also satisfies y(O0) = |, and y’(0) = 2. Substituting x = 0, and y’ = 2 into (10) we see that c; = 2, so that y=

at ot Bias epg,

Finally, substitutingx = 0, y = 1, we see that c, = | so that the required solution is yox+2x+1.

EXAMPLE 1.4 Consider the one-parameter family of curves x? -—3x’7y =e.

(12)

8

Chapter I

Definitions; Families of Curves

A differentiation of both sides of this equation with respect to x yields 3x7 — ax?

d

—oxy

=0

xX or

dy dx

x-—2y

=

‘

x

when x 40.

(13)

If we had started this example with equation (13) and tried to find the family of curves given by equation (12), we would face a more challenging problem than in the previous examples.

We will learn how to solve equation (13) in Chapter

2. Here we simply point out that the value x = 0 creates a difficulty for both the differential equation (13) and for its family of solutions _ xi-c¢ =

3x2

obtained from equation (12).

EXAMPLE

1.5

Consider the family of circles

(x — 2)? + (y+ 1)? =e’.

(14)

A simple differentiation with respect to x yields

dy

2(x —2)+2(y+ )— dx

=0

or dy

—(x — 2) = —__, dx y+1

h when y #

—l,

1 ay

We are forced here to think of the family of circles as consisting of two families of semicircles: the one family

y=-l1+ve?

—@ —2)?

(16)

y=—l—vc*?

— (x — 2).

(17)

and the other

In equation (16) we have a family of solutions of the differential equation for y > —1, whereas in equation (17) we have a family of solutions of (15) for y < —1. To solve the initial value problem

dy =_ —(x—2)

dx

y+

= 2,

wh

Wee

=-l,

18

(18)

1.3

Families of Solutions

we must choose the parameter c from equation (16), since 2 > —1. 2=-14+vVc?

—9orc

9

We have

= V18. The function

y=—1+18 — (x — 2) is therefore the solution we seek. Its graph is that of a semicircle of radius V 18.

Wi Exercises Solve the differential equations in Exercises | through 6.

dy

5

=x dx

dy _3 dx

d £Y

dx

—

4+ 2x. rex

dy

—= dx

5

&w__?

x

dx

cos 6x.

4

4.

6.

x*—-I1

;

244

@)

:

dx

x*+x

.

Solve the initial value problems in Exercises 7 through 12.

d

SF dx

ee

y = 6, when x = 0.

d i

dx

y = 2, when x = 0.

d oY = Ay, y = 3, when x = 0. dx d AY = 55, y =7, when x = 0. dx

dy—=4sin2?x, ; dx

d

y=2,

OY a x2 43-40%, dx

whenx = 7/2.

y =—l,

whenx =0.

Show that the family of circles (x + 1)? + (y — 3)? = c? can be interpreted as two families of solutions of the differential equations dy

dx 14.

_ “e+

I)

y-3

Show that the family of parabolas y = ax? can be interpreted as two families of solutions of the differential equation

dy _ 2y dx Xx

,

10>)

Chapter1

Definitions; Families ef Curves

then find the solution of the initial value problem d

2:

dx

x

= aa

y = 2, whenx = —1.

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 = 0.

Geometric Interpretation In Section 1.3 we saw that a first-order equation usually has a family of solutions. A useful technique in understanding 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 — Ix = §sin4x, sin 4x

(1)|

Recall from Section 1.3 that the family of solutions is y = —2cos4x +c.

(2)

Graphing the solutions corresponding to c = 2, 1, 0, —1 we obtain Figure 1.1. It is not difficult to imagine what the rest of the family looks like.

fa

Figure 1.1

1.4

Geometric Interpretation

11

The one-parameter 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 will formalize this fact in Section 1.6, but for now, we claim that it is true for the solutions of any suitably restricted first-order differential equation. If we specify a point in the plane, by the property above there will be exactly one solution passing through that point. The unique curve that results is the solution curve of the initial value problem. This is the geometric version of the process described in Section 1.3. If we extend these ideas to higher order equations, we find that only part of the geometric interpretation carries over. EXAMPLE 1.7 Graph several members of the family of solutions of the equation dy

at = 12x". As we saw in Section

(3)

1.3, the family of solutions is

yo=xt tex

ten.

(4)

Graphing the solutions corresponding to the pairs of constants c; = 2, cz = 1; cy = 0, ce: = 0; and c; = 0, c2 = 1 we obtain Figure 1.2.

Clearly these solutions do not satisfy the uniqueness property of the first-order case. There are at least two solutions passing through the points (0, 1) and a point near (—1/2, 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 second-order equation, we could specify both a point through which the solution y

Figure 1.2

12.

Chapter!

‘Definitions; Families of Curves

passes, and the slope at that point. Given this information there is, in this case, a unique solution. The second-order 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 will be discussed in greater detail in Chapter 6. @ Exercises

1. 2.

For Exercises 1 through 6 of Section 1.3, draw a representative sample of solution curves. For Exercises 7 through 10 of Section 1.3, draw the graph of the solution of the initial value problem.

The Isoclines of an Equation In Section 1.4 we noted some of the geometric properties of families of solutions which we had found by the analytic methods of Section 1.3. In this section we see that we can use geometric methods to actually find solution curves. Consider the equation of order one

aD a fy). ax

(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 (1) 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 points to indicate the direction associated with each of those points. This can be done rather systematically by first drawing curves called isoclines, that is, curves along which the direction indicated by equation (1) is fixed. EXAMPLE 1.8 Consider the equation dy

dx

_

"

(2)

The isoclines are straight lines 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 = 1, equation (2) determines a direction of slope 1. In Figure 1.3 we have drawn several of these

1.5

The Isoclines of an Equation

TT J

m

TTT a Voi

13

c=3

r

Figure 1.3

isoclines, indicating the direction associated with each isocline by short markers. If one starts at any point in the plane and moves along a curve whose direction is always in the direction of the direction marks, then a solution curve is obtained,

Several solution curves have been drawn in Figure 1.3.

gi EXAMPLE 1.9 Use the method of isoclines to sketch some of the solution curves for the equation dy 2 2 ey? Ae x+y

3 (3)

Here the isoclines will be the circles x* + y* =c, withe > 0. Whenc = 1, the isocline has radius 1; for c = 4, radius 2. In Figure 1.4 we have drawn these isoclines, marking each of them with the appropriate direction indicator,

and finally, sketching several curves that represent solutions of equation (3).

WB Exercises For each of the following differential equations, draw several isoclines with appropriate direc-

tion markers, and sketch several solution curves for the equation. d

a dx

j,

dy _y Se ak

6

d

2

4

228,

4,

dy “age

dx

dx

x

a

14

Chapter !

Definitions; Families of Curves

Figure 1.4

5

i 7.

dy

= dx dy dx dy dx

eryt

ee a

8.

i.

dy dx d

—=y—x?. us

g, Bak y dx

1

10.

—=2x-y. ee

dy

== dx

—-x

—. y

An Existence Theorem It should be clear even to the casual reader that drawing isoclines is not a practical tool for finding solutions to any differential equations other than those involving the simplest of functions. Before discussing some of the analytic techniques for finding solutions, we shall state an important theorem concerning the existence and uniqueness of solutions, a theorem discussed in detail in Chapter 13.

Consider the equation of order one dy

_

(1)

gl Let T denote the rectangular region defined by

|x —xo| Sa

and

a region with the point (xo, yo) at its center.

continuous functions of x and y in T.

ly—yol

(diff (y(x),x)=x*2+y%2, DEplotl {[0,2],[0,0],[0,1]}, y(x), w=-2..2,

produces Figure 1.6.

y=-2..2

);

)}

1.7 Lon Lf

ek £ tf]

Lk

dt

Lao

dg

FEY

ft fhe

Lb

Pe

Ea

bh

fof

f

Lod bog

fof fuk

Ek

t

ik

ie? ta L f

Tob ¢ € tT

t

8

£6

OE f £

ff

Foe

of

FoR

tw

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#&

#5

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fa

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PFS

©

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4,

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LF

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“of

At

Ff

Be OL Awe Lf

¢

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oe

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Look

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dt

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£ Of

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te t #

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17

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gh

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tal

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noe

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€ & F tf fd

Computer Supplement

OL

fp

EF tb

to OF

fb

Ff

ob FT

Rodeo

bt t

et Pad

‘igure 1.6

WB Exercises 1.

For Exercises | through 10 of Section 1.5 use the computer software package of your choice to produce slope fields and representative solution curves.

2.

Consider the problem ysin(y +x).

of drawing

isoclines for the equation

This is a very hard problem.

dy/dx

=

Now use the computer to draw

the slope field and representative solution curves. 3.

What can you say about the curves drawn

in Exercise 2 as x

x —» —oo? Do your answers depend on the initial conditions?

—

ox, as

is Equations of Order One

Separation of Variables In this chapter we study several elementary methods for solving first-order differential equations. We begin by studying an equation of the form Mdx+Ndy=0O,

where M and N may be functions of both x and y. Some equations of this type are so simple that they can be put in the form A(x) dx + B(y)dy =0; that is, the variables can be separated,

()

Then a solution can be written at once.

For it is only a matter of finding a function 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

forx > Oand y > 0.

x

(2)

We note that for the function in equation (2), the theorem of Section 1.6 applies and assures the existence of a unique continuous solution through any point in the first quadrant. By separating the variables we can write

dy _ 2dx y

Xx

Hence we obtain a family of solutions

In|y| = 2In|x| +c

(3)

or, because we are in the first quadrant,

y = ex’, 18

(4)

2.1

Separation of Variables

19

If we now put c; = e*, we can write y=

ex’,

cy, > 0.

(5)

EXAMPLE 2.2 Solve equation (2) of Example 2.1 for x 4 0. The argument now must be taken in two parts. First, if y 4 0, we can proceed as before to equation (3). However, equation (5) must be written

ly] =c1x?,

¢, > 0.

(6)

Second, if y = 0, we see immediately that since x 4 0, y = O is a solution of the differential equation (2).

As a matter of convenience the solutions given by equation (6) are usually written

y = c2Xx’,

(7)

where ¢3 is taken to be an arbitrary real number. Indeed, this form for the solutions incorporates the special case y = 0. Several representative solution curves are

shown in Figure 2.3. We must be cautious, however. The function defined by

gtx) =e",

x>0

= —4x?,

x 0, y*e~” is an integrating factor for equation (11) and for y < 0, —y%e~” serves as an integrating factor. In either case we are led to the exact equation

yee dx + y?(3 — y)e~’'x dy = —2y’e7” dy, from which we get

xye?

=

2 | ye

dy

= 2y*e +4ye + 4e" +c. Thus a family of solutions is defined implicitly by xy? = 2y?+4y +44 ce’.

The General Solution of a Linear Equation In Section 1.6 we stated an existence and uniqueness theorem for first-order differential equations. If the differential equation in that theorem happens to be a linear equation, we can prove a somewhat stronger statement.

Consider the linear differential equation dy

ig tt ey = OG), Xx

(1)

2.6

The General Solution ofa Linear Equation

39

Suppose that P and Q are continuous functions on the interval a < x < b, and that x = xo is any number in that interval. If yo is an arbitrary real number, there exists a unique solution y = y(x) of differential equation (1) that also satisfies the initial condition

y(%o) = Yo. Moreover, asimplify("); e°x*

2

As a second example consider equation (7) in Example 2.7 of Section 2.4,

3x(xy — 2) dx + (x° + 2y) dy = 0. Here the first step is to check whether the equation is exact. Maple can do this as follows: >M}=3*X*

(K*¥-2)

;

M

:=3x (xy — 2)

>N:=(x%3+2*y);

N:=xi+2y >diff(M,y);

55° >diff(N,x);

32°

44°

Chapter2

Equations of Order One

We could then use the computer to assist in the

Thus the equation is exact.

remaining steps in the process. Fortunately, most symbolic 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;

“ys

= -»

This can be solved in one command: >dsolve(",y(x));

y(x) = x?_Cl The second example is just as easy: > (3*x* (x*y-2)) + (x*34+2*y) *di ft (y (x) ,x) =0; 3x (xy —2)+

d (x? + 2y) —y(x) dx

=0

>dsolve(",y(x));

y(x)x? — 3x? + (ya)? = -C1 Finally, the computer can also handle initial value problems.

For example,

consider equation (8) in Example 2.3 of Section 2.1:

(l+y?)dx + (1+x7) dy =0, with the “initial condition” that when x = 0, y = —1. The equation is entered as >difft(y(x),x)=-(l+y(x)%*2) / (L+x%2);

d dx?

1+ (y@)) ~

Pe x?

and then solved by entering >dsolve({",y(0)=-l},y(x)); y(x) = tan (-

arctan(x) — =).

Exercises 1.

Use a Computer Algebra System to solve a representative sample of prob-

lems 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,

ACAS is capable of solving even equations as general as dy/dx + P(x)y = Q(x). Try it on your system.

BD Numerical

os.

Methods

aS

General Remarks There is no general method for obtaining an explicit formula for the solution of a differential equation. Specific equations do occur for which no known attack yields a solution or for which the explicit 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 clear grasp of good

numerical methods requires time-consuming practice and the availability of an adequate computer. This chapter is restricted to a fragmentary

discussion of some

simple and

moderately useful methods. The purpose here is to give the student a concept of the fundamental principles of numerical approximations to solutions. We shall take one problem, which does not yield to the methods developed earlier, and apply to it several numerical processes.

Euler's Method We seek to obtain that solution of the differential equation

yaya?

(1)

for which y = | when x = 0. We wish to approximate the solution y = y(x) in the interval 0 < x < 5. Equation (1) may be written in differential form as dy = (y* —x7)dx.

(2)

Figure 3.1 shows the geometrical significance of the differential dy and of Ay, the actual change in y, as induced by an increment dx (or Ax) applied to x. In

Suppose that we choose Ax = 0.1; then dy can be computed from

Nile

calculus it is shown that near a point where the derivative exists, dy can be made to approximate Ay as closely as desired by taking Ax sufficiently small. We know the value of y at x = 0; we wish to compute y forQ < x < dy = (y* — x7) Ax.

45

Chapter3

Numerical Methods

Indeed, dy = (1 — 0)(0.1) = 0.1. Thus for x = 0 + 0.1, the approximate value

ofy is 1 + 0.1. Now we havex = 0.1, y = 1.1. Let us choose Ax = 0.1 again. Then

dy = [(1.1)? — (0.1)7] Ax, sody = 0.12. Hence atx = 0.2, the approximate value of y is 1.22. The complete computation using Ax = 0.1 is shown in Table 3.1. The computations are carried out to six decimal places and then rounded off to three decimal places. The increment Ax need not be constant throughout the interval. Where the slope is larger, it pays to take a smaller increment. For simplicity in computations, equal increments are used here. It is helpful to repeat the computation with a smaller increment and to note the changes that result in the approximate values of y. Table 3.2 shows a computation with Ax = 0.05 throughout. In Table 3.3 the value of y obtained from the computations in Tables 3.1 and 3.2 and also the values of y obtained by using Ax = 0.01 (computation not shown) are exhibited beside the values of y correct to three decimal places. J i

46

(x, y)

ays |

|

Figure 3.1

TABLE 3.1

Ax

= 0.1

x

y

y?

x7

(y?—x7)

0.0 0.1 0.2 0.3 0.4

1.000 1.100 1.220 1.365 1.542

1.000 1.210 1.488 1.863 2.378

0.000 0.010 0.040 0.090 0.160

1.000 1.200 1.448 1.773 2.218

0.5

1.764

= dy 0.100 0.120 0.145 O77 0.222

3.2 TABLE

3.2

Ax

=

Euler's Method

0.05

2

y

¥

x?

(v2 = x7)

dy

0.00 0.05 0.10

1.000 1.050 1.105

1.000 1.102 1.221

0.000 0.002 0.010

1.000 1.100 [211

0.050 0.055 0.061

0.15 0:20 0.25

1.166 1.232. 1.306

1.359 1519 1.706

0.022 06,040 0.062

1.336 1.479 1.644

0.067 0.074 0.082

0.30 0.35 0.40 0.45

1.388 1480 1.584 1.701

1.928 2.192 2.508 2.894

0.090 0.122 0.160 0.202

1.838 2.069 2.348 2,692

0.092 0.103 0.117 0.135

0.50

1.836

TABLE When

Ax

=O0.1

Ax

47

3.3

=0.05

Ax

=0.01

Correct

x

¥

y

y

¥

0.0 0.1 0.2 0.3 0.4 0.5

1.000 1.100 1.220 1.365 1.542 1.764

1.000 1.105 1.232 1.388 1.584 1.836

1.000 1.110 1.244 1.411 1.625 1.911]

1.000 L111 1.247 1.417 1.637 1.934

The correct values are obtained by the method of Section 3.7. Their availability is in a sense accidental. Frequently, we know of no way to obtain the y value correct to a specified degree of accuracy. In such instances it is customary to resort to decreasing the size of the increment until the y values show changes no

larger than the errors we are willing to permit. Then it is hoped that the steadying down of the y values is due to our being close to the correct solution rather than (as is quite possible) to the slowness of convergence of the process used. For the more general initial value problem

d

=

=

FI,

y);

whenx

= Xo,

YS

Ms

(3)

the sequence of approximations described above can be expressed in terms of the recurrence relations

Chapter3

Numerical Methods

Xe =X th Vert = Ye thf Oe, Ye), fork

=0,

1, 2,

(4)

.... Here we have used / for the value Ax.

to be known

The technique described above has come

as Euler’s method,

although it involves nothing more than the linear approximations of elementary calculus.

Exercises In each of the following exercises, use Euler’s method with the prescribed Ax to approximate the solution of the initial value problem in the given interval. In Exercises 1 through 6, solve the problem by elementary methods and compare the approximate values of y with the correct values.

whenx=0,y=1;

DEplot {[0,

(diff (x(t)

«5 ] »

LO. LJ

,t)=%*

[0,2]

,

(3-2*x)

,x(t),t=-0.5..2,

[0,2.5)},6=-0.5.

~3)7

We see that the population will stabilize at 3/2 or b/a.

If a small disturbance

either increases or decreases the population, it will return to this stable value. This is an example of a sustainable population.

WH Exercises 1.

2.

Use a computer to produce Figure 4.3. Now assume that there is a constant level of harvesting of the population, so that the equation becomes dx 7) = x(b— ax) —h.

Chapter4

Elementary Applications

m

74

Figure 4.3

For small levels of harvesting, that is, when / is near zero, we would expect

that the population would not be greatly affected. Modify your solution to the above by inserting a small value for / and plot the results. 3.

If the population is a food source for humans, we would want to maximize the harvesting without endangering the health of the population. Gradually increase /: and see what happens. What is the maximum harvesting level that can safely be maintained?

4.

Next, suppose that the harvesting is seasonal. For example, replace the h by A(sin (t) + 1). See what happens to the population for different values of h.

5.

Can you finda value for/ and initial conditions so that the population survives for one “season”, but dies out in the second?

BW Additional Topics on Equations of Order One

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 discussion of tests for the determination of integrating factors. At present we are concerned with equations that are simple enough to enable us 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(xy)

=xdy+ydx,

a(=) _ oe x

d(arctan *) =

ene

(1)

(2)

J

Sr

(4)

Note the homogeneity of the coefficients of dx and dy in each of these differentials.

A differential involving only one variable, such as x~? dx, is an exact

differential. EXAMPLE 5.1 Solve the equation

ydx +(x+x°y")dy =0.

(5)

Let us group the terms of like degree, writing the equation in the form

(ydx +xdy)+x3y? dy =0.

75

76°

Chapter 5

Additional Topics on Equations of Order One

Now the combination (y dx + x dy) attracts attention, so we rewrite the equation, obtaining

d(xy) + x*y?dy =0.

(6)

Since the differential of xy is present in equation (6), any factor that is a function of the product xy will not disturb the integrability of that term. But the other term contains the differential dy, and hence should contain a function of y

alone. Therefore, let us divide by (xy)? and write

4 9dy

d(x

(xy)

oy

The equation above is integrable as it stands. A family of solutions is defined by “Tee

+ In|y| = —In |e,

or 2x7 y? In|jey| = 1.

i EXAMPLE 5.2 Solve the equation

yt — y)dx — x0? + y)dy =0.

(7)

Let us regroup the terms of (7) to obtain x*(y dx — x dy) — y(ydx +xdy)

=0.

(8)

Recalling that (dsolve({diff

(y(x)

,x)-2*x*y=1,y(0)=1},y(x));

way

e* ./merf (x ——

+e.

2

It is easy to see that this is the same as equation (5) of Section 5.6. Mf Exercises 1.

Solve a selection of exercises from the chapter using a Computer Algebra System.

WW Linear Differential Equations

The General Linear Equation The general linear differential equation of order n is an equation that can be written

box)

d"

n—1

Fb

4 + bie) ax"

d

+ha@y =R@).

ax" dx If the value of the function R(x) is zero for all x, then the equation is called a homogeneous! linear differential equation. If the coefficient functions bo... +5 On 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

d + y =sinx (x -1)2 dx

is a first-order linear, nonhomogeneous, and normal equation on any interval that does not contain x = 1. On the other hand,

god2

xyv=

0

dx? y is a second-order linear, homogeneous, and normal equation on any interval.

We now prove that if y; and y2 are solutions of the homogeneous equation

bo(x)y™ + bi(x)yO"? +--+ + bn) y’ + Pn@)y = 9,

(2)

and if c; and c2 are constants, then

y=ciyi + €2y2 is a solution of equation (2). The statement that y, and y2 are solutions of (2) means that

ye + diay? +--+ + Bp); bo(x)

+ ny

= 0

(3)

! tis perhaps unfortunate that the word homogeneous as it is used here has a very different meaning from that in Sections 2.2 and 2.3.

99

100°

Chapter6

Linear Differential Equations and

box) ys? + by (xy? +--+ Dye)yh, + Bax)2 = 0.

(4)

Now let us multiply each member of (3) by c;, each member of (4) by co, and add the results. We get

bo(x)(ciy\” + cays”) + dilx)(cry™? + cay) +... + by (x)(ciy, + 0295) + One V(ery: +. e2y2) = 0.

(5)

Since cyy; + czy, = (ciy1 + €2y2)’, and so on, equation (5) is neither more nor less than the statement that ¢) y; + ¢2y2 is a solution of equation (2). The proof is

completed. The special case c. = 0 is worth noting; that is, for a homogeneous linear equation, any constant times a solution is also a solution. In a similar manner, or by iteration of the result above, it can be seen that if y;, with? = 1, 2,...,k, are solutions of equation (2), and ifc;, withi = 1,2,...,k ? are constants, then

Y=HeCyyy +eaye tess + KE

(6)

is a solution of equation (2). The expression in equation (6) is called a linear combination of the functions Yi, Y2,---, Ye. The theorem just proved can thus be stated as follows: Theorem

6.1

Any linear combination of solutions of a homogeneous linear differential equation is also a solution.

An Existence and Uniqueness Theorem In Section 2.6 we stated an existence theorem for an initial value problem involving a first-order linear differential equation. The generalization of this theorem to nth-order linear equations can be stated as follows:

Theorem 6.2

Given an nth-order linear differential equation d" y

bo(x) Fon + iG)

qd’! y

$e tb C2dx +o@y= RO)

that is normal on an interval I. Suppose that xo is any number on the interval f and yo, Y1,-.+, Yn—1 are n arbitrary real numbers.

Then a unique function

y = y(x) exists such that y is a solution of the differential equation on the interval and y satisfies the initial conditions y(xo) = yo,

y'(xo) = yi,

a

ye

a) = rads

An Existence and Uniqueness Theorem

6.2

101

The proof of this theorem for n = 1 was given in Section 2.6 and was a result of showing that every normal first-order linear equation can be made exact by introducing an integrating factor. Unfortunately, no such method of proof is available for n > 1. We do not prove Theorem 6.2 in this book, but in Chapter 13 prove an existence and uniqueness theorem for first-order equations in general.

EXAMPLE 6.1 Find the unique solution of the initial value problem

(2)

y'(0) = 1.

y(O) = 0,

y’+y=0,

We observe that sinx and cos x are solutions of the differential equation in (2),

so that for arbitrary c; and c2, y = cy; sinx +c. cosx

is also a solution by the theorem of Section 6.1. Because of the initial conditions c, sin0 + c2 cos0 = 0 and c; cos0 — way, namely by choosing c; = | and a solution of the initial value problem

in (2), we must choose c, and cz so that cz sin = 1. This can be done in only one cz = 0. We find that the function sinx is (2). Moreover, since the problem satisfies

the conditions required in Theorem 6.2, for any interval that contains x = 0, sin x is the only solution of the problem given in (2).

EXAMPLE 6,2 Consider the initial value problem

yd) =4,

= 90, + 2xy' — 12y x*y"

y(t) =5.

The differential equation is normal on either x > 0 or x < 0.

(3)

Since the initial

conditions are stated for xy = 1, we let J be the interval x > 0. It is a simple

matter to show that x? and x~* are solutions of the differential equation in (3), so

that for arbitrary c; and co, y=

ex? + e9x4

is also a solution. The two initial conditions now require that cpta=4

and

3c); —4c. =5.

It follows that c; = 3 and c) = | and therefore the function y=

3x3

satisfies the initial value problem for x > 0.

x74

102

© Chapter6

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, determine all intervals on which the equation is normal.

lL

@-—Dy’+xy'’+y =sinx.

3.

x?y" + ey =Inx.

2.

(x7

4.

(cotx)y”+y=0.

-1)y” + 6y =e".

In Exercises 5 through 8, determine the unique solution of the initial value problem following

the examples of this section.

5.

6. 7.

y”’—y=0, y(0) = 4, y’(0) = 2. Use the fact that e* and e~ are solutions of the differential equation. y”’+4y =0, y(0) = 2, y'(0) = 4. Use the fact that sin2x and cos 2x are solutions of the differential equation. y" —2y'’+y = 0, y(0) = 7, y'(0) = 4. Use the fact that e* and xe* are solutions of the differential equation.

8.

x?y” + xy’ —9y =0, y(1) = —1, y’(1) = 15. Use the fact that x3 and x3 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 J and yo = yy = --- = yn_y = 0, then y = O is the only solution.

Linear Independence Given the functions f, f2,..., fn, if constants c), c2,..., C,, not all zero, exist such that

er fila) + en fa(x) ++ +++ cy fax) = 0

(1)

for all x in some interval a < x < b, then the functions f,, fo, ..., f, are said to be /inearly dependent on that interval. If no such relation exists, the functions are said to be linearly independent. That is, the functions f\, fo,..., fy are linearly independent on an interval when equation (1) implies that

sad ko Me

Uinearly

eperdent

Ccjp=2=-:-=c,

= 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.

64

The Wronskian

103

The Wronskian now obtain a sufficient an interval a < x < D. is differentiable at least equation

With the definitions of Section 6.3 in mind, we shall condition that n functions be linearly independent on Let us assume that each of the functions f), f2,..., /, (n — 1) times in the interval a < x < b. Then from the efi

()

=9,

tofet--+erfn

it follows by successive differentiation that

aff taht: af,

aff?

tafe

+0

ee

t+

tof, =9,

RSE

+

tn fi,

= 0,

Be

wasn afer

.

—0.

For any fixed value of x in the interval a < x < b, the nature of the solutions of these n linear equations inc), C2, ..., Cn Will be determined by the value of the determinant

fi) f{@)

Jo@) Ro)

«+ se

fr) f(x)

Pe)

OR)

as

FPO)

W(x) =

.

(2)

Indeed, if W(xo) #4 0 for some xo on the interval a < x < }, it follows that Cc) = C2 = ++: = ¢, = O, and hence the functions f\,..., fy are linearly independent ona

y, yields the result d"y

By

Og

diy

At ed

d

Ot

a

y

ey

(2)

The coefficients ao, a), ...,@, 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 which 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 differential operators always exists and is a differential operator. For operators with constant 3 The function y is assumed to possess as many derivatives as may be required in whatever operations take place.

110°

Chapter 6

Linear Differential Equations

coefficients, but not usually for those with variable coefficients, it is true that AB=BA. EXAMPLE 6.7 Let A= D+2and B = 3D — I.

Then d

fy

OD Dy = Fy dx

and A(By)

= (D +2)

2

dx?

ae

=idy dx?

G2

dx

_ r)

dx ©

dx

_dy 43-2 dx

*

y

= (3D? +5D—2)y. Hence AB = (D + 2)(3D — 1) = 3D? + 5D —2. Now consider BA. Acting upon y, the operator BA yields

B(Ay) = BD —1) (2d a 2y) —

3-—

J

dx?

dx

d’y | =3—-+5—

dy

dx? t

d

dx

-2

= (3D? +5D—2)y. Hence

BA =3D?+5D—2= AB. io EXAMPLE 6.8 Let G =xD+2,and H = D—1. Then

d G(Hy) = (xD +2) (2 ” v)

6.8

The Fundamental Laws of Operation

111

so

GH =xD?+(2—x)D—-2. On the other hand,

H(Gy) = (D—1)(x—dy +2y dx

dy

= —

d

(x—+42y)—(x—

dx (x24 d’y

dy

tpt We

») (x4 dy

ae

r 42 ») y

“ae

d*y dy =x—— -2y:y x72 +t (3 —x)— ale that is,

HG =xD?+(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 agD" +a@,D"!

4.6.4 an—|D + ay

and adding corresponding coefficients. For instance, if

A=3D?—_D+x-2 and

B=x’D? +4D +7, then

A+B=(34x°)D?+3D+x45. Differential operators are linear operators; that is, if A is any differential operator, c; and c2 are constants, and f; and f> are any functions of x each possessing

the required number of derivatives, then

ACeifi + c2f2) = c Afi + QAfs.

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 follows that differential operators satisfy the following:

112

Chapter6

Linear Differential Equations

(a)

The commutative law of addition:

A+B=B-+A. (b)

The associative law of addition:

(A+ B)+C=A+4+(B+C). The associative law of multiplication:

(c)

(AB)C = A(BC). The distributive law of multiplication with respect to addition:

(d)

A(B+C)=AB+AC.

If A and B are operators with constant coefficients, they also satisfy the commutative law of multiplication:

(ec)

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 myn DED?

m-+-n =D",

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 multiplications indicated in Exercises | through 4.

1. 2.

— 2). (4D+1)(D 3). +D (2D —3)(2

3. 4.

—2D +5). (D+2)(D? (D—2)(D+1)’.

In Exercises 5 through 16, factor each of the operators.

5. 6. 7. 8.

2D? +3D—-2. 2D?-—5D-—12. Di —2D?—-5D+6. *+6. —- 11D 4D? —4D

9. 10. ll. 12.

D* —4p?. D3 —3pD*+4. D?—21D+20. 2D3-— D?-13D—6.

6.9

Some Properties of Differential Operators

13.

2D*+11D°?+18D?+4D-8.

14.

8D*+36D3—66D?+35D—6.

15. 16.

113

D*+ D3—2p?44p~24. — D3 —11D—20.

Perform the multiplications indicated in Exercises 17 through 22.

17. 18. 19.

[ 6.9|

(D—x)(D+x). (D+x)(D —x). D(xD—1).

20. 21. 22.

(xD—1)D. (xD+2)(xD—1). (xD—1)(xD +2).

Some Properties of Differential Operators Since for constant m and positive integral k, Dkem

_—

m* gmt.

(1)

it is easy to find the effect that an operator has upon e’”"*. Let f(D) be a polynomial in D,

f(D) = aD" +a, D"" +--+ +ay_-1D + ay.

(2)

Then

f(D)e™ = agm"e™ + aym""e™ +... + a,_\me™ + a,e™, so

f(D)e"™ = e™ fm). Ifm

(3)

is a root of the equation f(m) = 0, then in view of equation (3),

f(Dye™ = 0. Next consider the effect of the operator D — a on the product of e“* anda function y. We have (D

—a)(e"y)

=

Die“

—

e"

y)

_

ae

y

Dy

and (D

_

a)*(e"y)

_—

(D

—

e

_

a)(e“*

Dy)

D*y,

Repeating the operation, we are led to (D

_

ay"

(e** y)

i ef

D"y.

(4)

114.

Chapter 6

Linear Differential Equations

Using the linearity of differential operators, we conclude that when f(D) is a polynomial in D with constant coefficients, then

e f(D)y = f(D —a)e“y].

(5)

The relation (5) shows us how to shift an exponential factor from the left of

a differential operator to the right of the operator. This relation has many uses, some of which we examine in Chapter 7. EXAMPLE

6.9

Let f(D) = 2D? + 5D — 12. Then the equation f(m) = Ois Im? + 5m — 12 =0, or

(m + 4)(2m — 3) =0, of which the roots are m, = —4 and m2 = 3.

With the aid of equation (3) it can be seen that

(2D + 5).—e

=0

and that

(2D? + 5D — 12) exp ($x) =0. In other words, y; = e~* and y. = exp ($x) are solutions of

2D? +5D —12y= 0.

EXAMPLE Show that

6.10

=0 (D—m)"(xke™)

fork =0,1,...,@—1).

(6)

In equation (5) we let f(D) = D” and y = x*. Then using the exponential shift, we obtain (D

_ m)" (xke™)

—

el D"x*.

But D’x* =0 fork =0,1,2,...,2 —1, which gives us equation (6) directly. The results obtained in equations (3), (5), and (6) are of fundamental impor-

tance to the solving of linear differential equations with constant coefficients, which we consider in Chapter 7.

a

6.10

Computer Supplement

115

EXAMPLE 6.11 As an example of the use of the exponential shift, we solve the differential equation

(D +3)*y =0.

(7)

First we multiply equation (7) by e** to obtain

e*(D+3)*y =0. Applying the exponential shift as in equation (5) leads to D*(e*y) =0. Integrating four times gives us ery

= ¢, + eon + 9x7 + c4x?,

and finally,

y = (cy teox +.03x* +eaxe™.

(8)

Note that each of the four functions e~**, xe", x?e73*, and x3e7>* 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 functions are linearly independent, equation (8)

gives the general solution of equation (7). See Exercise 5.

& Exercises In Exercises solution.

l. 2.

| through 4, use the exponential shift as in Example 6.11 to find the general

(D—2)y=0. (D+1)*y=0.

3. 4.

(2D-1)*y =0. (D+7)*y =0.

To show that the four functions in Example 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 obtained in Exercise | of Section 6.4. 6.

Prove that the set of functions e C

xe

j

x 2 el ax

—T

x” - I ax2

is a linearly independent set on any interval. See Exercise 5.

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 D? — 3D? + 4 can be factored with the Maple command

116

Chapter 6

Linear Differential Equations >factor

(D*3-3*D*2+4) ;

(D + 1)(D - 2) A second computer application is illustrated in showing that the four functions exp (—3x),

x exp (—3x),

x? exp (—3x),

and x? 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)]); [e>*,

>Ans:=det

xe*

x*e*,

xte*]

(Wronskian(y,x));

Ans := 12 (e

—aegyidl

3x)

We can easily see that the Wronskian is never zero, and hence the functions are linearly independent. @ Exercises

1.

Do aselection of the factoring exercises in Section 6.8 using a computer.

2.

Do Exercises 2, 3, 6, and 7 in Section 6.4.

Linear Equations with Constant Coefficients

Introduction Several methods for solving differential equations with constant coefficients are presented in this 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 use in solving linear differential equations. In Chapter 12 we study matrix techniques for solving linear equations with constant coefficients. Each method has its advantages and its disadvantages. Each is theoretically sufficient: all are necessary for maximum efficiency.

The Auxiliary Equation:

Distinct Roots

Any linear homogeneous differential equation with constant coefficients, d"y 40h

ge" Tye

dy FT

Aa

+ aay

= 0,

(1)

may be written in the form

f(D)y = 0,

(2)

where f(D) is a linear differential operator. As we saw in the preceding chapter, if m is any root of the algebraic equation f(m) = 0, then f (D)e"* = 0, which means simply that y = e’”* is a solution of equation (2). The equation

f(m) =0

(3)

is called the auxiliary equation associated with (1) or (2). The auxiliary equation for (1) is of degree n. Let its roots be 771,..., m,. these roots are all real and distinct, then the n solutions yi =exp(mx),

yo =exp (mex),

...,

If

Vy = exp (in, x)

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 = cy exp (m Xx) + cz exp (mx) + +++ + Cy EXP (nx), in which ¢), C2, ..., Cy are arbitrary constants.

Repeated roots of the auxiliary equation will be treated in the next section. Imaginary 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

dy —

d’y | dy

dx?

—4——~+—

dx? r dx wmy

+6y

=0.

First write the auxiliary equation

m> —4m? +m+6=0, whose roots m = —1, 2, 3 may be obtained by synthetic division.

Then the

general solution is seen to be

y=cye*

+ ee" +. c3e™.

EXAMPLE 7.2 Solve the equation

(3D* +5D? —2D)y = 0. The auxiliary equation is 3m? + 5m? — 2m =0 and its roots are m

=

0, —2,

I

By using the fact that e®*

3.

solution may be written

y =e, +ere* +3 exp (4x).

EXAMPLE 7.3 Solve the equation ad

“~ —4x =0 dt with the conditions that when ¢t = 0, x = 0 and dx/dt = 3.

=

1, the desired

7.2

The Auxiliary Equation: Distinct Roots

119

The auxiliary equation is m> —4=0, with roots m = 2, —2. Hence the general solution of the differential equation is

x=cqet+oe. It remains to enforce the conditions at t = 0. Now

a

= 2cje" — 2ee7”.

Thus the condition that x = 0 when t = O requires that 0=c,

+ ¢2,

and the condition that dx /dt = 3 when ¢ = 0 requires that 3 = 2c, — 2. From the simultaneous equations for c; and cz we conclude that c) OQ=

—3.

=

- and

Therefore,

x = 3(e"% —e%), which can also be put in the form

x = 3 sinh (2r).

Exercises In Exercises | through 22, find the general solution. When the operator D is used, it is implied that the independent variable is x.

1,

(D?+2D-3)y =0.

2.

(D?+2D)y

3.

(D?+D-6)y=0.

4.

10.

=0.

(D*-—5D+6)y =0.

(4D? — 13D —6)y =0. d*x

d*x

eta dex

die

5.

(D3 43D2—4D)y =0.

12, ap

6.

(D> —3D*—10D)y =0.

13.

(99D?-7D+2)y =0.

7.

(D?+6D*4+11D+6)y=0.

14.

(4D?—21D—-10)y =0.

8. 9. 17.

(D34+3D?-4D-12)y=0. 15. (4D? -—7D+3)y =0. 16. (4D* — 8D? —7D* + 11D + 6)y = 0.

18.

(4D* — 16D? +7D* +4D —2)y =0.

19.

(4D*+4D> — 13D? -7D + 6)y = 0.

7

dx

7a”

ay + 30x = 0.

(D'-14D+8)y =0. (D?— D?-4D—2)y =0.

120 >

Chapter 7

Linear Equations with Constant Coefficients

20.

(4D° — 8D* — 17D3 + 12D* +. 9D)y = 0.

21.

(D* —4aD + 3a7)y =0;

22.

[D* —(a+b)D+ ably =0;

areal 4 0.

a and b real and unequal.

In Exercises 23 and 24, find the particular solution indicated.

23. (D? —2D —3)y =0; whenx =0, y =0, y’ = —4. 24, (D? — D —6)y =0; when x = 0, y = 0, and whenx = 1, y =e’. In Exercises 25 through 29, find for x = | the y value for the particular solution required.

25.

(D* —2D

26. 97. 28. 29.

(D* (D* (D? (D3

—3)y

y=4, Y= =0, nx whe

=0;

—4D)y =0; whenx =0, y=0, y’ =0, y”= = 0; when x = 0, y =3, y’ = —1. —D—6)y x = 2, y= 1. x =0, y =0, and when +3D —10)y =0; when —2D* —5D +6)y =0; whenx =0, y=1, y’=-7, y"=-l.

The Auxiliary Equation: Repeated Roots Suppose that in the equation

f(D)y =0

(1)

the operator f(D) has repeated factors; that is, the auxiliary equation f(m) = 0 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; = b, my = b, m3 = b. The corresponding part of the solution yielded by the method of Section 7.2 is y=

oe"

+ oe”

+ c3e*,

y=(qotateaje™.

(2)

Now (2) can be replaced by

y = cqe™

(3)

with c4 = cj +c2+c3. Thus, corresponding 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,; = m2 = m3 = b are not linearly independent.

What is needed is a method for obtaining n linearly independent solutions corresponding to n equal roots of the auxiliary equation. Suppose that the auxiliary equation f(m) = 0 has the n roots

7.3

The Auxiliary Equation: Repeated Roots

121

my, =m,=---=m,=b. Then the operator f(D) must have a factor (D — b)". We wish to find # linearly

independent y’s for which (D — b)"y =0.

(4)

Turning to result (6) near the end of Section 6.9 and writing m = b, we find that

(D—b)"(x*e"*)=0

for

k=0,1,2,...,(n—1).

(5)

The functions y, = x*e?* where k = 0, 1,2,..., (n—1) are linearly independent because, aside from the common factor e’*, they contain only the respective powers x°, x!,x?,...,x"7!. (See Exercise 5 of Section 6.9.) The general solution of equation (4) is 6) exe, y =cye™ + egxe® +--+ Furthermore, if f(D) contains the factor (D — b)", then the equation

(1)

f(D)y =0 can be written

a

g(D)(D —b)"y =0,

where g(P) contains all the factors of f(D) 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 solution 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. Corresponding to a root m; distinct from all others, there is the solution

yi = c exp (m;x), and corresponding to n equal roots mj), m2, solutions ce,

(8)

..., M#,, each equal to b, there are

coxe™, 2.2, Cg

eP*.

(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 Constant Coefficients

EXAMPLE 7.4 Solve the equation

(D* — 7D? + 18D* — 20D + 8)y = 0.

(10)

With the aid of synthetic division, it is easily seen that the auxiliary equation

m' —Tm3 + 18m* — 20m + 8 = 0 has the roots m = 1, 2,2,

2. Then the general solution of equation (10) is

y=ce

+ coe

+ cgxe™ + cgx7e™,

or

y =cye™ + (co + egx +c4x7Je™.

fet EXAMPLE 7.5 Solve the equation 4 -

3 + 2

2 + =

= 0.

The auxiliary equation is m+

with roots m = 0, 0, —1,

2m? o m=

0,

—1. Hence the desired solution is

yey

tox

+c3e* +c4xe™.

BI Exercises In Exercises | through 20, find the general solution.

1. 2. 3. 7. 8.

(D?-6D+9)y =0. (D?+4D+4)y=0. (4D3+4D2+ D)y =0. (4D° -—3D+1)y=0. (D+ —3D3 — 6D? + 28D — 24)y =

.

9,

(D?+3D*4+3D+4+1)y =0.

10.

(D?4+6D*2+12D+8)y=0.

13.

(4D*+4D3 —3D* —2D + 1l)y =0.

5. 6.

(D? — 8D? + 16D)y = 0.

(D* + 6D? + 9D*)y = 0. (D3 —3D* + 4)y =0.

0.

11. = 12.

(D> — D*)y = 0.

(D° — 16D3)y = 0.

7.4

A Definition of exp z for Imaginary z

14. 15.

(4D* — 4D? — 23D? + 12D + 36)y = 0. (D*+3D3 — 6D? — 28D — 24)y =0.

16.

(27D* — 18D? +8D—1)y =0

17. 18.

(4D° — 23D? — 33D? — 17D —3)y =0 (4D°— 15D? —5D? +15D+9)y =0

19.

(D*—5D* —6D—2)y = 0.

20.

(D5 —5D*+7D> + D? -8D+4)y =0

123

In Exercises 21 through 26, find the particular solution indicated.

21. (D?+4D+4)y =0; whenx =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). 93. (D? —3D —2)y =0; whenx =0, y=0, y' =9, y” =0.

24. (D*+3D3+2D?)y =0; whenx =0, y=0, y' =4, y’ =—6, y” = 14. 25. The equation of Exercise 24 with the conditions that when x = 0, y = 0, y’ =3,

y"

=—5,

yt

=—9,

26. (D3? + D? — D — 1)y = 0; whenx = 0, y = 1, whenx = 2, y = 0, and also with the condition as x + co, y > 0. In Exercises 27 through 29, find for x = 2 the y value for the particular solution required.

27. (4D? —-4D + 1)y = 0; when x =0, y = —2, y’=2. 28. (D?+2D*)y =0; whenx =0, y = —3,y’ =0,y” = 12. y = —1, whenx = 1, y = 0, +3D —9)y = 0; whenx = 0, 29. (D3 +5D? and also with the condition as x —

oo, y >

0.

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 = a +if witha and f real. Since it is desirable to have the ordinary laws of exponents remain valid, it is wise to require that exp (a + if) = e%e"®,

(1)

To e* with a real, we attach the usual meaning. Now consider e’?, 6 real. In calculus it is shown that for all real x

x x x3 SaltSe Stott 2!

x" ote, n!

(2)

124)

Chapter 7

Linear Equations with Constant Coefficients or oo

x?

)

ar

n=0

ie

e=

If we tentatively put x = if in (2) as a definition of e'?, we get ; cap

5 1292 73.93 Ft

i”

Bp" bee,

(3)

Separating the even powers of $ from the odd powers of f in (3) yields

.

en

(22

“4 o4

a

tor

2k

Q2k

ee

tap

tt

Gert i

i?

1!

3

joke!

3!

2k+1

(2k + 1)!

or ip _

OO [2k 2k

= (—1)*, so we may write p*

er BB es=]

(4)

(2k)! +2, (Qk +1)!

° =) Now i

08 gehe 2k+1h gee B+t

p*

Pe ntrat

(—1)*

smell

oa

Ne ate

Qh)! B

ae p?

Hi]

(—1)* pk!

Feet

(Qk +1!

to],

or

oie= So DEBE oa (2k)!

+i>> —

EB (2k + 1)!

But the series on the right in (5) are precisely those for cos 6 and sin f

(5) as

developed in calculus. Hence we are led to the tentative result

e'? = cosh +isinB.

(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 (a + 76), namely, exp (a + if) = e*(cosB +i sin f)

when @ and f are real.

(7)

7.5

The Auxiliary Equation: Imaginary Roots

125

Replacing 6 by(—f) in (7) yields a result that is of value to us in the next section,

exp (a — if) = e*(cosB — isin B). It is interesting and important that with the definition (7), the function e* for

complex z retains many of the properties possessed by the function e* for real x. Such matters are often studied in detail in books on complex variables.! Here we need in particular to know that if y=exp(a+ib)x, with a, b, and x real, then

(D—a—ib)y =0. The result desired follows at once by differentiation, with respect to x, of the function

y =e" (cos bx +i sinbx).

The Auxiliary Equation:

Imaginary Roots

Consider a differential equation f(D)y

= 0 for which the auxiliary equation

f(m) = 0 has real coefficients. From elementary algebra we know that if the auxiliary equation has any imaginary roots, those roots must occur in conjugate

pairs. Thus if m;=a+ib

is a root of the equation f(m) = 0, with a and b real and b # 0, then m,=a—ib

isalso arootof f(m) = 0. It must be keptin mind that this result is a consequence

of the reality of the coefficients in the equation f(m) = 0. Imaginary roots do not necessarily appear in pairs in an algebraic equation whose coefficients involve imaginaries.

We can now construct in usable form solutions of

f(D)y =0

(1)

corresponding to imaginary roots of f(m) = 0. For since f (mm) is assumed to have real coefficients, any imaginary roots appear in conjugate pairs, m, =a+ib

and

mz =a— ib.

! For example, R. V. Churchill and J. W. Brown, Complex Variables and Applications, 6th ed, (New

York: McGraw-Hill, 1996).

126

Chapter?

Linear Equations with Constant Coefficients

Then, according to the preceding section, equation (1) is satisfied by y =c

exp[(a + ib)x] + cz exp [(a — ib)x].

(2)

Taking x to be real along with a and b, we get from (2) the result y = ce“ (cos bx + isinbx) + me (cos bx —i sin bx).

(3)

Now (3) may be written y = (cy +o)e™ cos bx + i(c, — co)e™ sin bx. Finally, let c} + cz = ¢3 and i(¢; — c2) = c4, where cz and c4 are new arbitrary constants. Then the equation (1) is seen to have the solutions y =c3e corresponding to the two roots my

cos bx + cse sin bx, = a+

ib and m2

(4)

= a — ib (b # 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 particular solution corresponding to those two roots.

EXAMPLE 7.6 Solve the equation

(D? — 3D? +9D + 13)y =0. For the auxiliary equation m one root, m,;

=

—3m* +9m + 13 =0,

—l, 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 m

—4m

+13

=0.

Those roots are found to be mz = 2+ 3i and m3 = 2 — 3. The auxiliary equation has the roots m = —1, 2 + 37. Hence the general solution of the differential equation is

y = cye* + cae” cos 3x + c3e”* sin 3x. Repeated imaginary roots lead to solutions analogous to those brought in by repeated real roots. For instance, if the roots m = a + ib 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 (cy + cox + c3x")e™ cos bx + (c4 + 65x + cox? )e

sin bx.

EXAMPLE 7.7 Solve the equation

(D* + 8D* + 16)y = 0. The auxiliary equation m+ + 8m? + 16 = 0 may be written (m? +4)? =0,

so its roots are seen to be m = +27, +2i. The roots m, = 2i and m2 = —2i occur twice each. Thinking of 2/ as 0 + 2i and recalling that e°* = 1, we write

the solution of the differential equation as y = (€) + cox) cos 2x + (c3 + c4x) sin 2x.

In such exercises

as those following

Section 7.6, a fine check can be ob-

tained by direct substitution of the result and its appropriate derivatives into the differential equation. The verification is particularly effective because the operations performed in the check are so different from those performed in obtaining the solution.

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

ithe x == ————; 2“ —* sinh 2

(1)

the hyperbolic cosine of x, written cosh x, is defined by x

coshx =

—Xx

—

From the definitions of sinh x and cosh x it follows that

sinh? x = 4(e* —2+e*) and cosh?x = i(e*

24. gy,

(2)

128

Chapter 7

Linear Equations with Constant Coefficients sO

cosh? x — sinh* x = 1,

(3)

an identity similar to the well-known identity cos” x +sin’ x = 1 in trigonometry.

Directly from the definition we find that y = sinhu

is equivalent to y=

i(e"

_

ey,

Hence, if uw is a function of x, then dy

wa

on ste

—u

du

+8") Tx’

that is, d du — Ix sinh sinhu uw == cos coshu—. i

(4) 4

The same method yields the result d du 1 a —x coshu u = sinhu—.

(5) 5

The graphs of y = coshx and y = sinh» are exhibited in Figure 7.1. Note the important properties: (a)

coshx

> | for all real x.

(b)

The only real value ofx for which sinhx = Ois x = 0.

(c)

cosh(—x) = coshx; that is, coshx is an even function of x.

(d)

sinh (—x) = —sinh.x; sinhx is an odd function of x.

The hyperbolic functions have no real period.

Corresponding to the period

27 possessed by the circular functions, there is a period 277i 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 D* coshax = a? coshax and D* sinhax = a’ sinhax, it follows that both cosh ax and sinh ax are solutions of

(D? — a*)y =0,

a0,

(6)

7.6

A Note on Hyperbolic Functions

129

y =cosh x

y =sinh x

Figure 7.1

Furthermore, the Wronskian of these two functions, Wa) =

cosh ax

sinh ax

asinhax

acoshax

=

4,

is not zero, so that coshax and sinhax are linearly independent solutions of equation (6). Hence the general solution of (6) may be written y =c, coshax + c) sinhax

instead of using the form y=cse™

+ c4e™.

It is very often convenient to use this alternative form for representing the general solution of (6).

EXAMPLE 7.8 Find the solution of the problem (D? —4)y = 0;

= 0, y=0, whenx

y’ =2.

The general solution of the differential equation (7) may be written y =e, cosh 2x + co sinh 2x, from which

y’ = 2c; sinh 2x + 2c. cosh2x.

(7)

130

9 Chapter 7

Linear Equations with Constant Coefficients

The initial conditions now require that 0 = c; and 2 = 2c, so finally,

y = sinh 2x, Note that if we were to choose the alternative form y= ce” + ce" for the general solution of (7), we would obtain the same result with a little more

fuss in determining c3 and cy. Indeed, one major reason for using the hyperbolic functions is that cosh ax and sinhax have values | and 0 when x = 0, a fact that is particularly useful in solving initial value problems.

fal i Exercises Find the general solution except when the exercise stipulates otherwise.

1.

Verify directly that the relation y = c3e cos bx + case“ sin bx

(A)

satisfies the equation

[(D — a)’ +b*]y =0. 2. 3.

(D?—2D+4+5)y =0. (D?-2D+2)y =0.

4. 5.

(D?+9)y =0. (D?—9)y=0.

9, 10.

(D*++2D? + 10D*)y =0. (D*-2D?+2D°—-2D+1)y =0.

6.

(D?+6D+4 13)y =0.

11.

(D4+18D*+81)y =0.

12. 13. 14.

7. 8.

(D*?-—4D+7)y =0. (D?+2D?+D+4+2)y =0.

(2D*+ 11D? —4D* — 69D + 34)y =0. (D°+9D* + 24D? + 16)y =0. (2D3 — D? + 36D — 18)y =0.

15.

(D*? —1)y

16.

(D?+1)y =0; whenx =0, y = yo, y’ =0.

=0;

whenx

=0, y = yo, y’ = 0.

17. 18.

(D3 +7D? + 19D + 13)y =0; whenx =0, y=0, y’ =2, y” = —12. (D° + D* —7D? — 11D? — 8D — 12)y =0.

19,

ds ip

20.

y whenx =0, y=0, y =—-1, y”=5. (D3 + D?+4D+4+4)=0;

a,

d ? 2 pop 4x dt? dt

dx = 0, x = 0, ap = Me + k?x = 0, k real; when

=0,k > b> 0; whent =0,x =0,

4

=H. dt

7.6

A Note on Hyperbolic Functions

131

i Miscellaneous Exercises

-

&

WwWwWNNNNNYNNNYNNYY

WY

SSE S&FRESESESSSVRNRARESHA

SS gd SI

Gy be eee

be

Obtain the general solution unless otherwise instructed.

10. (D* + 3D)y = 0. ll. (9D* + 6D? + D*)y =0. 12. (D? + D—6)y =0. 13. (D? + 2D* + D+2)y =0. 14. (D3 — 3D? +4)y =0. 15. 0. = (D? —2D? —3D)y 16. (4D? —-3D + 1)y =0. 17. (D3 +3D?2—4D—12)y=0. 18. (D? +3D? +3D+ 1)y =0. (D4 — 11D3 + 36D? — 16D — 64)y = (D? +2D+5)y =0. (D* + 4D3 + 2D? — 8D —8)y =0.

(4D? — 21D — 10)y = 0. (4D3-7D+3)y =0. =0. (D?—14D+8)y (8D? -—4D? —-2D + 1)y=0. =0. (D*+ D3 — 4D? —4D)y (D*-2D3+5D*-8D+4)y = 0. (D+ +2D*+4+1)y=0. =0. (D*+5D?+4)y (D*+3D? —4D)y =0. 0.

(4D* — 24D? + 35D* + 6D —9)y = 0. (4D* + 20D? + 35D? + 25D + 6)y = 0. (D* —7D? + 11D? +5D-— 14)y =0.

(D3 +5D?+7D+3)y =0.

33.

=0. (D*— D'—3D°+D+2)y

(D> — 2D? + D—2)y =0.

34.

(D? — 2D? —3D+

— D? + D—-1)y=0. (D3 (D? +4D*? + 5D)y =0. (D* — 13D* + 36)y = 0.

35. 36. 37.

(D+ D*—6D%)y =0. (4D>+28D?+61D+37)y = 0. (4D3+12D7+13D+10)y = 0.

(D*—5D3+5D?4+5D—6)y =0.

38.

(18D3—33D?+20D—4)y = 0.

39. (4D? +8D2—11D+3)y=0. 40. (D?+D?-16D-—16)y=0. (D° — 15D? + 10D? + 60D — 72)y = (4D* — 15D? +5D+6)y =0. (D4 +3D3 — 6D* — 28D — 24)y =0. (4D* — 4D? — 23D? + 12D + 36)y =

= 0. 10)y

(D°—2D?—2D*-—3D—2)y = 0. (D*-2D3+2D?-2D+l)y =0. 0.

0.

(4D° — 23D? — 33D? — 17D — 3)y =0. y=2, yw = 1. nx = 0, =0; whe (D? — D—6)y (D* + 6D? + 9D*)y = 0; when x = 0, y = 0, y’ = 0, y” = 6, and as x — oo, y’ > I. For this particular solution, find the value of y when = 1,

132.

Chapter 7

Linear Equations with Constant Coefficients

48.

(D?+6D*+12D+8)y=0;

whenx =0, y=1, y’ = —2, y” =2.

49. 50. 51.

(D> + D* —9D° — 13D? + 8D +4 12)y =0. (4D° + 4D* — 9D} — 11D? + D+3)y =0. (D5 + D*—7D3 — 11D* — 8D — 12)y = 0.

Computer Supplement The techniques described in the Computer Supplement to Chapter 2 extend easily

to higher-order equations. We can illustrate this with Example 7.6 in Section 7.5

(D? — 3D* +9D + 13)y =0. If we add the initial conditions, y(O) =

1, y’(0) = 2, y"(0) = 3, Maple solves

the problem with the commands >diff(y(x),x$3)-3*diff(y(x) +9*diff(y(x)

,x$2)

,x)+13*y(x)=0;

d°

ae

d?

—3 Ge)

>dsolve({",y(0)=1,D(y)

yx) =

4e*

d

+9 a”

(0)=2,D(D(y)) = 5e?* cos(3x)

9

9

+13 y(x) =0 (0)=3},y(x)); 4e** sin(3.x)

9

We can also use Maple to plot the resulting solution with the command splot

(rhs (") ;xX=-2a082)

3

See Figure 7.2.

Figure 7.2

7.7 Computer Supplement

Exercises 1. 2.

Solve a variety of problems from the chapter. Use a plotting routine to display your results.

133

a Nonhomogeneous Equations: Undetermined| Coefficients Construction of a Homogeneous Equation from a Specific Solution In Section 6.6 we saw that the general solution of the equation

(bp D" + by DD"! +--+ + by

D + by)y = RQ)

(1)

is Y=¥erYps where y,, the complementary function, is the general solution of the homogeneous

equation

(by D" + b)D" | +++ +b,1D+b,)y =0

(2)

and y, is any particular solution of the original equation (1). Various methods for getting a solution of (1) when the bo, bj, ..., &, are con-

stants 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 c,e“* occurred only when the auxiliary equation f(m) = O had a root m = a, and then the operator f(D) had a factor (D — a). In like manner, coxe“*

appeared only when f(D) contained the factor (D — a)’, c3x?e* only when f(D) contained (D — a)*, and so on. Such terms as ce“* cos bx or ce“* sin bx correspond to roots m = a + ib, or to a factor [(D — a)? +b’).

EXAMPLE 8.1 Find a homogeneous

linear equation, with constant coefficients, that has as a

particular solution y =7e™* + 2x. First note that the coefficients (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= ce** + c>x no matter what the constants c, and c2 may be.

134

81

Construction of a Homogeneous Equation

135

A term c)e** occurs along with a root m = 3 of the auxiliary equation. The term c2x will appear if the auxiliary equation has m = 0, 0, that is, a double root m = 0. We have recognized that the equation

D*(D —3)y =0, or

(D? —3D*)y = 0, has y = cye** + crx + c3 as its general solution, and therefore that it also has

y = 7e** + 2x as a particular solution.

EXAMPLE 8.2 Find ahomogeneous linear equation with real, constant coefficients that is satisfied

by y =6+3xe*

—cosx.

(3)

The term 6 is associated with m = O, the term 3xe* with a double root m = 1, 1, and the term (— cos x) with the pair of imaginary roots m = 0 +7. Hence

the auxiliary equation is man — 1)° (mn? +1)=0,

or m> —2m* + 2m? — 2m? +m

= 0.

Therefore, the function in (3) is a solution of the differential equation

(D> —2D* +2D? —2D* + D)y =0.

(4)

That is, from the general solution

y=cy

+ (eo + c3x)e* + c4 cos x + cs sinx

of equation (4), the relation (3) follows by an appropriate choice of the constants: c|

= 6,09

=

0,3

=

3.

a

=

—1,

¢5

=:

EXAMPLE 8.3 Find a homogeneous linear equation with real, constant coefficients that is satisfied

by y = 4xe" sin 2x. The desired equation must have its auxiliary equation with roots m=1l+2i,1+42i.

136

Chapter 8

Nonhomogeneous Equations: Undetermined Coefficients The roots m = 1+2i correspond to factors (m— 1)? +4, so the auxiliary equation must be

[om — 1)? +4] =0, or

m* — 4m? + 14m* — 20m + 25 = 0. Hence the desired equation is

(D* — 4D? + 14D? — 20D + 25)y =0. & Note that in all such problems, a correct (but undesirable) solution may obtained by inserting additional roots of the auxiliary equation.

be

@ Exercises In Exercises 1 through 14, obtain in factored form a linear differential equation with real, constant coefficients that is satisfied by the given function.

1.

y=4e%* +3e.

2.

y=7—2x+he*.

3.

ys —2x+ sets

8

y=e*sin2x.

9,

y=xe™* sin2x + 3e~* cos 2x.

10.

y =sin2x + 3cos 2x.

4, y=x?—Ssin3x.

Il. y= coskx.

5.

y =2e* cos3x.

12,

y=xsin2x.

6.

y = 3e

13.

y=4sinhx.

7.

y=

14.

y =2cosh2x

sin3x.

—2e* cosx.

— sinh 2x.

In Exercises 15 through 34, list the roots of the auxiliary equation for a homogeneous linear equation with real, constant coefficients that has the given function as a particular solution.

15.

y= 3xe.

25.

yoo cos 2x.

16.

y=x%e* + 4e*.

26.

yoe* cos3x.

17.

y=e*cos4x.

27.

y=xcos2x

18.

y =3e*cos4x + 15e~* sin4dx.

28.

y= e-**(cos3x + sin3x).

19.

y=x(e* +4),

29.

y=sin’x = }(3sinx — sin3x).

20.

y=442x7-e*,

30.

y =cos* x.

21.

y=xe*.

31.

y=x*—x+e*(x

22. 23.

y=xe* +5e*. y=4cos2x.

32. 33.

you sine y=x?sinx + xcosx.

24.

y=4cos2x — 3sin2x.

34.

y = 8cos4x + sin3x.

—3sin2x.

+c0sx).

8.2

Solution of a Nonhomogeneous Equation

137

Solution of a Nonhomogeneous Equation Before proceeding to the theoretical basis and the actual 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

D?(D — 1)y = 3e* + sinx.

(1)

The complementary function may be determined at once from the roots

m=O, 0, |

(2)

of the auxiliary equation. The complementary function is Ye = Cy tex

+ c3e".

(3)

Since the general solution of (1) is Y=Ve+Yp;

where y, is as given in (3) and y, is any particular solution of (1), all that remains for us to do is to find a particular solution of (1). The right-hand member of (1),

R(x) = 3e* +sinx,

(4)

is a particular solution of a homogeneous linear differential equation whose auxiliary equation has the roots

m' = 1, +i.

(5)

Therefore, the function R is a particular solution of the equation

(D —1)(D? + D)R=0.

(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 — 1)(D? + 1) will annihilate the right member of (1).

Therefore we apply that operator to both sides of equation (1) and get

(D — 1)(D* +1) D?(D — 1)y = 0.

(7)

Any solution of (1) must be a particular solution of (7). The general solution 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 = cy + 2x + c3e" + caxe” +5 c0SX + €6 sinx.

(8)

But the desired general solution of (1) is

Y=Yet Vp, where Ye =e

+ ox fese*,

(9)

138

Chapter8

Nonhomogeneous Equations:

Undetermined Coefficients

the c), ¢2, 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

yp1 = Axe“ + Bcosx +C sinx.

(10)

We now have only to determine the numerical coefficients A, B, C by direct use of the original equation

D*(D — 1)y = 3e* +sinx.

(1)

From (10) it follows that Dyp = A(xe* + e*) — Bsinx + Ccosx,

D*y, = A(xe* + 2e*) — Bcosx — Csinx, D*y, = A(xe* + 3e*) + Bsinx — Ccosx. Substitution of y, into (1) then yields Ae’ +(B4+C)sinx

+ (B — C)cosx

= 3e* + sinx.

(LD

Because (11) is to be an identity and because e*, sinx, and cos. are linearly independent, the corresponding coefficients in the two members of (11) must be equal; that is,

A=3 B+C=1 B-C=0. Therefore, A = 3, B = + c= solution of equation (1) is

i Returning to (10), we find that a particular

Yp = 3xe* + 4 cosx + $sinx. The general solution of the original equation,

D?(D— l)y = 3e* + sinx,

(1)

is therefore obtained by adding to the complementary function the y, found above:

y =c + cox +.c3e* + 3xe" + 5 cosx + 5 sinx.

(12)

A careful 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 m and m’ as exhibited in (2) and (5).

(b)

From the values of m and m’ write y, and y, as in (3) and (10).

8.3 (c)

The Method of Undetermined Coefficients

Substitute y, into (1), equate corresponding coefficients,

139

and obtain the

numerical values of the coefficients in y,. (d)

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 f(D) be a polynomial in the operator D, Consider the equation

f(D)y = R(&).

()

Let the roots of the auxiliary equation f (mm) = 0 be m=

M4, M19, ..., Mp.

(2)

Y=YetYp,

(3)

The general solution of (1) is

where y, 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 itselfa particular solution of some homogeneous linear differential equation with constant coefficients,

g(D)R = 0,

(4)

whose auxiliary equation has the roots ma! = tty My, ccs IMs

(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 =0

(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 y, of (3) and so is of the form Y=Ye + Yq:

But also any particular solution of (1) must satisfy (6). Now, if

f(D) Qe + Yq) = R(x), then f(D)y, = R(x) because f(D)y. = 0. Then deleting the y,. from the general solution of (6) leaves a function y, that for some numerical values of its coefficients must satisfy (1); that is, the coefficients in y, can be determined

so that yy = yp,. The determination of those numerical coefficients may be accomplished as in the following examples.

140)

Chapter8

~Nonhomogeneous Equations: Undetermined Coefficients

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 itself a particular solution of some homogeneous linear differential equation with constant coefficients. EXAMPLE 8.4 Solve the equation

(D? + D —2)y = 2x — 40 cos 2x

(7)

Here we have

m=

1,—-2

and

m' = 0, 0, £23. Therefore, we may write i

—2x

ye = ce +cje”, yp = A+ Bx+Ccos2x + E sin2x, in which

c, and c2 are arbitrary constants,

whereas

A,

B, C, and

E

are to be

determined numerically, so that y, will satisfy equation (7). Since

Dy, = B —2C sin2x + 2E cos 2x and D’y, = —4C cos 2x — 4E sin2x, direct substitution of y, into (7) yields —4C cos2x —4E sin2x + B —2C sin2x +2E cos2x —2A — 2Bx —2C cos2x —2E sin2x = 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 functions cos 2x, sin 2x, x, and | appearing in the identity. Thus it follows that —6C + 2E = —40, —6E —2C =0, —2B = 2, B-2A=0.

83

The Method of Undetermined Coefficients

141

The equations above determine A, B, C, and Z. Indeed, they lead to — _! A=-1,

_— C=6,

B=-l,

E = -2.

Since the general solution of (7) is y = ye + yp, We can now write the desired result, y=cje?

+ me"

— 5 —x+6cos2x

— 2sin2x.

i EXAMPLE 8.5 Solve the equation

(D? + l)y =sinx.

(9)

At once m = +i and m’ = +i. Therefore, Ye = c, cosx +c

sinx,

Yp = Ax cosx + Bx sinx. Now

yy = A(—xcosx — 2sinx) + B(—x sinx +2cos x), so the requirement that y,, is to satisfy equation (9) yields —2Asinx +2Bcosx

= sinx,

from which A = —5 and B = 0. The general solution of (9) is

y =c) cosx +c¢2 sinx — 5X COS X.

EXAMPLE 8.6 Determine y so that it will satisfy the equation yf?

_

y’

=

4e*

i

3e2*

with the conditions that when x = 0, y = 0, y’ = —1, and y” = 2. First we note thatm = 0,1, —1 and m’ = —1, 2. Thus

Yo = € + e2e* + c3e™*, Vp = Axe * + Be.

x

(10)

142

Chapter8

Nonhomogeneous Equations:

Undetermined Coefficients

Now

y, = A(—xe™ +e) + 2Be™, yh = A(xe™* — 2e™*) + ABe™, Vp

= A(—xe™* + 3e*) + 8Be™*.

Then

yy’ — yl, = 2Ae™ + 6Be™, so that from (10) we may conclude that A = 2 and B = 5 The general solution of (10) is therefore y=ey toe’ +e3e 7% + 2xe% + ce,

(11)

We must determine c;, co, c3 So (11) will satisfy the conditions that when x = 0, y =0,y’ =—l, and y” =2.

From (11) it follows that y’ = coe" — c3e* — 2xe* +20 * + e*

(12)

and y" — ce"

+ eye * +2xe*

—4e*

+ ae.

(13)

We put x = 0 in each of (11), (12), and (13) to get the equations for the determination of cy), c2, and c3. These are O=c

+o.

+0344,

—-l=o-@t+3,

= og bag2, from which c, = —3, co = 0, c3 = 4. Therefore, the final result is

j= -3 +4e* +2xe* + se",

fa] An important point, sometimes overlooked by students, is that it is the general solution, the y of (11), that must be made to satisfy the initial conditions.

i Exercises In Exercises | through 35, obtain the general solution.

1. 2.

(D?+ D)y =—cosx. (D?-6D+4+9)y =e".

3. 4.

(D7 43D + 2yy = 12x". (D?+3D4+2)y =143x +2".

Go ol

oy or

8.3

10. 17.

The Method of Undetermined Coefficients

(D* + 9)y = 5e* — 162x.

ll.

y”—4y’+3y = 20cosx.

143

(D? + 9)y = 5e* — 162x?.

12.

y"—4y'+3y = 2cosx+4sinx.

y” —3y’ — 4y = 30e*.

13,

y’+2y'’+y =74 75sin 2x.

y” — 3y’ —4y = 30e*. (D? — 4)y = e* +2. (D? — D —2)y = 6x + 6e~*.

14. 15. 16.

(D?+4D+4+5)y = 50x + 13e?*. (D?+1)y =cosx. (D?-4D+4+4)y =e.

(D? — l)y =e (2sinx

+ 4cosx).

18.

(pe

19. 20. 2Ag

(D> — D)y =x. (D> — D? + D—1)y = 4sinx. (D? + D? — 4D — 4)y = 3e-* — 4x — 6.

22, 23.

(D* —1)y

l)y = 8xe*.

= 7x’.

24,

(D*—1)y =e™*. (D? — 1)y = 10sin’ x. Use the identity sin?x = 4(1 — cos 2x).

25.

(D* + 1)y = 12 cos* x.

26. Au

(D* +4)y =4sin’ x.

30.

y’—y=e* —4,

y”—3y'—4y = l6ox—S50cos2x.

31.

y”’—y'—-2y =6x+6e.

28.

(D3 —3D—2)y=100sin2x.

32.

y’+6y'’+ 13y = 60cosx 4+ 26.

29.

33.

(D3-3.D?+4)y = 6480 cos 2x.

34.

y’ +4y' +3y = 15e%* +e", (D3 + D — 10)y = 29e**.

35.

(D3 + D? —4D —4)y = 8x +8 + 6e7*.

In Exercises 36 through 44, find the particular solution indicated.

36. ais

(D? + 1)y = 10e**; whenx = 0, y = 0, y’ = 0.

38,

(D? + 3D)y = —18x; whenx = 0, y =0, y’ =5.

39.

(D*? +4D + 5)y = 10e~**; whenx = 0, y = 4, y’ = 0.

(D? — 4)y = 2 — 8x;

d?x

AQ. aa Al.

dx tg t%

X+4x+5x

whenx =0, y =0, y’=5.

= 10:

= 8sint;

wheat b= 0.38

On

d. x

ae D

whent = 0, x = 0, x = 0. Note that the notation

* = dx/dt and ¥ = d?x/dt? is common when the independent variable is time.

42. 43. 44,

y” + 9y = 81x? + 14cos4x;

whenx

=0, y =0, y’ =3.

(D?+4D?+9D+10)y = —24e*; when x = 0, y =0, y’ = —4, y” = 10. y” + 2y’ + 5y = 8e*; whenx = 0, y=0, y’ =8.

144

Chapter8

Nonhomogeneous Equations:

Undetermined Coefficients

In Exercises 45 through 48, obtain, from the particular solution indicated, the value of y and the value of y’ atx = 2.

45, 46. A], 48.

49, 50.

51.

52. 53.

-—l. x; 3, andatx=1,y= +y= atx =0, y= y"+2y’ y’+2y' +y =x; atx =0, y = —2, y' =2. 4y"+y=2;atx=n,y=0,y'=1. 2y" —5y! —3y = —9x? — 1; atx =0, y=1, y’ =0. (D*+D)y =x+1; whenx = 0,y = 1,andwhenx = 1l,y = 5 Compute the value ofy atx = 4. (D? + Dy =x?; when x = 0, y = 0, and whenx = 2, y = 0. Show that this boundary value problem has no solution. (D? + 1)y = 2cosx; whenx = 0, y = 0, and whenx = m7, y = 0. Show that this boundary value problem has an unlimited number of solutions and obtain them. For the equation (D* + D*)y = 4, find the solution whose graph has at the origin a point of inflection with a horizontal tangent line. For the equation (D? — D)y = 2 — 2x, find a particular solution 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 frequently easy to obtain a particular solution of a nonhomogeneous equation

(bp D" + by D"! + +++ + dy-1D + bn)y = R(x)

(1)

by inspection. For example, if R(x) is a constant Ro and if b, # 0,

Ro

(2)

Yp= by

is a solution of

(bpD" + b,D" | +--+ +b, 1D+bp)y = Ro,

bn #0, Roconstant,

(3)

because all derivatives of y, are zero, so (by D" + b, Dp"!

+e

+b, D+

br)¥p

Ro_

= by b,

Ro.

Suppose that b, = 0 in equation (3). Let D*y be the lowest-ordered derivative

that actually appears in the differential equation. Then the equation may be written (bo)D" +--+ + Dye D*)y = Ro,

by-~ #0,

Ro constant.

(4)

8.4

Solution by Inspection

145

Now D*‘x* = k!, a constant, so that all higher derivatives of x“ are zero. Thus it becomes evident that (4) has a solution Ro x*

ye = tip’

(5)

for then (by D" + +++ + by, D*)y, = by, Ro k!/k! bn_-~ = RoEXAMPLE 8.7 Solve the equation

(D* —3D +2)y = 16.

(6)

By the methods of Chapter 7 we obtain the complementary function, Ye = cye* +.e,€**, By inspection a particular solution of the original equation is

pa

ps8.

Hence the general solution of (6) is

y= ce’ +ege" +8.

Ea EXAMPLE 8.8 Solve the equation 5 = From the auxiliary equation m

3 3 +4m3

=7, = 0 we get m

(7) = 0, 0, 0, +27. Hence

Ve = Cy +eox + 3x7 + 4008 2x + cs sin 2x. A particular solution of (7) is

ee

eS Bra

2

24"

As acheck, note that

7x3 Jiah (D°? BD? +4D\— + a =044.—_=7 + a

146

Chapter8

Nonhomogeneous Equations:

Undetermined Coefficients

The general solution of equation (7) is y=cy

tex

t+ 3.x" + ax

+ c4c0s 2x + cs sin 2x,

in which the c),...,¢s5 are arbitrary constants.

Examination of

(D? + 4)y = sin 3x

(8)

leads us to search for a solution proportional to sin 3x because if y is proportional to sin 3x, so is D’y. Indeed, from

y = Asin3x

(9)

we get

D°y = —9Asin 3x, so (9) is a solution of (8) if

(-9+4A=1

A=-t. Thus (8) has the general solution y =c, cos 2x + co sin2x — i sin 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,

and so to write yp = Asin3x + B cos ox.

(10)

When the y, of (10) is substituted into (8), it is found, of course, that

A=-z

B=0.

In contrast, consider the equation

(D? +4D + 4)y = sin 3x.

(11)

Here any attempt to find a solution proportional to sin 3x is doomed to failure because although D’y will also be proportional to sin 3x, the term Dy will involve cos 3x. There is no other term on either side of (11) to compensate for this cosine term, so no solution of the form y = Asin3x is possible. For this equation, m = —2, —2, m' = +33, and in the particular solution

84

Solution by Inspection

147

Yp = Asin3x + Bcos 3x,

it must turn out that B + 0. No labor has been saved by the inspection. In more complicated situations, such as

(D? + 4)y = xsin3x — 2cos 3x, the method of inspection will save no work.

For the equation

(D? +4)y = e™,

(12)

we see, since (D* + 4)e* = 29e>*, that %p

_—

Hen

is a solution. Finally, note that if y; is a solution of

f(D)y = R(x) and y3 is a solution of

f(D)y = Ra(x), then Yp = Yi + y2

is a solution of

f(D)y = Ri (x) + Ro(x). It follows readily that the task of obtaining a particular solution of

f(D)y = RQ) may be split into parts by treating separate terms of R(x) independently, if convenient. See the examples below. This is the basis of the “method of superposition,” which plays a useful role in applied mathematics.

EXAMPLE 8.9 Find a particular solution of

(D? — 9)y = 3e* +x — sin4x. Since (D? — 9)e* = —8e*, we see by inspection that yak

Yl =

—3e

(13)

148

Chapter 8

Nonhomogeneous Equations:

Undetermined Coefficients

is a particular solution of (D* — 9)y, = 3e*. In a similar manner, we see that v2. = —ix satisfies

(D* —9)y2 =x and that

3 = + sin 4x satisfies

(D? —9)y3 = —sin4x. Hence

Vp = ge" — gx + # sin 4x is a solution of equation (13).

EXAMPLE 8.10 Find a particular solution of

(D* + 4)y = sinx + sin 2x.

(14)

At once we see that y; = i sinx is a solution of

(D? + 4)y, = sinx. Then we seek a solution of

(D? + 4)y2 = sin 2x by the method of undetermined coefficients.

(15)

Because m = +2i and m’ = +21,

we put yo = Ax sin2x + Bx cos 2x into (15) and determine that 4A cos 2x —4B sin2x = sin2x,

from which A = 0, B = —{. Thus a particular solution of (14) is det Yp == = 3 Sinx —— 14x cos 2x.

8.4

Solution by Inspection

149

EXAMPLE 8.11 Find a particular solution of (D? +.a’)y = cos bx.

(16)

If b ¥# a, then a particular solution of the form y = A cos bx will exist. It follows from (16) that (—b*A +.a7A) cos bx = cos bx

and A = (a? — b*)~!. A particular solution of (16) is a

(a? — b*)~' cos bx.

If b = a, then equation (16) becomes

(D? + a*)y = cosax,

(17)

and no function of the form A cos ax is a particular solution, since the operator D? + a? will annihilate the function A cosax.

However,

a solution of the form

Ax cosax + Bx sinax exists. Upon substitution into (17) we require that —2aAsinax

+ 2aBcosax

= cosax,

an equation that is satisfied only if A = 0 and B = y=

I 2a*

5 sinax

Therefore, (18)

ad

is a particular solution of (17).

a We have seen in this example an important distinction between the cases b 4 a and b = a. Ina physical application considered in Chapter 10, the presence of

a solution of the form given in (18) results in a phenomenon called resonance. At this point we need only notice that the solution in (18) will be oscillatory in character, but the amplitudes of the oscillation will become increasingly large as x increases.

Ml Exercises 1.

Show that ifb 4 a, then (DP + a’)y = sinbx

has a particular solution y = (a? — b?)~' sin bx. 2.

Show that the equation (DB? +4 a’)y = sinax

has no solution of the form y = A sinax, with A constant. Find a particular solution of the equation.

Chapter 8

~Nonhomogeneous Equations: Undetermined Coefficients In Exercises 3 through 50, find a particular solution by inspection. Verify your solution,

r LY onan

150°)

9. 10. Ws 12. 13. 14. 15. 16. 17. 18, 19. 20. Qs 22. 23. 24,

(D* + 4)y = 12. (D* +9)y = 18.

2s 28. 29, 30.

(D?+4D+4)y =8. (D? + 2D — 3)y =6. (D? —3D4+2)y = —-7.

Bla 32.

(D* + 4D? + 4)y = —20. (D? (D? (D? (D? (D* (D* (Do (D> (D? (D?

+4D)y = 12. —9D)y = 27. + 5D)y = 15. + D)y = —8. — 4D?)y = 24. + .D*)y = —12. — D?)y = 3A. — 9D?)y = 27. + 4)y = 6sinx. + 4)y = 10cos 3x.

33. 34. 35. 36. 3 38. 39. AQ, Al.

(D? + 4)y = 8x + 1 — 15e*. (D? + D)y = 6+ 3e™. (D? + 3D —4)y = 18e?". (D? +2D +5)y = 4e* — 10. (D2 — 1)y = 2e**.

25;

(D? — 1)y = 2x +3. (D? — 1)y = cos 2x.

26,

(D*— 1)y = sin 2x.

42. 43. 44, 45, AO, 47. 48.

(D? + l)y = e* + 3x. (D* + 1)y = 5e7**. (D? + l)y = —2x + cos 2x.

(D* + 1)y = 4e7**. (D? + 1)y = 10sin 4x. (D*+ l)y = —6e**. (D? +2D +1)y = 12e*. (D?+2D+1)y=Te. (D? — 2D + 1)y = 12e~. (D? — 2D + 1)y = 6e~**.

(D? -2D —3)y =e". (D? — 2D —3)y =e”. (4D? + 1)y = 12sinx. (4D? + 1)y = —12cosx.

(4D? +4D + 1)y = 18e* — 5. (4D? +4D + l)y = 7e* +2. (D?—l)y =e™. (D3— 1)y = 4 — 3x?. (D3— D)y = e*. (D4 + 4)y = Se”.

49. 50.

(D* + 4)y = 6sin 2x. (D* + 4)y = cos 2x.

(D> — D)y = 5sin2x. (D? — D)y = 5cos2x.

Computer Supplement The algebraic manipulations needed to solve a differential equation using the method of undetermined coefficients are easily performed using a Computer Algebra System. As an example, consider Example 8.4 of Section 8.3.

(D* + 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) + F sin(2 x)

8.5

Computer Supplement

151

>yp (x) :=diff(y(x),x); yp(x) =

B-—2Csin2x)

+2 Fcos(2x)

>ypp (x) :=diff(yp(x),x); ypp(x)

:= —4C cos(2x) —4 F sin(2 x)

>ypp (x) typ (x) -2*y (x) -2*x+40*cos (2*x) =0; —6C cos(2x) —6F sin(2x)

+ B—2Csin(2x)

+2 Fcos(2x) —2A—2Bx >collect(",

—2x

+40

cos(2x) =0

[x,cos(2*x),sin(2*x)]);

(-2B-—2)x+(-6C+2F 440) cos(2 x) +(-6F —2C)snx)+B-2A=0 >solve({-2*B-2=0,-6*C+2*F+40=0,

-6*F-2*C=0,B-2*A=0},

{A,B,C,F}); {C =6,A =—-1/2, B =-1,F =-2}

We have only to substitute the resulting coefficients back into the original solution to obtain the desired result. i Exercises 1.

Use a computer to solve a selection of exercises from the chapter.

aa Variation of Parameters

Introduction In Chapter 8 we solved the nonhomogeneous linear equation with constant coefficients

(boD" + by D" | + +++ + bp-1D + bn)y = RO)

(1)

by the method of undetermined coefficients. We saw that this method would be applicable only for a certain class of differential equations: those for which R(x) itself was a solution of a homogeneous linear equation with constant coefficients. In this chapter we study two methods that carry no such restrictions. In fact, much of what we do will be applicable to linear equations with variable coefficients. We begin with a procedure by D’ Alembert that is often called the method of reduction of order.

Reduction of Order Consider the general second-order linear equation

y" + py’ +qy=R.

(1)

Suppose that we know a solution y = y, of the corresponding homogeneous equation

y” + py +qy =0.

(2)

Then the introduction of a new dependent variable v by the substitution (3)

y=yiv

will lead to a solution of equation (1) in the following way. From (3) it follows that y" y”

152

=yv —

yu"

+yv, +2y\v" ;

+

yv,

9.2

Reduction of Order

153

so substitution of (3) into (1) yields

yu" +2y\v' + ylu + py’ + py t+qyiu = R, or

yiv" + (2y, + pyiu’ + Or! + py + ayi)v = R.

)

But y = y; is a solution of (2). That is,

yi + py, +49y1 = 0 and equation (4) reduces to

yu" + (2y, + py)u' = R. Now

(5)

let v’ = w, so equation (5) becomes

yw + (2y, + pyi)w = R,

(6)

a linear equation of first order in w. By the usual method (integrating factor) we can find w from (6). Then we can get v from v’ = w by an integration. Finally, y = y,v. Note that the method is not restricted to equations with constant coefficients.

It depends only upon our knowing a single nonzero solution of equation (2). For practical purposes, the method depends also upon our being able to effect

the integrations. EXAMPLE 9.1 Solve the equation

yi -yse.

(7)

The complementary function of (7) is Ye = cye* +e2e ~.

We shall take the particular solution e* and use the method of reduction of order by setting y=uve. Then

y =ve*+v'e and

y” = ve* +20'e* + 0"e". Substituting into equation (7) gives vo’ +2

= 1,

(8)

154)

Chapter9

Variation of Parameters

Equation (8) is a first-order linear equation in the variable v’. Applying the integrating factor e** yields

e*(y" + 2u') =e. Thus

ery’ = 5e* +0,

(9)

where c is an arbitrary constant. Equation (9) readily gives uo = 7 + ce,

and hence v=ce

+a+

5x,

where c; and c2 are arbitrary constants. Remembering that y = ve*, we finally have -y=cje*

+coe* + ixe*. 2

Of course, the solution 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? + 1)y =csex.

(10)

The complementary function is Yo = €, coSx + cz sinx. We may use any special case of (11) as the y; in the theory above.

(11) Let us then

put = vsinx. We find that

y’ =v’ sinx + vcosx and

y” =v" sinx + 2v' cosx — vsinx. The equation for v is

v” sinx +2v’cosx =cscx, or v” + 2v’ cotx = csc? x.

(12)

9.2

Reduction ef Order

155

Put v’ = w; then equation (12) becomes w’ + 2wecotx = esc’ x, for which an integrating factor is sin? x. Thus sin’ x dw + 2w sinx cosx dx = dx is exact. From (13) we get w sin? x = x,

and if we seek only a particular solution, we have Ww

= x csc” x,

or , p= # case” 2 x.

Hence 9

v = [ xese? dx, or v=

—xcotx +In|sinx|,

a result obtained using integration by parts. Now y =

vsinx,

so the particular solution which we sought is Vp = —xX cosx + sinx In| sin x]. Finally, the complete solution of (10) is seen to be y =c, cosx +c? sinx — xcosx + sinx In| sin x|.

i Exercises

YS

Use the method of reduction of order to solve the equations in Exercises | through 8.

(D? -l)y=x—-1. (D* —5D + 6)y = 2e*.

So

(D? + l)y = secx.

(D? + 1)y = sec? x. Use y = vsinx.

(D? +2D + l)y = (e* — 1). (D? —-3D+2)y =(1+e*)!”,

3. 4.

(D* —4D+4)y =e". (D*+4)y =sinx.

(13)

156

Chapter 9

Variation of Parameters

9.

Use the substitution y = vcos.x to solve the equation of Example 9.2.

10.

Use y = ve

11.

(D*?+1)y = csc? x. Take a hint from Exercise 6.

to solve the equation of Example 9.1.

12.

Verify that y = e* is a solution of the equation

(x — ly” — xy’ +y=0. Use this fact to find the general solution of

(x — ly" —xy’ +y=1. 13.

Observe that y = x is a particular solution of the equation

2x7y"+xy'—y=0 and find the general solution. For what values of x is the solution valid? 14.

In Chapter 19 we shall study Bessel’s differential equation of index zero

xy’ +y+xy

=0.

Suppose that one solution of this equation is given the name Jo(x). Show that a second solution takes the form

dx

Jotx) | ———F x[Jo(x)] 15.

One solution of the Legendre differential equation

(1—x?)y" — xy’ +2y =0 is y = x. Find a second solution.

Variation of Parameters In Section 9.2 we saw that if y; is a solution of the homogeneous equation

y" + p(x)y' + 4(x)y = 0,

(1)

we Can use it to determine the general solution of the nonhomogeneous equation

y" + pix)y' +¢(x)y = RO). In using the method of reduction of order, y; 1s a solution of (1), the function ¢; y; is also c,. We replaced the constant c, by a function of the existence of a solution of equation (2)

(2)

we proceeded as follows. Because a solution for an arbitrary constant v(x) and considered the possibility of the form v - y;. This led us to a

first-order linear equation in the variable v’ that we were able to solve.

9.3

Variation of Parameters

157

Suppose now that we know the general solution of the homogeneous equation (1). That is, suppose that

(3)

Ye = C1¥1 + C2y2

is a solution of (1), where y; and y> are linearly independent on an interval

a < x .

(5)

Rather than becoming involved with derivatives of A and B of higher order than the first, we now choose some particular function for the expression A’y, + B’y.

Technically, we could let this function be sin x, e*, or any other suitable function. For simplicity we choose

A’y, + B’y2 = 0.

(6)

y" = Ay| + By + A’y, + B'y3.

(7)

It then follows from (5) that Because y was to be a solution of (2), we substitute from (4), (5), and (7) into equation (2) to obtain

Ay + py, +ay1) + Byy + py: + ay2) + A’y, + Blyy = RG). But y; and y2 are solutions of the homogeneous equation (1), so that finally

Aly, + B'y, = R(x).

(8)

Equations (6) and (8) now give us two equations that we wish to solve for A’

and B’. This solution exists provided that the determinant yi yy

2 t

Yo

/

158

Chapter9

Variation of Parameters

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 < 6b. 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 B 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 differential equation involved from having variable coefficients.

EXAMPLE 9.3 Solve the equation

(D? + 1)y = secx tanx.

(9)

Of course, Ve = Cy COSX + €2 Sin x.

Let us seek a particular solution by variation of parameters. Put

y = Acosx + Bsinx,

(10)

from which y =—Asinx + Bcosx + A’cosx + B’ sinx. Next set

A’ cosx + B’sinx = 0,

(11)

so that

y' = —Asinx + Bcosx. Then

y” =—Acosx — Bsinx — A’ sinx + B’cosx. Next we eliminate y by combining

(12)

equations (10) and (12) with the original

equation (9). Thus we get the relation

—A’sinx + B’cosx = sec x tanx.

(13)

From (13) and (11), A’ is easily eliminated. The result is

B’ =tanx, so that B=In|secx|,

(14)

9.3

Variation of Parameters

159

in which the arbitrary constant has been disregarded because we are seeking only a particular solution to add to our previously determined complementary function ye. From equations (13) and (11) it also follows that

A’ = —sinx secx tanx, or A’ = —tan’ x. Then A=

= [ tan? xx

= [a

— 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| sec x|,

or yp = xX cosx —sinx + sinx In| sec x]. Then the general solution of (9) is y =e, cosx +c; sinx +x cosx + sinx In| sec x|,

(16)

where the term (— sin x) in y, 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* —3D+2)y=

l+e%"

Here

ie = tye" + ee™,

(17)

160

Chapter9

Variation of Parameters

so we put

y = Ae* + Be™*,

(18)

Because

y’ = Ae* + 2Be™ + A’e* + Ble™*, we impose the condition

A’e* + Ble* = 0.

(19)

y’ = Ae* +2Be*,

(20)

y” = Ae* + 4Be™ + A’e* + 2B'e™,

(21)

Then

from which it follows that

Combining (18), (20), (21), and the original equation (17), we find that .

.

1

A'e* +2B'e* =

(22)

l1+e>

Elimination of B’ from equations (19) and (22) yields Px

Ag

=-— f

_

1 I+e-*

:

e*

l+ew" Then A=In(l+e™).

Similarly, Be

_—

I

l+e* so that

o= | l+ec

—2x

x= f

—x

et l+e—_ ©

dx,

or

B=-e*+In(1 +e"). Then, from (18),

yp =e" In(l +e") —e* +e" In(l +e).

9.4

Solutionofy"+y= f(x)

The term (—e*) in y, can be absorbed into the complementary function.

161 The

general solution of equation (17) is

y=cj3e" +.ene™ + (e* + e*)In(l+e*).

al Solution of y’ + y = f(x) Consider next the equation

(1)

(D* + ly = 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 function with only a finite number of finite discontinuities on the interval axxy:=vector([cos(x),sin(x)]); y t= [cos(x), sin(x)] >M:=Wronskian(y,x);

M:=

cos(x)

— sin(x)

—sin(x)

cos(x)

Note that Maple uses the word Wronskian for the matrix itself rather than for its determinant. >ApBp:=linsolve(M,

ApBp

:=|-

[0,sec(x)*tan(x)]);

sin(x) sec(x) tan(x)

— cos(x) sec(x) tan(x)

(sin(x))° + (cos(x))*” (sin(x))? + (cos(x))?

Here Maple has found both A’ and B’ simultaneously but has not simplified its results. In the following we have it do that before integrating. >A:=int

(simplify (ApBp[1]),x);

Ai=x-— >B:=int

sin(x) cos(x)

(simplify (ApBp[2]),x);

B := — In(cos(x))

We can now use these values for A and B to construct the general solution. WB Exercises 1.

Use a computer to solve a selection of exercises from the chapter.

a

Applications

Vibration of a Spring Consider a steel spring attached to a support and hanging downward. Within certain elastic limits the spring will obey Hooke’s law: If the spring is stretched or compressed, its change in length will be proportional to the force exerted upon it and, when that force is removed, the spring will return to its original position

with its length and other physical properties unchanged. There is, therefore, associated with each spring a numerical constant, the ratio of the force exerted to the displacement produced by that force. If a force of magnitude Q pounds (1b) stretches the spring c feet (ft), the relation

Q=ke

()

defines the spring constant k in units of pounds per foot (Ib/ft). Let a body B weighing w pounds be attached to the lower end of a spring, and brought to the point of equilibrium where it can remain at rest, as on the left in Figure 10.1. Once the weight B is moved from the point of equilibrium £ as on the right in Figure 10.1, the motion of B will be determined by a differential equation and associated initial conditions. Let f be time measured in seconds after some initial moment when the motion begins. Let x, in feet, be distance measured positive downward (negative upward) from the point of equilibrium, as in Figure 10.1. We assume that the motion of B takes place entirely in a vertical line, so the velocity and acceleration are given by the first and second derivatives of x with respect to f. In addition to the force proportional to displacement (Hooke’s law), there will in general be a retarding force caused by resistance of the medium in which the motion takes place or by friction. We are interested here only in such retarding forces as can be well approximated by a term proportional to the velocity because we restrict our study to problems involving linear differential equations. Such a retarding force will contribute to the total force acting on B a term bx'(t), in

which b is a constant to be determined experimentally for the medium in which the motion takes place. Some common retarding forces, such as one proportional to the cube of the velocity, lead to nonlinear differential equations.

165

166

Chapter 10

Applications

Figure 10.1

The weight of the spring is usually negligible compared to the weight of B, so we use for the mass of our system the weight of B divided by g, the constant acceleration of gravity.

If no forces other than those described above act on the

weight, the displacement x must satisfy the equation

W x"(t) + bx'(t) + kx(t) = 0.

(2)

&

Suppose that an additional vertical force, due to the motion of the support or to the presence of a magnetic field, and so on, is imposed upon the system. The new, impressed force will depend upon time and we may use F'(r) to denote the

acceleration that it alone would impart to the weight of B. Then the impressed force is (w/g) F(t) and equation (2) is replaced by

ZrO +b

ix) = 7 FO.

(3)

At time zero, let the weight be displaced by an amount x9 from the equilibrium point and let the weight be given an initial velocity vp. Either or both xo and vp

may be zero in specific instances. The problem of determining the position of the weight at any time ¢ becomes that of solving the initial value problem consisting

of the differential equation

ar

+ bx'(t) + kx) = 2 F(t)

fort >0,

(4)

and the initial conditions

x(0) = x0,

x'(0) = vo.

(5)

It is convenient to rewrite equation (4) in the form

x"(t) + 2yx'@) + B°x(t) = FO), in which we have put

bg -=2y, Ww

kg —= Ww

6.

(6)

10.2.

Undamped Vibrations

167

We may choose 8 > 0 and we know y = 0. Note that y = 0 corresponds to a

negligible retarding force. A number of special cases of the initial value problem contained in equations (5) and (6) will now be studied.

Undamped Vibrations If y = 0 in the problem of Section 10.1, the differential equation becomes

x(t) + Px) = FC),

(1)

a second-order linear equation with constant coefficients in which

B° =kg/w. The complementary

function

associated

with the homogeneous

equation

x(t) + B2x(t) = Ois Xe = c, sin Bf + c2 cos fr, and the general solution of equation (1) will be of the form x =c,sin Bt +¢2 cos Bt+ Xp,

(2)

where x, is any particular solution of the nonhomogeneous equation. We now look at a number of examples of the motion described by equation (2) for different functions F'(f) in equation (1).

EXAMPLE 10.1 Solve the spring problem with no damping but with F(t) = Asin, where B # w. The case B = a leads to resonance, which will be discussed in Section 10,3. The differential equation of motion is w

“a

—x'(t)tkx(t) g

Ww

:

= — Asinwt 4g

and may be written

x(t) + x(t) = Asinet,

(3)

with the introduction of B* = kg/w. We shall assume initial conditions

x(0) = x0,

x'(Q) = vo.

A particular solution of equation (3) will be of the form X, =

Esinet,

and we may obtain £ by direct substitution into equation (3). We have

—Ew* sinwt + BE sinwt = Asinat,

(4)

168

= Chapter 10

Applications

an equation that is satisfied for all ¢ only if we choose E=

A fp?

ep?

The general solution of (3) now becomes A

x(t) = c, sin Bt + co cos Bt + ———~ p2 — w

sinwt

(5)

with derivative

x'(t) =c;Bcos Bt —c2B sin Bt +

A

p?

a —

COs wt.

wy?

The initial conditions (4) now require that Aw Xy

=

C2

and

vo

=

cB

+ Be — ww

and

c2

and force us to choose Cy

=

vo =

B

Aw Boo

Ce

B(B> —@-)

=

Xo.

From (5) it follows at once that U x(t) = zon

Bt + x9 cos Bt —Ae

The x of (6) has two parts.

Aw ay

. SP

A ae

.

6 (

)

The first two terms represent the natural simple

harmonic component of the motion, a motion that would be present if A were zero. The last two terms in (6) are caused by the presence of the external force

(w/g)A sine.

EXAMPLE 10.2 A spring is such that it would be stretched 6 inches (in.) by a 12-lb weight. Let the weight be attached to a spring and pulled down 4 in. below the equilibrium

point. If the weight is started with an upward velocity of 2 ft/sec, describe the motion. No damping or impressed force is present. We know that the acceleration of gravity enters our work in the expression for the mass. We wish to use the value g = 32 feet per second per second (ft/sec*) and we must use consistent units, so we put all lengths into feet.

First we determine the spring constant & from the fact that the 12-lb weight

stretches the spring 6 in., 4 ft. Thus 12 = $k, so that k = 24 Ib/ft. The differential equation of the motion is therefore

Sx") + 24x(1) = 0.

(7)

10.3°

Resonance

169

At time zero the weight is 4 in. GG ft) below the equilibrium point, so x(0) = ; The initial velocity is negative (upward), so x'(0) = —2. Thus our problem is

that of solving

x(t) + 64x(t) =0; = x(0) = §, x’) = -2.

(8)

The general solution of equation (8) is x(t) = c, sin 8f + co cos 8r,

from which x'(t) = 8c) cos 8f — 8c sin 8F. The initial conditions now require that i =c,

and

—2=

8c),

so that finally,

x(t)= —j sin 8 + 5 cos 8.

(9)

A detailed study of the motion is straightforward once (9) has been obtained. The amplitude of the motion is

¥ G4)? + G? = 3: that is, the weight oscillates between points 5 in. above and below E. The period el 1S qu

sec.

Resonance In Example 10.1 of Section 10.2 we postponed the study of the special case, 6 =a. In that case, the differential equation to be solved is

x(t) + B’x(t) = Asin Bt,

(1)

where we had let 6? = kg/w. The complementary x(t) + B?x(t) solution x, will The method ticular solution

function

associated

with the homogeneous

equation

= 0 will be the same as it was before, but the previous particular not exist because 6 = ow. of undetermined coefficients may be applied here to seek a parof the form

X» = Prsin Bt + Otcos Bt, where P and Q are constants to be determined.

(2)

Direct substitution of the x, of

(2) into equation (1) yields 2P6 cos Bt — 208 sin Bt = Asin Pr,

170

Chapter 10

Applications

an equation that can be satisfied for all ¢ only if P = 0 and Q = —A/28. Thus —At Xp = ap

COS Ph

(3)

and the general solution of (1) is

At

x(t) = c; sin Bt + co cos Bt — —cos pt,

(4)

2B

from which we obtain

, . At , A x(t) =c)P cos Bt — coB sin Bt + ry sin Br — op cos Br. The initial conditions x(0) = xo and x'(0) = vo now force us to take C2 = Xo

and

qg=

Uo A B + 2p

The final solution may now be written

x(t) = xocos Bt + — sin Br + A sin Bt — Btcos Bt). B

2p

(5)

That (5) satisfies the initial value problem is readily verified. In the solution (5) the terms proportional to cos fr and sin fr are bounded, but

the term with 6f cos Bt can be made as large as we wish by proper choice of f. This building up of large amplitudes in the vibration is called resonance. i Exercises

1.

A spring is such that a 5-lb weight stretches attached, the spring reaches equilibrium, then in. below the equilibrium point and started off ft/sec. Find an equation giving the position of

it 6 the with the

in. The 5-lb weight is weight is pulled down 3 an upward velocity of 6 weight at all subsequent

times.

2.

3,

A spring is stretched 1.5 in. by a 2-lb weight. Let the weight be pushed up

3 in. above E and then released. Describe the motion. For the spring and weight of Exercise 2, let the weight be pulled down 4 in. below £ and given a downward initial velocity of 8 ft/sec. Describe the motion.

4.

Show that the answer to Exercise 3 can be written x = 0.60 sin (16r + @), where @ = arctan 5

5.

A spring is such that a 4-lb weight stretches it 6 in.

An impressed force

; cos 8f is acting on the spring. If the 4-lb weight is started from the equilibrium point with an imparted upward velocity of 4 ft/sec, determine the position of the weight as a function of time.

10.3

Resonance

171

A spring is such that it is stretched 6 in. by a 12-lb weight. The 12-Ib weight is pulled down 3 in. below the equilibrium point and then released. If there is an impressed force of magnitude 9 sin 41 lb, describe the motion. Assume that the impressed force acts downward for very small f. Show that the answer to Exercise 6 can be written i=

10.

12. 13,

j

2cos (8f + 2/4) + + sin 4.

A spring is such that a 2-lb weight stretches it 5 ft. An impressed force t sin 8r is acting upon the spring. If the 2-lb weight is released from a point 3 in. below the equilibrium point, determine the equation of motion. For the motion of Exercise 8, find the first four times at which stops occur and find the position at each stop. Determine the appropriate position to be expected if nothing such as breakage interferes, at the time of the 65th stop, when ¢ = 87 (sec), in Exercise 8. A spring is such that a 16-Ib weight stretches it 1.5 in. The weight is pulled down to a point 4 in. below the equilibrium point and given an initial downward velocity of 4 ft/sec. An impressed force of 360 cos 4 Ib is applied. Find the position and velocity of the weight at time t = 7/8 sec. A spring is stretched 3 in. by a 5-lb weight. Let the weight be started from E with an upward velocity of 12 ft/sec. Describe the motion. For the spring and weight of Exercise 12, let the weight be pulled down 4 in. below E and then given an upward velocity of 8 ft/sec. Describe the motion.

14. I5.

16.

Vhs

Find the amplitude of the motion of Exercise 13. A 20-lb weight stretches a certain spring 10 in. Let the spring first be compressed 4 in., and then the 20-lb weight attached and given an initial downward velocity of 8 ft/sec. Find how far the weight would drop. A spring is such that an 8-lb weight would stretch it 6 in. Let a 4-lb weight be attached to the spring, which is then pushed up 2 in. above its equilibrium point and released. Describe the motion. If the 4-Ib weight of Exercise 16 started at the same point, 2 in. above E, but with an upward velocity of 15 ft/sec, when will the weight reach its lowest point?

18.

A spring is such that it is stretched 4 in. by a 10-Ilb weight. Suppose the 10-Ib weight is to be pulled down 5 in. below E and then given a downward velocity of 15 ft/sec. Describe the motion.

19:

A spring is such that it is stretched 4 in. by an 8-lb weight. Suppose the weight is to be pulled down 6 in. below E and then given an upward velocity of 8 ft/sec. Describe the motion.

172 = Chapter 10

Applications

20.

Show that the answer to Exercise 19 can be written x = 0.96 cos (9.87 + ),

21.

where @ = arctan 1.64. A spring is such that a 4-lb weight stretches it 6 in. The 4-Ib weight is attached to the vertical spring and reaches its equilibrium point. The weight is then (¢ =

22.

23.

3 in. and released.

0) drawn downward

There is a simple

harmonic exterior force equal to sin 8¢ impressed upon the whole system. Find the time for each of the first four stops following t = 0. Put the stops in chronological order. A spring is stretched 1.5 in. by a 4-Ib weight. Let the weight be pulled down 3 in, below equilibrium and released. If there is an impressed force 8 sin 161 acting upon the spring, describe the motion. For the motion of Exercise 22, find the first four times at which stops occur and find the position at each stop.

Damped Vibrations In the general linear spring problem of Section 10.1, we were confronted with

x(t) + 2yx"(t) + B’x(t) = FO); in which 2y = bg/w and B? =kg/w,

x) = x0, x’) = v0,

(1)

B > 0. The auxiliary equation

m* + 2ym + p* =0

has roots —y + ./y? — B? and we see that the nature of the complementary function depends upon whether 6 > y, B = y, or B < y.

IfB > y, Bp? —y* > 0, so let us put

©)

py? =8. Then the general solution of (1) will be

x(t) =e “' (ce; cos dt + cz sin dt) + W(t),

(3)

in which y(t) is any particular solution of equation (1). The presence of the function e~”, called a damping factor, will cause the natural part of the solution, that is, the part independent of the external force (w/g)F (1), to approach zero as t —> oo. If in (1) we have B = y, the two roots of the auxiliary equation are equal and the general solution becomes

x(t) =e" (cy + cat) + ya(t),

(4)

in which w(t) is a particular solution of (1). Again the natural component has

the damping factor e~”’ in it.

If in (1) we have B < y and y? — B? > 0, then we can set

y? — Bp =o’,

o > 0.

(5)

10.4.

Damped Vibrations

173

Critically damped

Figure 10.2

Since a < y, the two roots of the auxiliary equation are both real and negative,

and we have

x(t) = cye PO! + eye

-O™ + aba (2),

(6)

Again w3(t) is a particular solution of (1), and we see that the damping factor e~”' causes the natural component of (6) to approach zero as f +

oo.

Suppose for a moment that we have F(t) = 0, so the natural component of the motion is all that is under consideration. If 8 > y, equation (3) holds and the motion is a damped oscillatory one. If 6 = y, equation (4) holds and the motion

is not oscillatory; it is called critically damped motion. If B < y, (6) holds and the motion is said to be overdamped; the parameter y is larger than it needs to be to remove the oscillations. Figure 10.2 shows a representative graph of each type of motion mentioned in this paragraph: a damped oscillatory motion, a critically damped motion, and an overdamped motion. EXAMPLE 10.3 Solve the problem in Example 10.2 of Section 10.2, with an added damping force of magnitude ().6|v|. Such a damping force can be realized by immersing the body B in a thick liquid.

The initial value problem to be solved is

Bx(t) +0.6x'(t) + 24x01) =0;

xO) =4, x0) =-2.

The auxiliary equation of (7) may be written

m + 1.6m + 64 = 0, an equation that has roots —0.8 + \/63.36i. Therefore, the general solution of (7) is x(t) =e

—0.81

(c; cos 8.0¢ + co sin 8.0f)

Chapter 10

Applications *

174

i a [\

\i a

Figure 10.3

and

x'(t) = e 8" [(—8e, — 0.8c2) sin 8.0f + (8c2 — 0.8¢)) cos 8.04].

The initial conditions in (7) now give us

tac,

and

—2=8c,—08c,

so that c) = 0.33 and cz = —0.22. Therefore, the desired solution is

(8)

x(t) = exp (—0.81) (0.33 cos 8.07 — 0.22 sin 8.0F), a portion of its graph being shown in Figure 10.3.

Exercises 1.

Accertain straight-line motion is determined by the differential equation dx

qt

dx

Yat

169x = 0

and the conditions that when t = 0, x = 0, and v = 8 ft/sec.

(a) Find the

value of y that leads to critical damping, determine x in terms of t, and draw a graph for 0 < t < 0.2. (b) Use y = 12. Find x in terms of ¢ and draw the

2.

3.

graph. (c) Use y = 14. Find x in terms of t and draw the graph. A spring is such that a 2-lb weight stretches it 5 ft. An impressed force i sin 8¢ and a damping force of magnitude |v| are both acting on the spring. The weight starts 7 ft below the equilibrium point with an imparted upward velocity of 3 ft/sec. Find a formula for the position of the weight at time f. A spring is such that a 4-Ib weight stretches it 0.64 ft. The 4-Ib weight is pushed up 5 ft above the point of equilibrium and then started with a downward velocity of 5 ft/sec. The motion takes place in a medium which furnishes a damping force of magnitude + |u| at all times. Find the equation describing the position of the weight at time f.

10.4

Damped Vibrations

175

A spring is such that a 4-Ib weight stetches it 0.32 ft. The weight is attached to the spring and moves in a medium that furnishes a damping force of magnitude 5 lv]. The weight is drawn down 5 ft below the equilibrium point and given an initial upward velocity of 4 ft/sec. Find the position of the weight thereafter. A spring is such that a 4-lb weight stretches the spring 0.4 ft. The 4-Ib weight is attached to the spring (suspended from a fixed support) and the system is allowed to reach equilibrium. Then the weight is started from equilibrium position with an imparted upward velocity of 2 ft/sec. Assume that the motion takes place in a medium that furnishes a retarding force of magnitude numerically equal to the speed, in feet per second, of the moving weight. Determine the position of the weight as a function of time. A spring is stretched 6 in. by a 3-lb weight. The 3-Ib weight is attached to the spring and then started from the equilibrium with an imparted upward velocity of 12 ft/sec. Air resistance furnishes a retarding force equal in magnitude to 0.03 |v|. Find the equation of motion. A spring is such that a 2-lb weight stretches it 6 in. There is a damping force present, with magnitude the same as the magnitude of the velocity. An impressed force (2 sin 8f) is acting on the spring. If at t = 0, the weight is released from a point 3 in. below the equilibrium point, find its position for ¢ > 0. A spring is stretched 10 in. by a 4-lb weight. The weight is started 6 in. below the equilibrium point with an upward velocity of 8 ft/sec. If a resisting medium furnishes a retarding force of magnitude flv |, describe the motion. For Exercise 8, find the times of the first three stops and the position (to the nearest inch) of the weight at each stop. 10.

A spring is stretched 4 in. by a 2-Ib weight. The 2-lb weight is started from the equilibrium point with a downward velocity of 12 ft/sec. If air resistance furnishes a retarding force of magnitude 0.02 of the velocity, describe the motion.

II.

For Exercise 10, find how long it takes the damping factor to drop to onetenth of its initial value.

12.

For Exercise 10, find the position of the weight at (a) the first stop and (b) the second stop.

Let the motion of Exercise 8 of Section 10.3, be retarded by a damping force of magnitude 0).6|v|. Find the equation of motion. Show that whenever ¢ > | (sec), the solution of Exercise 13 can be replaced (to the nearest 0.01 ft) by x = —0.05 cos 8r. Let the motion of Exercise 8 of Section 10.3, be retarded by a damping force of magnitude |v|. Find the equation of motion and also determine its form (to the nearest 0.01 ft) fort > I (sec),

176

= Chapter 10

Applications

16.

Let the motion of Exercise 8 of Section 10.3, be retarded by a damping force of magnitude 2|v|. Find the equation of motion.

Tes

Alter Exercise 6 of Section 10.3, by inserting a damping force of magnitude one-half that of the velocity and then determine x.

18.

A spring is stretched 6 in. by a 4-Ib weight. Let the weight be pulled down 6 in. below equilibrium and given an initial upward velocity of 7 ft/sec. Assuming a damping force twice the magnitude of the velocity, describe the motion and sketch the graph at intervals of 0.05 sec for0 < ¢ < 0.3

19.

An object weighing w Ib is dropped from a height h ft above the earth. At time t (sec) after the object is dropped, let its distance from the starting point be x (ft), measured positive downward. Assuming air resistance to be negligible, show that x must satisfy the equation w d?x

(sec).

gd 20.

as long as x < hf. Find x. Let the weight of Exercise 19 be given an initial velocity vp. Let v be the velocity at time ¢. Determine v and x.

Zi,

From

the results

in Exercise

20,

a relation

find

that

does

not contain

t

explicitly. 22.

If air resistance furnishes an additional force proportional to the velocity in the motion studied in Exercises 19 and 20, show that the equation of motion becomes

5

w d-x ——;

dx ob

g dt

24,

—

dt

= WwW.

(A)

Solve equation (A) given the conditions r = 0, x = 0, and v = up. Use a = be/w. To compare the results of Exercises 20 and 22 when a = bg/w is small, use the power series for e~“' in the answer for Exercise 22 and discard all terms involving a” for n > 3. The equation of motion of the vertical fall of a man with a parachute may be roughly approximated by equation (A) of Exercise 22. Suppose that a 180-Ib man drops from a great height and attains a velocity of 20 miles per hour (mph) after a long time. Determine the implied coefficient b of equation (A).

2D.

A particle is moving along the x-axis according to the law

5 Mt Oe dt? aT

952

seen on

2

fdu\’

ut dw

\ do}

Substituting the last two expressions into (3) and simplifying yields a linear differential equation with constant coefficients that we know how to solve, du qq

rM 1"

(4)

The general solution of equation (4) is TM

u = b,cos@ + bo sin@ + — c

To simplify the final polar equation of the orbit further, we now choose the direction of the polar axis so that r is a minimum when @ = 0. Since u is the reciprocal of r, we want u to be a maximum when 6 = 0. But this condition requires that du/d@ = 0 and d*u/dé? < 0 when @ = 0. Since du — dé

. = —b, siné + by cosé

and

du do2 _ —=_

=

_b 1

COS

cos6 — by 2 sind —

09

$1

,

we require that b) = 0 and b; > 0. Thus u=b,cosd+

rTM a Cc

and

I

c?/TM

= [M/c? + b, cosé

— fire (b}c?/ TM) cos@’

where all the constants are positive. For simplicity we let byc?

é

= — CM

and

B=

]

— b

(5)

182

Chapter 10

Applications to obtain Be

"T+

(6)

ecosé’

where e and B are positive constants.

Equation (6) is the equation in polar coordinates of the orbit of the planet where the sun is at the pole. It is also the standard equation in polar coordinates of a conic section with its focus at the pole and its directrix perpendicular to the polar axis. The number ¢ is called the eccentricity of the conic. If we presume that the orbit of the planet is bounded, then the conic must be an ellipse and 0 < e < 1. We have shown that the orbit of a planet is an ellipse with focus at the center

of mass of the sun. This is Kepler’s first law.

Kepler's Third Law It is possible to transform the polar equation r=

Be

1+ecosé@ into rectangular coordinates and obtain the equation of the elliptical orbit in the form (x+hy y? —a

+palh

(1)

where

Be? L—#

Be 7 or

=DEplot

(diff (x(t)

,t$2)+(8/5)*

(diff (x(t),t))+64*x=0,

(1/3) ,-2]},stepsize=.01); x(t),0..2,{[0,

will produce Figure 10.3. The idea of resonance discussed in Section 10.3 is illustrated by letting the constants w/g = 1,k =1,A = l,andw = 1,so8 =a. If we use x(0) = 0 and x’(O) = 0 as our initial conditions, Maple plots the solution with the command, SDESlot(dLff£({x(t),t$2)+x2sin (t), x(t),-1..120,{[0,0,0]},stepsize=.5);

the result of which is shown in Figure 10.5. i Exercises 1.

Use a computer to produce a graph of underdamped vibrations as shown above.

2,

Vary the constants to show overdamped vibrations.

3.

Vary the constants to show critically damped vibrations.

10.10

Computer Supplement

185

Figure 10.5

Use a computer to explore a variation of the idea of resonance. Rather than letting B = w, we look at the case where 6 — w is small. In this case, we see a phenomenon known as beats.

Gy =) gy

Use a computer to plot the resonance equation used above. Modify the equation so that w = 0.9. What can you say about the nature of the graph? How small does 8 — w need to be to produce beats?

BD Linear Systems of Equations

11.1

Introduction We shall see in the next chapter that certain problems in the studies of electrical circuits and arms races lead naturally to systems of linear differential equations with constant coefficients. Although the subject of systems of equations can be studied in a wider context involving coefficients that are not constant, we shall not do so in this book.

First-Order Systems with Constant Coefficients In the following sections we show how matrix algebra can be used to reduce the problem of solving systems of differential equations to an algebraic routine. Before we do that it is important to realize that a system of linear equations of order higher than the first can be written in terms of a first-order system. Consider, for example, the single equation

y"+2y'—yse".

(1)

If we let u = y’, then equation (1) becomes uo =y—Qut+e*. In other words, the single second-order equation (1) has been replaced by the first-order system

y =u, u’=y—2ut+e*.

2)

In a similar manner, the third-order equation

y" + 2y" —y +3y =x can be written as a system of first-order equations by choosing new variables

ge

186

and

vor =v.

(3)

11.3

Solution of a First-Order System

187

Then equation (3) becomes v =—2v+u—3y+x,

and we can consider the first-order system

y =u, i = %, v =u—2v—3y4+x,

(4)

as being equivalent to (3). The system of second-order equations y" -y+5v' 2y’ —y"

+4

=x, =

2,

(5)

can be replaced by a first-order system if we let w = v' and w = y’, so that ul = 4u + 2w — 2, y =u,

w =—Su+ytx,

(8)

y =w. @ Exercises In Exercises | through 5, replace the given equation by a system of first-order equations.

1. y”—6y'+8y =x 42. 2. y”+4y'+4y =e". 5,

3. 4.

y+ py’ +ay = f(). y+ py" +qy' +ry = f(x).

y¥O—y=0.

In Exercises 6 through 9, replace the given system by an equivalent system of first-order equations.

6.

vi —2vu+2w’ = 2 —4e”, 2v' —3u+3w’ —w=0.

7. 8. 9,

3D+2)u+(D — 6)w = 5e*, (4D + 2)v + (D — 8)w = Se* + 2x — 3. (D —2)u = 2. )y+ =0, (D+2)y+ Du4+6 (D? D?y—(2D—-1)v=1, (2D+1)y+ (D? —4)v =0.

Solution of a First-Order System Consider the first-order system dx —=y, dt

ay

(1)

188

Chapter I]

Linear Systems of Equations

We can rewrite this system in the form Dx—-—y=0,

2x + (D—3)y =0.

2)

Operating on the first equation with the operator D — 3 and adding the two equations eliminates the variable y to give

(D* —-3D+2)x =0.

(3)

In a similar manner, we can eliminate x from the system (2) to obtain

(D? —3D +2)y =0.

(4)

Thus we realize that the solutions of equations (1) are of the form wot

2t

a

xXx=ce"+oe,

y = c3e7 + eye’, where there are some relations among the four constants ¢;, c2, C3, ca that we can determine by substitution back into equations (1). An alternative way of viewing the nature of the solutions of system (1) is to expect from the beginning the existence of solutions of the form X=

ce

mae ,

y=coe”", me

(5)

where the constants c), c2, and m must be determined by substitution into (1). If we do this we find cyme”"™

me

= Ee

mt

and

come™ = —2c;e™ + 3c.e™, or mc,

+co = 0,

—2c; + (3 -—m)c2 = 0.

(6)

The system (6) can have nontrivial solutions for c; and c> only if the determinant —m —2

| 3-m

(7)

11.4

Some Matrix Algebra

189

is zero. That is,

m — 3m +2 =(m—

1)(m — 2) = 0.

Moreover, for the choice m = 1, the system (6) yields the condition c, = ¢}, and form = 2 we would be forced to take cr = 2c). Thus there would be two distinct

solutions of the form of equations (5), namely x=cye,

‘

(8)

21 Ge";

(0)

ySece

and X=

y = 2c\e",

A careful perusal of what we have done here should lead one to suspect that the elementary algebra problem of finding nontrivial solutions to the system (6) holds the entire key to the problem of solving the system of differential equations (1). The formalization of this procedure is best accomplished with the assistance of some vector and matrix notation. In the next section we summarize the minimum amount of matrix algebra required for our purposes.

Some Matrix Algebra We shall presume at this point that the student is familiar with the elementary calculus of vector functions. definition

The basic ideas involved are deduced

Bil pom 4 £ (fie

fale.

_ (ahi df, fu)

= ( dt’

dhedt )

” numbers, by vectors of and from the properties of vector addition, multiplication and the scalar product of vectors. Most elementary calculus texts and courses provide all of the algebra required. It is not always the case that students who have completed a course in elementary calculus are familiar with matrix algebra. We therefore include here a brief introduction suitable for our purposes. A matrix is a rectangular array of numbers.

dt’ '°

from the

For example, the following arrays

are matrices: 2

1

1 3

3

2

5 al

12

1

3 1 2

Each of the numbers in a matrix is called an element of that matrix.

190

Chapter1]

Linear Systems of Equations

A matrix is said to be of dimension n x m(n by m) if it has n rows and m columns.

Thus the general 7 x m matrix can be written ay)

G@]2

"ts

Aim

a2)

422

+++

Gam ;

Gn)

Gn2

+++

Anm

where aj; indicates the number in the ith row and jth column. Two matrices of the same dimension are said to be equal if the corresponding

elements are equal. Addition is defined only for two matrices of the same dimension, and is defined elementwise. For example, the sum of two 2 x 3 matrices is accomplished as

follows: abe de f

(8 jk

h

i\_ fat+g Up \d+j

b+h e+k

er+i fi)’

(2)

Any matrix may be multiplied by a number by multiplying each of its elements by that number. For example, a kle

b ka d|)=lke if ke

kb kd]. kf

(3)

The student should recognize that the algebra of matrices, which is dictated by the definitions in the preceding paragraphs, is essentially the same kind of algebra as the algebra of vectors. This comes from the fact that the operations of addition and multiplication by numbers are done elementwise for both the matrix algebra and vector algebra. Indeed, one can regard the vector (a, b, c) asa l x 3

matrix and then interpret the usual vector algebra as a special case of the matrix algebra described above. The only caution to be observed is that the matrices (a

b

c)

a b

and

Cc

are very different from the matrix point of view, although we may wish to identify them as the same vector in some physical or geometrical context. We shall call the first vector a row vector and the second a column vector. In what follows we shall often form the product of a row vector, a matrix of dimension | x 1, with a column vector, a matrix of dimension x 1. The product

is the familiar scalar product of elementary calculus

by by (a

(a

An)

.

Dy

= ab,

+ agby + +++ + aybn,

(4)

11.4

Some Matrix Algebra

191

with the insistence that the row vector always be written on the left and the column vector on the right. The product of two matrices can now be defined in terms of scalar products of row and column vectors. Ann x m matrix and a p x q matrix can be multiplied only if m = p; that is, the number of columns in the first matrix must equal the number of rows in the second matrix. The resulting product matrix has dimension n x q. The definition is most easily described as follows: A,B! A>: B!

A,-B? Ay-B*

«+» ++»

A, Bd Ay: BY

A,

- B!

A,

+ B?

(5)

:

.

A. B= se

Ag

«BA

where A; is the ith row vector of the matrix A and B/ is the jth column vector of the matrix B. Thus the element in the ith row and jth column of the product is the ordinary scalar product of the vectors A; and Bi, A few examples will help fix these definitions in our minds.

EXAMPLE 11.1 21

(4 )=2 )-Ci + 2-005); (0 5

4\_

2

2 1

2

1

EXAMPLE

™

a

oo

rw

hE

™

ll

ay Se

se

oN

“to

‘Se +

ll

t? 7

aS

ren

+t

ge 03

EXAMPLE 11.2

11.3

(i 3)()=(45) EXAMPLE

11.4 a e

b\ @

(Pp

q\_ ap+br rs} \cp+dr

aq+hbs eq+ds}°

192

Chapter 11

Linear Systems of Equations

EXAMPLE

11.5

(i =i) EXAMPLE

1)

2-34+1-4 1-3+(-1)-:4

2:14+1-1 1-14 (-1):1

(4 4):

11.6

The product 12

|

1 3)°)2 ! 3

I

is not defined because the number of columns in the first matrix is two and the number of rows in the second matrix is not two. On the other hand,

1

2 5

3 1

49

2 1)-(1 3)=[3 7

is defined, a fact that demonstrates that matrix multiplication is not commutative.

In elementary vector algebra we often find it convenient to designate a vector by a single symbol instead of expressing the vector in terms of its components.

We shall frequently use a single symbol to refer to a matrix, Particular caution must be observed when using such abbreviations to be sure that the dimensions of the objects involved are appropriate in terms of the definitions in our algebra. For example, the equation ge dt

=X" = AX

can be interpreted as a system of first-order linear equations with constant coef-

ficients if we interpret X as an n-dimensional column vector function of ¢ and A ann

x mn matrix of real numbers. Thus d

—

dy

= dxty,

—_—==x

dt

”

can be written

X’= AX,

;

[1.4

Some Matrix Algebra

193

where

¥= (aia)

= G)

a=(i a) (2

,__ (dx/dt

[x

_

1

We shall use capital letters here for matrices and lowercase letters for numbers. In particular, we shall use a large zero for a matrix, all of whose elements are zero wherever the dimension of that zero matrix is clear from the context. 11.7

Me

—

A=

Wh

EXAMPLE If

I I |

and X’— AX = O, then X must be a three-dimensional column vector function and O must be the three-dimensional column vector

0 0 0)

We shall single out one further kind of matrix for special attention. multiply any 2 x 2 matrix A by the matrix

If we

li) 1 0at we clearly obtain the results AI=A

and

IA=A.

A similar observation can be made for any 1 x n matrix provided that / is interpreted as the n x n matrix modeled after the two-dimensional case, namely I = (aij)

where

ajj =O,

ifi # j,

and

ajj

ifi

= 1,

= J.

Again, we shall use the symbol J wherever the dimension is clear from the context. The algebraic structure that one obtains as a consequence of the definitions above is called the algebra of matrices. Some of the basic theorems of this algebra are listed below. Most are easily proved in cases where the dimensions are low and rather tedious for large matrices. We shall ask the student to prove some theorems in the exercises that follow. In each theorem it is presumed that

194

Chapter II

Linear Systems of Equations

the matrices involved have dimensions for which the operations indicated are defined. (A+ B)+C=A+4+(B+C).

(6)

A+B=B+A.

(7)

A+O=A.

(8)

A+(-lDA=0O.

(9)

(AB)C = A(BC)

(10)

A(B+C)=AB+AC.

(11)

IA=AI=A.

(12)

kK(A+ B)=kA+KB.

(13)

kO=

@,

(14)

kK(AB) = (kKA)B = A(KB),

(15)

As a final theorem for this section, we state, without proof, a basic theorem from elementary algebra that we have often used in previous chapters.

Theorem

11.1

Let A be ann xn matrix of constant real numbers and let X be an n-dimensional column vector. The system of equations

AX=0O has nontrivial solutions, is zero.

that is, X

~

O, if and only if the determinant of A

Exercises In Exercises | through 8, find the matrix requested given the following matrices:

(3) G9) b>

1

A+2B.

2. 3. 4. 9.

2A+C. AB+2I. AC+ Bl. Prove that D + E = E + D is defined.

5.

C21,

6. AB+C, 7. AC —B. 8 ABand BA. for any matrices D and E for which the sum

10.

Prove the distributive law of equation (11) for 2 x 2 matrices.

11. 12.

Prove the theorem of equation (8). Prove the theorem of equation (9).

11.5.

First-Order Systems Revisited

13.

Prove the theorem of equation (13).

14. 15.

Prove the theorem of equation (15) for 2 x 2 matrices. Show that the system of equations

95

(; )()=9 |

2

7X

has no nontrivial solutions.

a C)O-°

eEDO-2

Find all the solutions of the systems in Exercises 16 through 21.

18.

19,

22

1

x y|/=O Zz

1-1 1 2 Oo -1

2 3 |

x y| Zz

23,

2 -1

=O

-1 5 1 1

x y| z

1 3 1 2 14

2

1 1 2

1 1 2

1 1

1

x y| z

l

=O

=O

In Exercises 22 through 25, write the given system of differential equations as a matrix equation.

1 24, =dt = 2x +3y,

26, Sda =2x-yte, .

dy

dt 25.

dy

dt

ae

dx af

y ay — —_—= x+2y-2+ at

A

dt

11.5]

as

.

—=tx+yt+z+sint,

27,

—=x-yrtett,

Bh ab

dx ai

y 9 —e=rxt+tyt+l, + x+ty A

l,

aa

+ete

dt

Be 4 wipe

ates

First-Order Systems Revisited We return now to the consideration of first-order linear systems of equations with constant coefficients. Let X be an n-dimensional column vector function of t and let A be ann x n matrix of real numbers. Further suppose that B(r) is a column vector whose components are known functions of tf. Then the vector equation X'=AX+B

(1)

196

9 Chapter I!

Linear Systems of Equations

represents a system of n equations in the n unknown component functions of X. If B =

O, we say that the system is homogeneous.

For the present we restrict

our attention to homogeneous systems and return to nonhomogeneous systems in the next chapter. From our past experience we have reason to believe that the homogeneous system

X'’=AX

(2)

may have solutions of the form xX

me

Ce”,

(3)

where C is a constant vector and #7 is some number that we wish to determine. Substitution of X into system (2) yields

Cme™

= ACe™,

which can be written

(AC —mC)e™

= O.

We can rewrite this equation again, remembering that C = /C, in the form (A—ml)Ce™

=O.

(4)

Equation (4) is to be satisfied for all real values of ¢t, a condition that can be satisfied only if

(A—mI)C =O.

(5)

The theorem at the end of the preceding section states that the algebraic system (5) has nontrivial solutions only if the determinant of A — mJ is zero; that is,

|A —ml|

=0.

(6)

Equation (6) is a polynomial equation of degree n in the unknown number m. We observe that the polynomial |A — m/| depends only on the matrix A. The polynomial |A — m/| is called the characteristic polynomial of the matrix A and equation (6) is called the characteristic equation

of the matrix A.

The roots of the characteristic equation of A are called eigenvalues of the matrix A.

A nonzero vector C;, which

is a solution of equation (5) for a particular

eigenvalue my, is called an eigenvector of the matrix A corresponding to the eigenvalue m,. Thus we see that the first step in solving the homogeneous system (2) is to find the eigenvalues and the corresponding eigenvectors of the matrix A. We might also suspect from our experience with the roots of the auxiliary equation in Chapter 7 that the nature of the solutions of equation (3) will depend

11.5

First-Order Systems Revisited

197

on whether the eigenvalues are real and distinct, complex, or repeated. We shall deal with these cases separately. EXAMPLE 11.8 Let us now reconsider the system of equations of Section 11.3. In matrix notation we have X

1 = AX

; _ 0 where A = &

| :) ‘

(7)

The characteristic equation of A is

(2 3)-(0 ‘)/-

ae

3

m[ =m

3m

+2=0.

The eigenvalues are distinct real numbers m; = 1 and m2 = 2. For m, = 1, equation (5) becomes

Ch (=o

e-(2)=()=0()

so that —c; + c2 = 0 and we have

Thus corresponding to the eigenvalue m, = 1, there is a set of eigenvectors whose elements are scalar multiples of the column vector (1): Similarly, for m2 = 2 equation (5) is

(3 i)(a)=° —2

|

ey)

c=(2)-(2) 900)

so that —2c, +c. = 0 and

—

fer)

_

cy\

]

That is, the eigenvectors are multiples of the vector (3).

Thus we have obtained two distinct sets of solutions for system (7), AGS by

(1)

e

and

xX1=

bo

(3)

en,

(8)

198

Chapter 11

Linear Systems of Equations

where }, and bz are arbitrary constants. It is a simple matter to verify that these vector functions are solutions of system (7). It is quite another matter to make the claim that every solution of system (7) is some combination of the two solutions we have found. The fact that this is true requires the support of several important

definitions and theorems. The student should observe that these definitions and theorems closely parallel the theoretical development in Chapter 6, We shall state the appropriate theorems without proof. A set of m constant vectors of dimension n, {X1,

is linearly independent

X2,

oy

Xm};

if eX,

+

2X9

ape

+ Cm Xin

=O

implies that cy) = c2 = ++: =c, = 0. A set of vector functions of f,

{X1 (7); Xo), os, XO}; is linearly independent on an interval a < ¢ < bif

cy X(t) + cgX2(t) + +++ + emXm(t) = O for all ¢ on the interval implies that Cp =o

Theorem

11.2

S++:

= cy, = 0.

/f X\(t),..., Xm(t) are each solutions ofa homogeneous linear system X' = AX, then eX \(t) + e2X2(t) + +++ + Cm Xm(t) is a solution of the same system for arbitrary constants C),.... Cm-

Theorem

11.3

JfAisannxn matrix of real numbers and{X,,..., X,}isalinearly independent set of solutions of the system X' = AX on the intervala < t < b, then any solution of the system is a unique linear combination of the set {X,,..., Xn). If the set of n-dimensional column vectors {X)(f),..., X,(t}} is considered as ann x m matrix

(X1(t)

X2(t)

«++

Xp(t)),

---

X,(t)|

then the determinant

|X, (t) Xo2(t)

is called the Wronskian of the set of vectors.

11.5

Theorem

11.4

The set of n-dimensional column

First-Order Systems Revisited

199

vectors {X\(t),..., Xn (t)} is linearly inde-

pendent at t = to if, and only if, the Wronskian of the set is not zero at t = to; that is,

W{X (to), Xa(to),..-, Xn(to)} = |X1 Mo)

Xo(to) +++ Xn (to)| #0.

Theorem 11.5 [f the vector functions X\(t),..., X(t) are solutions of the system X' = AX for all t on the interval a < t < b where A is ann x n matrix, then the set {X,(t),..., Xn(t)} is linearly independent on a < t < b if, and only if, W{X (to), X2(to), ..., Xn (to)} 4 0 for some to on the interval a < t < b.

Theorem

11.6

If X,(t),...,Xn(¢) are linearly independent solutions of the n-dimensional homogeneous system X' = AX on the intervala < t < band if X(t) is any solution of the nonhomogeneous system X' = AX + B(t) on the interval a = | —3 | e* as a second solution to the system (10).

—3 Finally, choosing m3 = —2 from equation (11), we have 3

-l1

-!l

Cc]

0

3

3

co)=O,

0

3

3

C3

0 1 | and the solution

which yields the eigenvector} —1

0 x34=

tle —1

To establish the linear independence of the three solutions X;, X2, X3, we compute the Wronskian; that is,

W{X1(t), X2(t), X30} Because

W

1 =|0

2 -3 _

is never zero, it follows from Theorem

0 lle* = Ge", 11.5 that the solutions are

linearly independent on any interval. Thus the general solution of system (10) is

X=

I 2 [Oli tal—3]}e +e] 0

-3

0 1]e%. —1

202

Chapter 1]

Linear Systems of Equations

EXAMPLE 11.11 We conclude this section with an example to illustrate why the use of the word Wronskian, in the context of a set of solutions of a system of first-order linear

differential equations, is consistent with the usage of the same word, made in Section 6.4 in the context of a set of solutions of a single nth-order linear differential equation.

Consider the second-order linear equation

[D? — (a +b)D +ab]x = 0,

axb,

(12)

in which D = d/dt. The operator factors into (D — a)(D — b) and the functions e“' and e” are therefore solutions of equation (12). The Wronskian of these solutions, as defined in Section 6.4, is the determinant at W{e

obty

_ }=

»€

ett!

et

ae"

:

be?!

(13)

where the functions in the second row are the derivatives of the functions in the first row.

Equation (12) may be converted to a system of first-order equations by setting Dx = y so that (12) becomes

D*x = Dy = (a+b)y — abx. Thus (12) is equivalent to the system f

x =y y

f

= -—abx+(at+b)y,

or

x\ (;)

0 ~

(2

1\

(x

a -1)

(;)

,

(14)

If we apply the technique of the current section, we write the characteristic equation of the matrix of system (14): —m

—ab

l

aek—n

=0,

which reduces to

m? —(a+b)m+ab=0.

(15)

It is important to observe that the characteristic polynomial of (15) and the operator polynomial of (12) have the same form, hence have the same zeros. From equation (15) we obtain the eigenvalues m,; = a and mz = b. Choosing the eigenvalue m, = a leads to

(5 5) (2) =

ILS

First-Order Systems Revisited

203

so that co = ac; and

fea) = Kea) #2) Thus (‘) e“ is one solution of (14). Choosing the eigenvalue m2 = b leads to —b

|

Cc}

—ab

a}

\o

O,

so that ce. = be; and cy)

Cl

co}

\bey J

_

|

Eb

ip

Thus @) e”' is a second solution of (14). In the context of the current section, the Wronskian of these two solutions is

W{Xi(t), X2(t)} = Thus

we see that the expressions

é

at

de

at

e

bt

be bt}:

( 16)

given in (13) and (16), while coming

from

entirely different contexts are the same, and the word Wronskian is used in both contexts for that expression.

The examples that we have considered have all involved matrices whose eigenvalues are distinct real numbers. In each case the eigenvectors corresponding to distinct eigenvalues turned out to be linearly independent. That was no accident. It is possible to prove a theorem to that effect.

Theorem

11.7

If m,,m,...,m,

are distinct eigenvalues of an n x n matrix

A and

if

X,, X9,..., Xs are corresponding eigenvectors, then the set {X,,...,

Xs}

is linearly independent. The definitions and theorems of this section have been stated without providing the proof required to understand them clearly, It is hoped that this will serve as a motivation

for a study of linear algebra,

where

the definitions and theorems

become more easily understood. @ Exercises In Exercises | through 7, find the general solution of the system X’ = AX for the given matrix

A, In each case check on the linear independence of solutions by examining the Wronskian.

204

Chapter 1)

Linear Systems of Equations

L. a=(ig 2.

3

a)

a=(_ . ah (J 4: 4)

5,

a=(4 3

6.

a=(j 2 5)3 a= (412-15 a

7,

2-1 4 azo -1 3). 0 0 2

')3

12 -1 g8 A=] 21 1 -1 1 0

Complex Eigenvalues In Section 11.5 we carefully avoided systems for which the eigenvalues were complex numbers. We now consider some examples in which complex numbers occur.

EXAMPLE 11.12 Solve the system

(3) -@ =)()-

am

The characteristic equation of the matrix in system (1) is |

2—m

| -4on[=”

2

_ +2m+2=0,

(2)

with eigenvalues m; = —1 +i and m2 = —1 —i. For m,; = —1 +i we must satisfy the system

(2 a2) (a)=° 3-1

—5

cy \

which requires that o= 2

3-i 5

Cj.¢!

One solution is obtained by choosing c, = 5. Thus an eigenvector corresponding to the eigenvalue m7, is (,

_

with the complex vector function i = (,

3) ec tOr

at least formally a solution of system (1).

(3)

11.6

Complex Eigenvalues

205

The second eigenvalue m2 = —1—/ leads ina similar way to a second solution, 5 X,==— (543):

(4)

,(-l-it

The two solutions can be combined to give X=

C|

6

i)

ef

tt)

+) (,

W)

ef lt,

(5)

The presentation of a solution in this form should be reminiscent of the situation in Chapter 7, where we were solving single linear equations with constant coefficients. We will proceed in much the same way as we did there by making use

of Euler’s formula

ett

— of! (cos bt + isin bt).

(6)

Formally changing the form of equation (5) gives us X= (,

5\

:) e ‘(cost +isint) +e (54; Je

_

i,

‘(cost —isint),

and after combining real and imaginary parts, we obtain ug

tse

ce

+ €2) (seosr

5cost runs)

Sees

TAKS

ccs

ry

©) (cose

5sint 4 sci) |

cen ()omr-( Same om| (Somes oo]

@

If we let b} = cy + 2 and by = i(c; — c2), equation (7) can be written finally as

(8)

The linear independence of the two solutions in (8) can be established by computing the Wronskian at ¢ = 0. The student should show that W(0) = —5.

We make the following observations from the example above: (a)

Since the matrix of system (1) is real, the eigenvalues occur in conjugate pairs.

(b)

The eigenvectors corresponding

to conjugate eigenvalues are also conjugates

of one another. (c)

The first eigenvector,

B= (371i) = (3) + (1) i=Res trims,

206

= Chapter 1!

Linear Systems of Equations

appears in the solution (8) in the form X =e ‘'[b\{Re Bcost —Im B sins} + b{Im

(d)

Bcost+Re

Bsinr}].

(9)

The Wronskian of the two solutions in (9) att = Ois given by the determinant W = |Re BIm B}.

In Exercises of observations two unknowns.

EXAMPLE

12 through

14 the student is asked to show the applicability (d) to the general

(a) through

of two

system

in

equations

11.13

Solve the system

a

()-(G 2)(): making use of the observations made in Example 11.12. The characteristic equation 2—m —4

l 2—m

=m?

Am

+8 =0

has conjugate roots m,; = 2+2i and m2 = 2—2i. An eigenvector corresponding

tomy is

(2) _ (:) . (:) i

If we accept the results of the observations in Example 11.12, observing that 1 0 WO=199 =240,

c= fof) so») en] +0) e>-()aef]

we conclude that the general solution of system (10) is

EXAMPLE 11.14 We now consider a system of three equations in three unknowns given by | X'= AX

2 1

whereA=]0 0

-]

-1 1}. I

1)

11.6

Complex Eigenvalues

207

The characteristic equation (1 —m)(m? — 2m +2) =0 has roots m, = 1, m2, = 1+i,andm3

The eigenvalue m,

=

= 1—i. ]

I has the vector | 0}

as an eigenvector, giving one

solution of (11) as

The eigenvalue m) = | +i yields an eigenvector 2-i O+i}=

—-140i

2 O}

—1 +i

|

—|

0

The general solution can be written I

2

X=c,JO0let+e

la

0

—|

0 | cost — —|

1 | sine 0 —|

+ 3

2 0 | sint

| | cost + 0

—]

The linear independence of the solutions at f = 0 is guaranteed by evaluating the Wronskian att = 0. Its value is

wo)=|0

1

2+ oO t/=140. 0-1 O i

Exercises In Exercises | through 7, find the general solution of the system X’ = AX for the given matrix A.

. a=(4 2) » an(4 2). 4 4

5

|

an($ 3) a(2 3). 4

—13

3

5

Linear Systems of Equations

Chapter II

9:

_ {12 a= (lf

-17 =A

1 7.

5 an(8 2),

0 1 2

A=|2 3

a)

0 -2

Following the example of Section 11.2, replace each of the following equations by a system of first-order equations. Solve that system using matrix techniques and check your answers by solving the original equation directly. Q

yO

9,

y"

—

-y=

0.

2y’

+. 2y

=

(0).

10.

3"

_

11.

y" -|-

3y" Ay

+4y’ aw. 2y =

—

0.

(0;

In Exercises 12 through 15, we consider the general homogeneous system with real coefficients

x\'

(:) 12.

13.

a

b\

(x

7 (( ‘) (:):

“

Find the eigenvalues for the matrix of (A) and show that complex eigenvalues occur only if (a—d)*+4be < 0. In particular, note that complex eigenvalues occur as conjugate pairs and that they occur only if b and ¢ are not zero. Forthe system (A) suppose that the eigenvalues are complex numbers p+qi and p — gi, where g # 0. Show that the corresponding eigenvectors are conjugate pairs.

14.

Show that the observation made in equation (9) is true in the complex case.

15.

Find the value of the Wronskian in the complex case for f = 0 and show that it is not zero.

Repeated Eigenvalues We

now

in which

consider an example

the characteristic equation

has re-

peated roots.

The characteristic equation of A is —m

—-4

i

4-—m

pe

ne

| a

lI

Re

ee

°

mal

Se

x’

I

EXAMPLE 11.15 Solve the system >

NI

208

(1)

11.7

Repeated Eigenvalues

209

For the eigenvalue mm, = 2 we obtain the solution

X,= (;) eg.

(2)

A second solution X, independent of X,, is not immediately available because

the eigenvalue m, is a double root of the characteristic equation.

From our

experience with repeated roots in Chapter 7, we may be tempted to guess that a

second solution has the form X=

_

fel C2

2t

te".

(3)

However, a substitution back into equation (1) quickly shows that the only solution of this form is the trivial solution with cy) = c. = 0. Another suggestion might be made from our previous experience, that is, to

try to find a second solution of the form

_ (eitt)\ X2= i)

a,

(4)

where we are essentially using a variation of parameters technique. Direct substitution of (4) into (1) gives

(03)2+ (Qn)e=(4 a) ae ci)\

5,

(en)

ar— ( 0 LV)

(ei)

We may rewrite this system of equations in the form

(203) = (2) (Gt). c(t)\

—

f-2

1

c(t)

System (5) can be rewritten

ce, (t) = —2e;(t) + c2(t), cy(t) = —4c)(f) + 2es(t),

(6)

6

from which we conclude that c(t) = 2c;(t). Integrating, we obtain c2(t) = 2c\(t) + a, for arbitrary constant a. Substituting back into the first equation of (6) gives

ei) =a

or

c(t) =at +b.

Thus we obtain a set of solutions of equations (5) c(t) =at+b

and

c(t) = 2at +2b+

a,

210

Chapter 1]

Linear Systems of Equations

with arbitrary constant values for a and b. Equation (4) now becomes X=

(3) ate” + (3) be + (7) ae”.

If we were to choose a = 0 and b = 1, this solution would be the same as X). Instead, we will choose a = | and b = 0 to give

X,= (3) te + (1) a. Att

= 0 the Wronskian

W(X, (0), X20} = |p | =1#0, so the two solutions are linearly independent. The general solution of system (1) is therefore

reo(eeo[eQe}

In retrospect we note that the guess we made earlier in equation (3), although incorrect, was nevertheless not very far from the truth. We are now in a position to make a more reasonable assumption about the nature of a second solution in the case of repeated roots. EXAMPLE

11.16

Solve the system

(3) =

33)()-

m

The eigenvalues are the roots of the equation s—m

4

—] = 0. 12—m|\= m2 — 20m + 100 = (m — 10)?

Therefore, one solution is given by the eigenvalue m, = 10. This solution is Sox

(3).

(8)

Guided by our experience in Example 11.15, we now seek a second solution of the form X,=—( (2)1)

(¢3\ te ot + (:)

10 go,

(9)

11.7

Repeated Eigenvalues

211

Substitution of (9) into system (7) yields

(1) ems ( 2)ems (8) l

10r

L\

10

C4

_{8%

108

—l

-(j

1\

tor

12) (3) 10r

We note that the terms involving te’’

tla

8

—1)\

12)

fe3\

io

he)? -

cancel each other, leaving us with

r Ca a) (e= (a)

or

(10)

3) @ * (.)

(“3

One solution of system (10) is cz = 0 and cy = —1. Therefore, Ro

_

(.;)

te!

op

(!)

lot

xaa( Jemeoen[(ea(9]

is a second solution of system (7). The general solution of system (7) is

el EXAMPLE 11.17 Solve the system

D?y+(D-1)v

=

0,

(2D—Ny+(D-lDw

=

0,

(D+3)y+(D-—4v+3w

=

0.

In order to reduce (11) to a system of first-order equations, we let Dy Then the system (11) can be written Du = Dv= Dw

= u.

u —3u+3w+3y, —u +4

= — 2u

Dy =

(qt)

—3w +

it.

w+

—3y, y,

(12)

212

Chapter I]

Linear Systems of Equations

The matrix of system (12) has the characteristic equation

m(m —4)(m — 1)? = 0. The eigenvalues m; = 0, m2 = 4, and m3 = | give rise to solutions

0 0 XxX

=

—|

12 —16 :

xX2=

0 1

7

1

a,

X3=

1

3

é.

0

If we assume that the repeated root m3 = | will yield a solution of the form 0

_

Xa

|

=

1

C5 te

t

C6

+

t

C7

0

ey

cg

a direct substitution into (12) will yield one set of values for the constants cs to CE, C5

—]

C6

_

C7

~

0

Cg

—]

Thus the desired solution is 0 X4

=

;

—l te’

-+-

’

O

e!

_

Finally, the solution of system (11) is v

|

wl=c|l)ete| y

0

1

0

|i }te +

L}e | teo]—-l]+e,[

0

—|

0

—16

]

—7]

e”.

3

The student is asked to fill in the details of this example in Exercise 10.

EXAMPLE

11.18

We now consider an example in which a repeated root of the characteristic equa-

tion of a matrix gives rise to two linearly independent eigenvectors of that matrix and thus avoids the complications encountered in Examples 11.15, 11.16, and 11.17.

11.7

Repeated Eigenvalues

213

The eigenvalues of the matrix of the linear system i

x Zz

0

1

|

x

1

1

O

Zz,

are the roots of the equation —m

]

1

|

—m

1] = —(m + 1)*(m — 2) =0.

1

1

-—m

For the repeated root m = —1, we seek nontrivial solutions of the system IL

1

1

Cl

]

1

|

Co

]

|

|

C3

=

7 ;

that is, solutions of ¢; + ¢2 +c¢3 = 0. Thus

C=

Cj

cl

|a}]=

co]

C3

—€)

—

|

=c|

0

Oy+

&&

—1

C2

| —|

It follows that both |

O}e!

and

lle’

—]

—|

are solutions of system (13).

The second eigenvalue, m = 2, requires us to find nontrivial solutions of the system of equations =;

I

— I

l

Cj

1

c)/=0O

—

C3

That is, —2¢e,

+

co+

cy —2c2

cy +

+

cz =0 3

=0

cy —2c3 = 0.

Elementary elimination of c. from the first and last equations and c, from the

second and third equations leaves us with eT =&

and

C2 = C3,

214

Chapter 11

Linear Systems of Equations

so that ey C=

1

qc},

=—c,

1

C3

1

Hence a third solution of system (13) is

The linear independence of the three solutions is established by examining their Wronskian at t = 0,

W{X,(0), X2(0), X3(0)} =|)

1 O -1

ol 1 Ij=3 40 -1 1

The general solution is therefore x

1

0

yl=leal

O)]+e.{

1}

Z

-1

1

]et+es]

—1

1]

e”.

1

|| EXAMPLE 11.19 Solve the system

x\o y}

(2.1 =]1

Zz

1

2\

fx

2

2]

]y

1

3

Zz

(14)

The characteristic equation

2—m

|

1

2

2-—m

1

2| = —(m — 1)?(m —5) =0 1

3-—m

has the two roots 1 and 5. The repeated eigenvalue | gives rise to the equation c) = —c2 — 2c3. Consequently, cl C=

[a] C3

Oandq > 0, it follows that if pg — ab > 0, the two eigenvalues are both negative, but if pq — ab < 0, the eigenvalues will have opposite signs. The presence of a positive eigenvalue is disturbing since it will lead to an exponential

function that becomes unbounded as time increases, a situation that may result in

a runaway arms race. We now examine several different examples to illustrate the possible consequences of Richardson’s model. EXAMPLE 12.3 Consider a situation in which the parameters in equations (2) area = 4, b = 2, p=3,qg=1,r=2,5

=2,x9

xX'= a

4)

=4, and yo = 1, that is,

and

X+ (3)

X(0) = eat

The characteristic equation of the matrix is ate

2

—-l-—m

4)

an? + ta

— $

=,

so the eigenvalues are m, = 1 and m2 = —5.

For m, = 1, we compute the nontrivial solutions of the system

(3 2)(a)-° Thus c;

One such solution is obtained by taking c;

= co.

eigenvector

=

| to give the

.

For m2 = —5, the system

G 3)(2)=° requires c; + 2c, = 0. Taking cy = —1 yields the eigenvector (1):

230

Chapter 12

Nonhomogeneous Systems of Equations

The general solution of the homogeneous system X' = AX is therefore

XH=c

@

e+ cy (.;) et,

The nonhomogeneous system X' = AX + B has a constant solution of the e be form ( ‘): Substitution into the system gives

(2 -)(7)+@)=¢ .

;

—2

a system with solution (3)

. Thus the general solution of the nonhomogeneous

system is AAS) Se]

(1) eto

The initial condition X (0) = (1)

(_;)

gp

(=)

.

now requires that

(i)=«(i) +e) +2): so that c; = 4 and cp = 1. The final solution is therefore _ iy xm=a(te

2\ +(e

31

|

—2 appa

or

x(t) = 4e! + 2e7* — 2, y(t) = 4e’ —e* — 2, We have a runaway arms race.

EXAMPLE 12.4 As a second example of an arms race, we take the following values for the parameters in equation (2): a=4,b=2, p=3,qg=1,r = —2,s = —2, x9 =2, yo= 5. The system of differential equations haz the same solution as in Example 12.3 except for the sign of the particular solution. Thus the general solution is X(th=c

(i)

eto

(_*)

gt

4 (5)

‘

12.2.

Arms Races

231

€)=a(()=(3)0)

The initial conditions now require that

3

from which we obtain c) = —1 and ce. = i.

The solution is

x(t) =—e! +e" 4-2,

y(t) = —e — Se" +2, and each party will eventually decrease its arms expenditure to zero, a condition of disarmament.

EXAMPLE 12.5 Let us now change the values of the parameters in system (2) toa = 3,b = 1, p=4,q =2,r = 6,s = 1 with initial conditions x9 = O and yo = 0. The system to be solved becomes

() = x\

—4

=)6)+(): S\

(a

6

A particular solution of this system is the vector

2

and the eigenvalues of the

matrix are —1 and —5 with corresponding eigenvectors (1) and (4)

()-a()eeoQ}er0)

The general solution of the system is therefore

3

¥

The initial conditions x9 = yo = 0 require

()=«(t) ea) +Q), 0

|

3

3

an equation that is satisfied only if ¢, = —3 and cz = —}.

Thus the solution of

the initial value problem is

)-TMeaGjenrG@

®

232

Chapter 12.

Nonhomogencous Systems of Equations We may now interpret equations (3) and (4) as an arms race with each party

starting with zero expenditure but with dx/dt = 6 and dy/dt = 1, both positive quantities. Because of the negative exponents, the rates at which the expenditures are changing will tend toward zero and the arms expenditures will approachx = 3 and y = 2. There will be a stabilized arms race.

Exercises For Richardson’s model as described by equations (2), solve the following special cases, noting in each exercise whether there will be a stable arms race, a runaway arms race, or disarmament.

@=2,b = 4, p = 5,.¢ = 3, f=

1,5 — 2) x9 = 8, Yo = 7.

2.

What effect does the changing of the initial values xo and yo have on the stability of the solution in Exercise |?

3.

a=4,b=4,p=2,¢

4.

Show that the solution in Exercise 3 will remain unstable if the initial values are changed to any other nonnegative values.

5.

Fora =4,b=4, p=2,¢g =2,r = —2, 58 = —2, show that there will be disarmament if xo -- yo < 2 and a runaway arms race if xq + yo > 2.

6.

Fora=4,b=4, p=2,q =2,r > 0,» > 0, show that there will be a runaway arms race for any nonnegative xp and yo.

7.

Pora=4,b=4,

8.

=2,r=8,5

=2,x9 =5, yo =2.

p=2,q

= 2,r 2

< 0, show that there will be a

Show thatif pg —ab > 0,r > 0, ands > 0, there will be a stable solution to the arms race.

9.

Show that if pg —ab < 0, r > 0, ands > 0, there will be a runaway arms Trace,

10.

Show

that if pg — ab > 0, r < 0, ands

< 0, there will be disarmament.

Electric Circuits The basic laws governing the flow of electric current in a circuit or a network will be given here without derivation. The notation used is common to most texts in electrical engineering and is:

12.3. t

(seconds)

Q I

Electric Circuits

233

= time

(coulombs) = quantity of electricity (e.g., charge on a capacitor) (amperes)

= current, time rate of flow of electricity

E

(volts) =

R

(ohms)

electromotive force or voltage

= resistance

L

(henrys) = inductance

G

(farads) = capacitance

By the definition of Q and J it follows that

I(t) = O'(t). The current at each point in a network

may be determined by solving the

equations that result from applying Kirchhoff’s laws: (a)

The sum of the currents into (or away from) any point is zero.

(b) Around any closed path the sum of the instantaneous voltage drops in a specified direction is zero.

A circuit is treated as a network containing only one closed path. Figure 12.1 exhibits an “RLC circuit” with some of the customary conventions for indicating various elements, For a circuit, Kirchhoff’s current law (a) indicates merely that the current is the same throughout. That law plays a larger role in networks, as we shall see later.

To apply Kirchhoff’s voltage law (b), it is necessary to know the contributions of each of the idealized elements in Figure 12.1. The voltage drop across the resistor is RJ, that across the inductor is LJ’(t), and that across the capacitor is C~' Q(t), The impressed electromotive force E(t) is contributing a voltage T1S€.

th

I(t) E(t)

Figure 12.1

234

Chapter 12.

Nonhomogeneous Systems of Equations

Assume that at time f = 0 the switch shown in Figure 12.1 is to be closed. At t = Othere is no current flowing, / (0) = O and, if the capacitor is initially without

charge, Q(0) = 0. From Kirchhoff’s law (b), we get the differential equation

LI'(t)+ RI) +C7'O@) = E@),

(1)

I(t) = O'(t).

(2)

in which

Equations (1) and (2), with the initial conditions

10)=0,

(0) =0,

(3)

constitute the problem to be solved. The function /(t) may be eliminated from (1), (2), (3) to obtain the initial

value problem

LQO"@) + RQ'(t)+C' OW) = E);

Q(0) =0, Q'(0) =0.

4)

It follows that the circuit problem is equivalent to a problem in damped vibrations

ofaspring (Section 10.4). The resistance term R Q'(t) corresponds to the damping term in vibration problems. The analogies between electrical and mechanical systems are useful in practice.

Initial value problems of the type given in equation (4) may be solved by the general theory of linear equations with constant coefficients. We present here an example using this technique. EXAMPLE 12.6 In the REL circuit shown in Figure 12.2, find the current / (f) if the current att = 0 is zero and £ is a constant. From equations (1) and (3) we have

LI'(t) + RI(t) = E:

WW

R

= nh Figure 12.2

Oy. =f.

(5)

12.4.

Simple Networks

235

This first-order linear equation may be written

p+k

,£ La

Es

for which the general solution is

re

| Here plea

=

—

é

>

——S

x,

Ne

The initial condition (0) = 0 requires that E

0=—+c1,RS! so that finally,

Majed

R

- exp (-F1)].

M=F

Simple Networks Systems of equations occur naturally in the application of Kirchhoff’s laws to

electric networks. We consider in this section two extremely simple networks to indicate how the techniques of this chapter can be applied. EXAMPLE

12.7

Determine the character of the currents /,(t), J(t), and /;(f) in the network

having the schematic diagram shown in Figure 12.3, under the assumption that when the switch is closed the currents are each zero. In a network, we apply Kirchhoff’s laws, Section 12.3, to obtain a system of equations to determine the currents. Since there are three dependent variables, 1, I, 13, we need three equations. R

WW ——_$_—_}

f

i.

—_>

|

ts

Ry

§ Ly

Figure 12.3

236

Chapter 12

Nonhomogeneous Systems of Equations

From the current law it follows that

h=b+h.

(1)

Application of the voltage law to the circuit on the left in Figure 12.3 yields (2)

Rili + Loh = E. Using the voltage law on the outside circuit, we get

RN

+R3b+

Ls

= E.

(3)

Still another equation can be obtained from the circuit on the right in Figure 12.35

R3lg + L3hi — Lal = 0.

(4)

Equation (4) also follows at once from equations (2) and (3); it may be used instead of either (2) or (3). We wish to obtain the currents from the initial value problem consisting of equations (1), (2), and (3) and the conditions /; (0) = 0, /2(0) = 0, and /;(0) = 0. One of the three initial conditions is redundant because of equation (1). If we eliminate /, from equations (1), (2), (3), we can write

ponyig * i

_

R a

Bye Ey De’ _

R

1 +

La

R

E

Is

Ls

.

Ls

or in matrix notation,

R

BY a

(7)

=

Ry

Ly

Lo

R

Ry + R3

Ly

L3

E

I

L»

es

(7)

Tl

&

a

©)

5

L3

The characteristic equation of the matrix of system (5) is therefore R,

Ey

Ry

ee AR by

_ At

La] _

_RitRs

_

—

Ly

or

A = Lo Lym + (Ri Lo + R3L2 + Ri L3)m + RK, R3 = 0.

(6)

12.4

Simple Networks

237

We are interested in the factors of the characteristic polynomial A. Equation

(6) has no positive roots. The discriminant of A is

(RiLz + R32 + RiL3)” — 4L2L3R\R3 and may be written

(Ri La)? + 2RiLo(R3L2 + RiL3) + (R3L2 + RiL3)? — 4LoL3R 1 R3, which equals

(Ri L)” + 2R) L2(R3L2 + Ri L3) + (R3L2 — RiL3)° and is therefore positive. Thus we see that equation (6) has two distinct negative roots. Call them —a,; and —ap. It follows that A=

LoL3(m

+ a,)(m +a)

and that the eigenvalues of the matrix of system (5) are —a, sponding to these eigenvalues, we obtain eigenvectors Ry

.

Lal, _ OQ

and

R

loli

_ 7)

and —a.

Corre-

.

(7)

It follows that the general solution of the homogeneous system associated with

(5) is

I,

=

(7).

It should

=o

Ry (ots

be clear in system

a ait

os i

e

: +2

=

=

R,

aa art

Ry

=

,

(5) that there exists a particular solution

of the

form

h\ q;

_

(Bi By}?

>

where 8, and B3 are constants. Direct substitution into (5) yields

(1) =a (0): We have therefore found the general solution of system (5) to be I,

_

Ry

—ayt

(2) =a (ote)?

.

Fel

Ry

ar —R)°

ant

E

TR

l

lo):

®

The initial conditions /3(0) = /2(0) = 0 now require

,

Ri “1

(nitsert)

+

Ri\_

62(ots)

Efi Ry,

(4)

(9)

238

= Chapter 12.

Nonhomogeneous Systems of Equations

as the system which must be satisfied by c; and cz. The solution of equation (9) is c=

E(ayL2 — Rj)

E(a;Lz — Ry)

and

R?L>(a2 — a)

co = —..

°

Ri L (a2 — ay)

(10)

The solution of the initial value problem is given by the insertion of these constants into equation (8). Finally, the current /; is easily obtained as the sum of /5 and /3:

he :

Eay(agh2 — Ri) Ri(a2 — a)

au

Eaz(a,L. — ee

Ri) Ri (ao — ay)

an 4 ze Ry

(11)

ka EXAMPLE 12.8 For the network shown in Figure 12.4, set up the equations for the determination of the currents /;, Io, /3, and the charge Q3. Assume that when the switch is

closed, all currents and charges are zero. Find the characteristic polynomial for the matrix of the resultant system. Using Kirchhoff’s laws, we write the equations

Rh=h+h,

(12)

dl Ri + Lo — = Esinat, dt |

Ril + Roly +

(13)

Os = Esinot;

(14)

3

and the definition of current as time rate of change of charge yields 1

i ao

(15)

Ry

MAN ——_>

I

—_—_—>

|

E sin wt

I,

g

Ry

is

ater is

NY Figure 12.4

12.4

Simple Networks

239

Our problem consists of the four equations (12) through (15) with the initial conditions that

I(0) = 0,

13(0) = 0,

Q3(0) = 0.

(16)

If we use equations (12) and (15) to eliminate J; and Q3 from the system, we obtain dl,

_

C3R,R3

dt

di

4

C3L2(Ri + Rs) | 4

dls —_—

+ Lo

=-—-

Ew

il

I

C3(R) +R)”

bo

ER;

not

cos wt + ————— sino, Ri + Rs L2(Ri + Rs) R, —/]

EE. —s

i

t.

The matrix of the associated homogeneous system is C3R,R3

+ Lo

1

C3Lo(R, + R3)

C3(Ry + R3)

—R Lo

0

Thus the characteristic polynomial is

C3R,R3 + Lo pp a eS C3L2(R; + R3)

1

C3(R, + R3)

R

Bs

. 3

CyR,R

i

i

ST

Pet

C3L2(R, + R3)

R

C3L2(R\ + R3)

. (17) Bl

WB Exercises

|.

For the RL circuit of Figure 12.2, find the current J if the direct-current element £ is not removed from the circuit.

2.

Solve Exercise | if the direct-current element is replaced by an alternatingcurrent element £ cos wt. For convenience, use the notation x

3.

—

R2

gE?

in which Z is called the steady-state impedance of this circuit. Solve Exercise 2, replacing E cos wt with E sinwt.

240

Chapter 12

Nonhomogeneous Systems of Equations

E sin or

Cc

Figure 12.5

4.

Figure 12.5 shows an RC circuit with an alternating-current element inserted. Assume that the switch is closed at ¢ = 0, at which time @ = 0 and

I = 0. Use the notation

Z? = R* + (wC)”, 5.

6. 7.

where Z is the steady-state impedance of this circuit. Find / fort > 0. In Figure 12.5, replace the alternating-current element with a direct-current element E = 50 volts and use R = 10 ohms, C = 4(10)~* farad. Assume that when the switch is closed (at f = 0) the charge on the capacitor is 0.015 coulomb. Find the initial current in the circuit and the current for ¢ > 0. In Figure 12.1, find /(t) if E(t) = 60 volts, R = 40 ohms, C = 5(10)~> farad, L = 0.02 henry. Assume that /(0) = 0, Q(0) = 0. In Exercise 6, find the maximum current.

In Exercises 8 through 11, use Figure 12.1, with E(¢) = notations used to simplify the appearance of the formulas: R

= OL’

DF

yet

ie

‘|

—=, wc

E

te

2

Esinwr

I

and with the following

° ~re7*

Z=R

2

+y’.

The quantity Z is the steady-state impedance for an RLC circuit. through 11, find /(¢) assuming that / (0) = 0 and Q(0) = 0.

In each of Exercises 8

8. 9, 10.

Assume that 4L < R°C. Assume that R?C < 4L. Assume that R?C = 4L.

11.

Show that the answer to Exercise 10 can be put in the form

1 = EZ~*(Rsinawt — y cos wt) + EZ?e “ly + (ay - Ro)t].

12.4

Simple Networks

241

12,

In Exercise 4, replace the alternating-current element FE sin wt with E cos 2or. Determine the current in the circuit.

13.

In Figure 12.6, let E = 60 volts, R} = 10 ohms, R3 = 20 ohms, and C> = 5(10)~* farad. Determine the currents if, when the switch is closed,

the capacitor carries a charge of 0.03 coulomb, 14.

In Exercise 13, let the initial charge on the capacitor be 0.01 coulomb, but

15.

leave the rest of the problem unchanged. For the network in Figure 12.7, set up the equations for the determination of the charge Q; and the currents /,, 2,

13. Assume all four of those quantities

to be zero at time zero. Use matrix algebra to show that the nature of the solutions depends on the zeros of the polynomial

C3Lo(Ry + R3)m® + [C3(RiRo + RoR3 + R3Rj) + Lom + Ry + Ro. 16.

For the network in Figure 12.8 set up the equations for the determination of the currents. Assume all currents to be zero at time zero. Use matrix algebra to discuss the character of J; (t) without explicitly finding the function.

ay h

Ry WW

oy I

|

E =

——

R3

hy Figure 12.6

Ry

WAN ——> |. t=

q

—> Iy

Ry

:

Ly ee Figure 12.7

fy

Ry Cs

242

Chapter 12.

Nonhomogeneous Systems of Equations

——>

Ry AMMAN

I 1

——> I2

Ry

I3

Rs

les

E —— Ly eo

Figure 12.8

Ls

ee The Existence

and Uniqueness of Solutions

13.1

Preliminary Remarks The methods of Chapter 2 are strictly dependent on certain special properties (variables separable, exactness, and so on), which may or may not be possessed

by an individual equation. It is intuitively plausible that no collection of methods can be found that would permit the explicit solution, in the sense of Chapter 2, of all first-order differential equations. We may seek solutions in other forms, employing infinite series or other limiting processes; we may resort to numerical approximations. Confronted with this situation, a mathematician reacts by searching for what is known as an existence theorem, that is, to determine conditions sufficient to ensure the existence of a solution that has certain properties. In Chapter 2 we stated such a theorem, and now we wish to examine it more closely.

An Existence and Uniqueness Theorem Consider the equation of order one d

= f(y).

(1)

x Let 7 denote the rectangular region defined by

Ix—xo]S@

and

|y—yol 0,

is the one to which this chapter is devoted.

252

14,3

Transforms of Elementary Functions

253

Definition of the Laplace Transform Let F(t) be any function such that the integrations encountered may be legitimately performed on F(t). The Laplace transform of F(t) is denoted by L{F(t)}

and is defined by “iron = |

0

e

(1)

F(t) dt.

The integral in (1) is a function of the parameter s; call that function f(s). We may write

L{F(t)} = [

(2)

e'F(t)dt = f(s).

0

It is customary to refer to f(s) as well as to the symbol L{F'(t)}, as the transform, or the Laplace transform, of F(t).

We may also look upon (2) as a definition of a Laplace operator L, which

transforms each function F(t) of a certain set of functions into some function

f(s).

It is easy to show that if the integral in (2) does converge, it will do so for all

5 greater than!

some fixed value so.

That is, equation (2) will define f(s) for

5 > so. In extreme cases the integral may converge for all finite s. It is important that the operator L, like the differential operator D, is a linear operator.

If F\(t) and F2(t) have Laplace transforms and if c, and cz are any

constants,

L{e F(t) +o Fa(t)} = a L{ Fi} + eL{ha@}.

(3)

Using elementary properties of definite integrals, the student can easily show the

validity of equation (3). We shall hereafter employ the relation (3) without restating the fact that the

operator L is a linear one.

14.3

Transforms of Elementary Functions The transforms of certain exponential and trigonometric functions and of polynomials will now be obtained. These results enter our work frequently. EXAMPLE 14.1 Find L{e“’}. We proceed as follows: oo

L{e} = /

0

oO

ee

dt = /

0

g OO

dp,

! If is not to be restricted to real values, the convergence takes place for all s with real part greater than some fixed value.

254

Chapter 14

The Laplace Transform

For s < k, the exponent on e is positive or zero and the integral diverges. s > k, the integral converges. Indeed, for s > k, oo

L{e"}= [

gg

For

Ne bt

0 j="

ii

S—k

Jo

1

=0

.

* s—k

Thus we find that

L{e"}

kt

I

= s—k —.,

s>k,

(1)

Note the special case k = 0:

]

L{l}=-,

s>0Q.

(2)

EXAMPLE 14.2 Obtain L{sin kr}. From elementary calculus we obtain ae

e““(asinmx — mcosmx)

e* sims dt =—_______,—— a“+m

Since oO

Lsin kr) = | it follows that

L(sinkr) =|

e

0

sinkt dt,

ot (—g (—s sinkt sin kt —— kcos k cos kt ]* 3? +k?

a

(3)

For positive s, e~*’ — 0 as t + oo. Furthermore, sin kr and cos kt are bounded as ¢ —> oo. Therefore, (3) yields .

1(O—k)

L{sinkt} = 0

ER”

or ink

L{sinkt} =

k

ae

ie

0,

(4)

The result

L{cos kt} = ae can be obtained in a similar manner.

s>0,

(5)

14.3.

Transforms of Elementary Functions

255

EXAMPLE 14.3 Obtain L{t”} for n a positive integer. By definition co

L{t"} -|

et" dt. 0

If we perform an integration by parts on this integral we obtain 0°

|

n,—st

0

Fors > Oandn

TO

=e

ews! p" dt

s

oo

nA “{

4

0

S

e

st yn ne l dt.

(6)

Jo

> 0, the first term on the right in (6) is zero, and we are left with oO

n

et"

i 0

oo

ete!

|

dt=—S

s > 0,

de,

Jo

or

Li} ="ae 1")

0

:

(7)

s>0.

From (7) we may conclude that for n > 1, Lie"

!7}

n—l

=

Li},

S so L{t"}

_—nn

=O)

2

(8)

ry,

Iteration of this process yields

eT L(t) = Mae

404,

From Example 14.1, we have

L{f} = LUj=set. Hence, for

a positive integer,

bien~

|

ght

2

s>0.

The Laplace transform of F(t) will exist even if the object function F(r) is discontinuous, provided that the integral in the definition of L{F(t)} exists. Little will be done at this time with specific discontinuous F(t), because more efficient

methods for obtaining such transforms are to be developed later,

Chapter 14

The Laplace Transform

jee es re

= 1 | 1 | | | [4

o|

a

8

tr

F(t)

OU

258

Figure 14.1

In elementary calculus we found that finite discontinuities, or finite jumps, of the integrand did not interfere with the existence of the integral. We therefore introduce a term to describe functions that are continuous except for such jumps. Definition. The function F(t) is said to be sectionally continuous over the closed interval a < ¢ < b if that interval can be divided into a finite number of subintervals c < ¢ < d such that in each subinterval: (a)

F(t) is continuous in the open interval ¢ < ¢t < d.

(b)

F(t) approaches a limit as ¢ approaches each endpoint from within the interval; that is, lim

tect

F(t) and

lim

t>d-

F(t) exist.

Figure 14.1 shows the graph of a function F(t) that is sectionally continuous over the interval 0 < t < 6.

The student should realize that there is no implication that F(t) must be sectionally continuous for L{F(t)} to exist. Indeed, we shall meet several coun-

terexamples to any such notion. The concept of sectionally continuous functions will, in Section 14.6, play a role in a set of conditions sufficient for the existence of the transform.

Functions of Exponential Order If the integral of e~*' F (t) between the limits 0 and fy exists for every finite positive

fo, the only remaining threat to the existence of the transform /

e “'F(t)dt 0

is the behavior of the integrand as tf + oo.

(1)

14.5

Functions of Exponential Order

259

We know that 00

|

edt

(2)

0

converges for c > 0. This arouses our interest in functions F(t) that are, for large

t (t > tp), essentially bounded by some exponential e”", so that the integrand in (1) will behave like the integrand in (2) for s large enough. Definition. The function F(t) is said to be of exponential order as t > co if constants M and b and a fixed ft-value fo exist such that

|F(t)| < Me” — fort > to.

(3)

If b is to be emphasized, we say that F (1) is of the order of e”' ast — also write

F(t) = O(e"’),

t—> 0,

oo. We

(4)

to mean that F(t) is of exponential order, the exponential being e”'

ast > oo.

That is, (4) is another way of expressing (3).

The integral in (1) may be split into parts as follows: oo

|

To

Co

e' F(t) di = | 0

e' F(t)dt +f 0

e*' F(t) dt.

(5)

to

If F(t) is of exponential order, F(t) = O(e"'), the last integral in equation (5) exists because from inequality (3) it follows that for s > b,

oo

~

le F(t)dt < u |

/

Mexp[—to(s —b

eo. bl dt = Mepis

(6)

5

fo

to

For s > b, the last member of (6) approaches zero as fg —- oo. Therefore, the last

integral in (5) is absolutely convergent” for s > b. We have proved the following

result.

Theorem 14.1

Jf the integral of e~*' F(t) between the limits 0 and to exists for every finite positive to, and if F(t) is of exponential order, F(t) = O(e”) as t > oo, the Laplace transform

Liroy= |

oO

0

e“ F(t)dt = f(s)

(7)~

exists for s > b. We

know

that a function that is sectionally continuous

over an interval

is

integrable over that interval. This leads us to the following useful special case of Theorem 14.1. 2

if complex s is to be used, the integral converges for Re (s) > b.

260

Chapter 14

Theorem

The Laplace Transform

14.2 /f F(t) is sectionally continuous over every finite interval in the range t = 0, and if F (t) is of exponential order, F(t) = O(e"") ast + 00, the Laplace transform L{F(t)} exists for s > b. Functions of exponential order play a dominant role throughout our work. It is therefore wise to develop proficiency in determining whether or not a specified

function is of exponential order. Surely, if a constant b exists such that

limle "FOI

(8)

exists, the function F(t) is of exponential order, indeed, of the order of ec” To see this, let the value of the limit (8) be K 4 0. Then, for ¢ large enough, |e~"' F(t)|

can be made as close to K as is desired, so certainly

le" F(t)| < 2K. Therefore, for f sufficiently large,

|F(t)| < Me™,

(9)

with M = 2K. If the limit in (8) is zero, we may write (9) with M = I.

On the other hand, if for every fixed c, lim [e~' | F(t) |] = 00,

(10)

foo

the function F(t) is not of exponential order. For assume that b exists such that

|F(t)| < Me",

t>t;

(11)

then the choice c = 2b would yield, by (11),

Je F(t)| < Me™, soe? F(t) + 0 as t — oo, which disagrees with (10). EXAMPLE 14.5 Show that f° is of exponential order as f —

oo, We consider, with b as yet

unspecified, i

lim (et?) = too lim —. t—00 ebt

(12)

If b > O, the limit in (12) is ofa type treated in calculus. In fact,

e

lim —

too

ebt

=

a

lim —

too

bebt

_

of

too

p2ebt

= lim ——

6

= lim —— i300

Therefore, t* is of exponential order, r=

for any fixed positive b.

O(e"),

t —>

oo,

Pp ebt

='0,

74.6

EXAMPLE 14.6 Show that exp (t7) is not of exponential order as t — 2

Functions of ClassA

261

oo. Consider

say Ce (t") .

(13)

tcc exp (bf) Ifb < 0, the limit in (13) is infinite. Ifb > 0, exp @) jim. ep = Jim exp [t(t — b)] = co.

Thus, no matter what fixed b we use, the limit in (13) is infinite and exp (t7)

cannot be of exponential order.

The exercises at the end of Section 14.6 give additional opportunities for practice in determining whether or not a function is of exponential order.

14.6|

Functions of Class A For brevity we shall hereafter use the term “a function of class A” for any function that is: (a)

Sectionally continuous over every finite interval in the range ¢ > 0

(b)

Of exponential order as f —

co

We may then reword Theorem 14.2 as follows.

Theorem 14.3

Jf F(t) is a function of class A, L{F (t)} exists. It is important to realize that this theorem states only that for L{F (¢)} to exist, it is sufficient that F(t) be of class A. The condition is not necessary. A classic

example showing that functions other than those of class A do have Laplace transforms is

FQ) =17!”,

This function is not sectionally continuous in every finite interval in the range

t > 0, because F(t) > co ast — O*. But r7!” is integrable from 0 to any positive fo. Also, -!/? > O0ast > oo, so t~!/* is of exponential order, with M = 1andb = Oin the inequality (3) of Section 14.5. Hence, by Theorem

L{t~"/?} exists. Indeed, for s > 0, Co

Li)

= |

gat? aie 0

in which the change of variable st = y* leads to Luo} ss2s?

co

| 0

exp (5

dy,

s 2 0.

14.1,

262

Chapter 14

The Laplace Transform oo

In elementary calculus we found that [ pe

Ay

—

exp (— y)dy

0

9571/2

= 50 . Therefore,

. iM

= eae

s>0,

(1)

Ss

even though 1~'/? — oo as t > O*. Additional examples are easily constructed and we shall meet some of them later in the book. If F(t) is of class A, F(t) is bounded over the range 0 is of order of exp (O-t) but F; is not of exponential order. From

Example 14.6 of Section 14.5, exp (t7) im

i>oo exp (br) for any real b. t —

='0oO

Since the factors 2t cos [exp (t7)] do not even approach zero as

oo, the product FF exp (—cr)

cannot be bounded

as tf —

oo no matter how

large a fixed c is chosen. Therefore, in studying the transforms of derivatives, we shall stipulate that the derivatives themselves be of class A. If F(t) is continuous for ¢ > O and of exponential order as f — F'(t) is of class A, the integral in LiF’(t)} =|

e' F'(t)dt 0

oo, and if

(1)

264

Chapter 14

The Laplace Transform

may be simplified by integration by parts. We obtain for s greater than some fixed SO; so |

00 e'

F(t)

dt=

0

je"

oo +s

|

9

e

F(t) dt,

i

or

L{F'(t)} = —F(O) + sL{F(t)}. Theorem

14.5

(2)

If F(t) is continuous for t = 0 and is also of exponential order as t > o©, and if F'(t) ts of class A , then

L{F'(t)} = sL{F(t)} — FO).

(3)

In treating a differential equation of order n, we seek solutions for which the highest-ordered derivative present is reasonably well behaved, say sectionally continuous. The integral of a sectionally continuous function is continuous.

Hence we lose nothing by requiring continuity for all derivatives of order lower than n. The requirement that the various derivatives be of exponential order is forced upon us by our desire to use the Laplace transform as a tool.

purposes, iteration of Theorem makes sense.

For our

14.5 to obtain transforms of higher derivatives

From (3) we obtain, if F, F’, F” are suitably restricted,

L{F"(t)} = sL{F'()} — FO), or

L{F"(t)} = s° f(s) —sF (0) — FO),

(4)

and the process can be repeated as many times as we wish.

Theorem 14.6

/f F(t), F’(t),..., F~(t) are continuous for t > 0 and of exponential order ast > oo, and if F(t) is of class A, then from L{F(t)} = f&) it follows that

L{F

n-1

(t)} = 8" f(s) — yi

,

Ty po

/

OO).

k=0

Thus

L{F (1)} = 83 f (s) — s°?F (0) — s F'(0) — F"(0),

L{F (t)} = 54 f(s) — 8° FO) — 8° F'(0) — sF"(0) — F (0), ete.

(3)

14.7

Transforms of Derivatives

265

Theorem 14.6 is basic in employing the Laplace transform to solve linear differential equations with constant coefficients. The theorem permits us to transform such differential equations into algebraic ones. The restriction that F(t) be continuous can be relaxed, but discontinuities in F(t) bring in additional terms in the transform of F’(t). As an example, consider

an F(t) that is continuous for ¢ > 0 except for a finite jump at tf = f), as in Figure 14.2. If F(t) is also of exponential order as f —> oo, and if F’(f) is of class A, we may write

L{F'(t)} =i

e'F'(t)dt 0 fy

-|

co

eras

|

0

e' F'(t) dt. q

Then integration by parts applied to the last two integrals yields 4

LiFe

= Jere]

0

+sf

0

| e'F(t)dt

00 +

Je" reo]

60 + 7

fy

e “'F(t)dt fy

or

L{F'()} =s |

e' F(t) dt +e" F(t;) — FO) +0 —e 0

sL{F(t)} — F(0) — exp (—st)LF(qt) — F(ty1. In Figure 14.2 the directed distance AB is of length [F(t,) — F(t, ji. F(t) cea |

|4 iw

| | | |

|

t

| | | !

Figure 14.2

F(t)

266

Chapter 14

Theorem

14.7

The Laplace Transform

Jf F(t) is of exponential order as t ~

060 and F(t) is continuous for t = 0

except for a finite jump at t = t,, and if F’(t) is of class A, then from

L{F(t)} = f(s), it follows that

L{F'(t) = sf (s) — F(O) — exp (—st)[F (4) — Fy )I-

(6)

If F(t) has more than one finite discontinuity, additional terms, similar to the last term in (6), enter the formula for L{F’(t)}.

14.8]

Derivatives of Transforms For functions of class A, the theorems of advanced calculus show that it is legitimate to differentiate the Laplace transform integral. That is, if F(t) is of class A, from

roy=

[

0

e 'F(t)dt

(1)

it follows that

f's)= / 0 (—te™ F(t) dt.

(2)

The integral on the right in (2) is the transform of the function (—r) F(f).

Theorem 14.8

If F(t) is of class A, it follows from

L{F(t)} = f(s) that

f(s) = L{-tF (0).

When ger k.

(3)

F(t) is of class A, (—f)‘ F(t) is also of class A for any positive inte-

Theorem 14.9 Jf F(t) is of class A, it follows from L{F(t)} = f(s) that for any positive integern, HW

ds”

f(s) = L{(-t)" Fo}.

(4)

14.9

These theorems are useful in several ways.

The Gamma Function

267

One immediate application is to

add to our list of transforms with very little labor. We know that

k

.

Cue

= L{sinkt},

(5)

and therefore, by Theorem 14.8, —2ks

.

(+k?

=

L{—tsinkt}.

Thus we obtain

.

i,

——— (s2 + k?)2

|. dade

— sin 2k

(6)

.

From the known formula s e+e

=

L{cos

kt}

we obtain, by differentiation with respect to s, Ie

_

gs?

oe

=

L{-tcos kt}.

Let us add to each side of (7) the corresponding member

ae

=,

ti

(7)

of

sin ka}

to get

+e

+k —s*

(s2 + k*)?

| = 7 {sink

= reoskr

from which it follows that

| (92 +k)?

I

——

Es

(sinkt — kt cos

|

(8)

The Gamma Function Por obtaining the Laplace transform of nonintegral powers of t, we need a function not usually discussed in elementary mathematics. The gamma function I’(x) is defined by

ray =f

e 'B*—! dp, 0

x>0,

(1)

268

Chapter 14

The Laplace Transform

Substitution of (x + 1) for x in (1) gives

ra+=

|

e ? B* dB.

(2)

0

An integration by parts, integrating e * df and differentiating B*, yields

Tixe+l= [ 0, B* B — oo. Thus

ra+n=s/

(3)

9

—+

a B*- 1 qe = aT).

Oas

(4)

0

Theorem 14.10

Forx > 0, T(x +1) =2xT (x). Suppose that 7 is a positive integer. Iteration of Theorem 14.10 gives us Pint+1)=nl(n) =n(n—

1)

(n—1)

= n(n—

1)(n —2)---2-1-TC)

= Alhl CL).

But by definition,

rd) = [#6

ap = [- O and ¢ as the new variable of integration. This yields, since t + O as B + Oandt — coas

po, oO

Tiwa+l=

/

co

e'ts* ts dt = s**! [

0

0

e's dr,

which is valid forx + 1 > 0. We thus obtain Ec

ee se

1

OO

=| 0

is

e '?r* dt,

s>0O,x>-l,

(5)

14.10

Periodic Functions

269

which in our Laplace transform notation says that

Lips r23

1

9s Gam of

(6)

gt

If in (6) we put x = —,

we get ré

Lge

52at.

But we already know that L{t~!/?} = (2/s)'/?. Hence

TG) = Vn.

(7)

Periodic Functions Suppose that the function F(t) is periodic with period w:

F(t +m) = F(t).

(1)

The function is completely determined by (1) once the nature of F(t) throughout one period, 0 < t < a, is given. If F(r) has a transform,

L{F(t)} =|

e' F(t) dt,

(2)

0

the integral can be written as a sum of integrals, oO

L{F(t)} = yf n=0

(n+lw

e' F(t) dt.

(3)

Aa

Let us put ¢ = nw + f. Then (3) becomes L{F(@)}=

vf n=0

exp (—snaw — sB)F(B + nw) dB. 0

But F(B +nw) = F(f), by iteration of (1). Hence

L{F(t)} = Lexp (sna) [

exp (—sB) F (B) dB.

(4)

0

n=0

The integral on the right in (4) is independent of n and we can sum the series

on the right:

> exp (—snw) = = n=0

n=0

[ exp (—sw)|" =

| l=

ese"

270

Chapter 14

The Laplace Transform

Theorem 14.12 /f F(t) has a Laplace transform and if F(t + w) = F(t), feo

F

(B)

dp

0

L{F(t)} =

1

—

(5)

ese

Next suppose that a function H(r) has a period 2c and that we demand that H(t) be zero throughout the right half of each period. That is, H(t + 2c) = H(t),

H(t) = g(t),

(6)

O

289

2. Hence we

write

y=t—ra(t —2). This gives ¢? for ¢ < 2 and zero for f > 2. Then we add the term 6a(t — 2) and finally arrive at

(7)

+ 6a(t — 2). —Pa(t — 2)

y(t) = 0?

The y of (7) is the y of our example and, of course, it can be written at once after

a little practice with the aw function. Unfortunately, the y of (7) is not yet in the best form for our purpose. theorem we wish to use gives us L{F(t —c)a(t —c)} =e

The

“ f(s).

Therefore, we must have the coefficient of a(t — 2) expressed as a function of (tf — 2). Since

—P?+6=—-(t? —4¢+4) —40 -2) +2, y(t) = 0? — (t — 2)?a(t — 2) — 4(¢ — 2)a(t — 2) + 2a(t — 2), from which it follows at once that Lyi}

=

?

2

Qe-*5

de

s3

s2

~ 53

De

8 5

EXAMPLE 15.11 Find and sketch a function g(t) for which

1 {3 g(t) =L

4e% . 4e7*s

rr!

gs?

s?

We know that L~!{4/s7} = 4t. By Theorem 15.3 we then get

is

_,

[4e> af

4]

4e—3s 2

a4

Det =)

and

L

= A(t — 3)a(t — 3).

(8)

290)

Chapter 15

Inverse Transforms

We may therefore write g(t) =3—4(t — Da(t — 1) + 40 — 3)ar(t — 3).

(9)

To write g(t) without the w function, consider first the interval O — 6yp? +x* =0.

19.

p?—xp—y=0.

pe —xpty =0.

20.

2p?+xp—2y =0.

y= px+kp’.

21.

2p?+xp—2y =0.

x8p? +3xp+9y = 0.

22.

pi+2xp—y=0.

41.

xt p? + 2x?yp —4=0.

23.

4xp?-—3yp+3=0.

10.

xp? —2yp+4x

24.

pi—xp+2y=0.

12.

3x4p* —xp—y =0. xp?>+(x-y)pt+tl—y=0.

25. 26.

5p*+6xp —2y =0. 2xp?+(2x—-y)p+l—y=0.

=0.

13; 14, 15.

pixp —- y+k)+a = 0.

27.

Sp? +3xp—y=0.

x®°p? — 3xp —3y

28.

p?+3xp—y=0.

29.

y=xp+xp?.

16.

xp* — 2yp? + 12x? =0.

30.

8y = 3x?+ p’.

y =x°p?

=0.

— xp.

334

Chapter 16

Nonlinear Equations

Dependent Variable Missing Consider a second-order equation,

fey x),

(1)

which does not contain the dependent variable y explicitly. Let us put

y=p. t

Then

» _ ap

y=

dx

and equation (1) may be replaced by

F (4 Pa)

=9

(2)

an equation of order one in p. If we can find p from equation (2), then y can be obtained from y’ = p by an integration. EXAMPLE 16.5 Solve the equation

xy” —(y'P—y' =0.

(3)

Because y does not appear explicitly in the differential equation (3), put y’ = p. Then

yadx’

: so equation (3) becomes

2

— p—p=0.

Separation of variables leads to dp

_ dx

p(p? +1)

x

or

dp

pdp

p

peti

_ dx

x’

16.9

Independent Variable Missing

335

from which

In|p| — 3 In(p? + 1) + In|cy| = In |x|

(4)

follows. Equation (4) yields

cp(p?+1)!"

=x,

(5)

which we wish to solve for p. From (5) we conclude that

cip’ = «(1+ p’), 9

2

x

e

ce? — x?

But p = y’, so we have

pen

(6) Vici — x?

The solutions of (6) are

yc, =F (cf — x)'”, or

x + (y— oc)? =e.

(7)

Equation (7) is the desired general solution of the differential equation (3). Note that in dividing by p early in the work we might have discarded the solutions y =k (ie., p = 0), where & is a constant. But (7) can be put in the form

c3(x? + y’) Feay +1 =0, with new arbitrary constants cz; and cy. yields the solution y = k.

Then the choice cz =

(8) 0, cy =

—1/k

a independent Variable Missing A second-order equation,

fy. y,y") = 9,

(1)

in which the independent variable x does not appear explicitly can be reduced to

a first-order equation in y and y’. Put

y =p,

336

= Chapter 16

Nonlinear Equations then

,»

ap _dydp_ “dx

dxdy

dp dy’

so equation (1) becomes d

= 0,

f (>. Pp, r?) y

(2)

We try to determine p in terms of y from equation (2) and then substitute the result into y’ = p. EXAMPLE 16.6 Solve the equation

(3)

yy” +0’ +1=0.

Since the independent variable does not appear explicitly in equation (3), we put y’ = p and obtain

_

ap

=

Pay

s

as before. Then equation (3) becomes d ype

dy

(4)

+ pp? +1=0,

in which the variables p and y are easily separated. From (4) it follows that

pépd

, dy @

pe+l1-

iy

9

from which

1 In (p? + 1) +Inly| = Infei), so p+i=

cy’,

We solve (5) for p and find that pS

(gay? yo

(5)

16.9

Independent Variable Missing

33'7

Therefore, dy

_ rac

dx

_

yy

y

or

+y(ct — yy’)! dy = dx. Then

Fei — y)'? =x - 0, from which we obtain the final result,

(x — OP +y? = ctf. i Exercises In Exercises 1 through 24, solve the differential equation.

1. y=x(y')*. 2. yy" +P =0.

3. yy" +O =O. 4. (y+ Dy" =".

5.

Jay” + (y’)? =0.

6.

Do Exercise 5 by another method.

7.

y" = 2yG").

8

yy” + (yr

9.

- (’? = 0.

12,

y"” = (y’).

yy’ +(y’ =0.

13.

y” =e*(y’)?*.

10.

y”cosx = y’.

14.

x?y"4+ (y')? =0.

ll.

xy" —x?y' =3 — x2.

15.

y*=1+(0').

16.

Do Exercise 15 by another method.

17,

(+y)y"+Q%+y=0,

9 21.

("PP —xy"+y'=0.

18.

x?y” = y'(3x —2y’).

22.

(y")? = 12y'(xy" — Dy,

19.

xy" = y'(2—3xy’).

23.

3yy’y” = (yy -1.

20.

y’=2x +(x? -y’)?.

24.

4y(y’)?y” = (y’)4 +3.

In Exercises 25 through 43, solve the equation and find a particular solution that satisfies the given boundary conditions.

— 2xy’ =0;

whenx

25.

xy" =F Cr

26.

x7y" + (y')? — 2xy’ = 0; whenx

=2,y=5,

y =

—4,

= 2, y =5, y’ =2.

Chapter 16

Nonlinear Equations

21.

xy”

28.

xy’ +y'+x=0;

aos

y” + By

30.

y=

Bs 32.

y” =x(y’)?; whenx =0,y=1,y'= z: y” = —e?”; whenx = 3, y= 0, y’ = 1.

33.

y” = —e~”; when x = 3, y=0, y =—1.

34, 35.

2y"” = sin2y; 2y”’ =sin2y;

36.

= y’ +x:

whenx=l,y=

5, y Sil.

whenx =2,y=—1, y' =}.

= 0. Check your result by solving the equation in two ways.

x(y'?;

whenx = 2, y=

47

y=

—j.

whenx =0,y =27/2,y' =1. whenx = 0, y = —7/2, y' = 1. Show that if you can perform the integrations encountered, you can solve any equation of the form y” = f(y).

a 38,

2y” = (y' sin2x; whenx =0,y=1,y’=1. y” = [lL +(’/P”. Solve in three ways, by considering the geometric

39.

yy” = (y')[1 — y’sin y — yy’ cos y].

40. Al. 42.

Dy’ +14+0'"P =14+ 09. x?y" = y'(2x — y’); whenx =—-l,y=5. y’=1. xty” = y'(y’ +23); whenx = 1, y = 2, y’ = 1.

43.

(y")? — 2y" + Gy)? — 2xy’ + x? =0;

significance of the equation, and by the methods of this chapter.

whenx =0. y= 5 and y’ = 1.

The Catenary Let a cable of uniformly distributed weight w (lb/ft) be suspended between two supports at points A and B as indicated in Figure 16.4. The cable will sag and there will be a lowest point V as indicated in the figure. We wish to determine the curve formed by the suspended cable. That curve is called the catenary. Choose coordinate axes as shown in Figure 16.5, the y-axis vertical through the point V and the x-axis horizontal and passing at a distance yo (to be chosen B

a

338

Figure 16.4

16.10

The Catenary

339

Figure 16.5

later) below V. Let s represent length (ft) of the cable measured from V to the variable point P with coordinates (x, y). Then the portion of the cable from V to P is subject to the three forces shown in Figure 16.5. Those forces are: (a)

The gravitational force ws (lb) acting downward through the center of gravity

of the portion of the cable from V to P (b)

The tension 7, (Ib) acting tangentially at P

(c)

The tension 7 (Ib) acting horizontally (again tangentially) at V

The tension 7) is a variable, the tension 7> is a constant. Since equilibrium is assumed, the algebraic sum of the vertical components of

these forces is zero and the algebraic sum of the horizontal components of these forces is also zero. Therefore, if @ is the angle of inclination, from the horizontal, of the tangent to the curve at the point (x, y), we have

T; sin@ — ws = 0

(1)

T, cos@ — 7, = 0.

(2)

and

But tan @ is the slope of the curve of the cable, so tang an

=

dy

—. ax

(3)

We may eliminate the variable tension 7, from equations (1) and (2) and obtain Ws

tan@ = —., an B

4 (4)

340

Chapter 16

Nonlinear Equations

The constant 7/w

has the dimension of a length.

Put 7;/w

= a (ft).

Then

equation (4) becomes

tan@ = -. a

(3)

From equations (3) and (5) we see that

s

dy

a

dx

Now we know from calculus that since s is the length of are of the curve, then ds —=,/1

dy\? —).

dx

7

v (3)

7)

From (6) we get lds

_ d’y

adx

dx?’

so the elimination of s yields the differential equation

dy adx*

|

dy\?

—_~=-,/]

—).

«a

8

me (=)

‘®)

The desired equation of the curve assumed by the suspended cable is that solution of the differential equation (8) which also satisfies the initial conditions

whenx=0,

y=ypand

| dx

=0.

(9)

Equation (8) fits into either of the types studied in this chapter. It is left as an exercise for the student to solve the differential equation (8) with the conditions (9) and arrive at the result x

y =acosh—+ a

yo -4a.

(10)

Then, of course, the sensible choice yy = a is made, so the equation of the desired curve (the catenary) is x

y =acosh-.

a

@ Miscellaneous Exercises In Exercises | through 27, solve the equation.

1.

x3p*+xtyp+4=0.

2.

6xp* — (3x +2y)p+y=0.

16.10

5. 6.

x°p* —2xp

—4y

5p? + 6xp —2y

= 0. = 0.

Do Exercise 6 by another method. y*p? —y(x+1l)p+x=0.

10.

Ay?p? —2xp + y =0.

eo

ae

Op? + 3xytp+y? =0. 4y3 p* —4xp+y =0.

The Catenary

4x° pp? + 12xtyp +9 =0.

Lb

p'+xp—3y=0.

12.

Do Exercise |1 by another method,

x? p> —2Axyp? ++ y*p+1=0.

LS;

xp? —(x?+1)p+x=0.

14.

l6xp* + 8yp + y® =0.

16.

p?>—2xp—y =0.

1}.

Do Exercise 16 by another method.

18. 19. 20.

9xy* p? —3y°p—-1=0.

22,

(p+ 1)*(y — px) = 1.

x*p* —(Qxy+)l)pt+ty?+1=0.

23.

p—

x°p? = 8(2y + xp). ep=@—yy.

24, 25,

xp? +y(1—x)p—y?

2ly 206. ab.

xp? +(k—-x—y)pty=0. xp — 2yp? + 4x? = 0. See Exercise 10.

p*+xp—y=0.

=0.

yp? —(x+y)pt+y=0.

341

BB Power Series Solutions

Ly .d

|

Linear Equations and Power Series The solution of linear equations with constant coefficients can be accomplished

by the methods developed earlier in the book. The general linear equation of the first order yields to an integrating factor as was seen in Chapter 2. For linear ordinary differential equations with variable coefficients and of order greater than one, probably the most generally effective method of attack is that based upon the use of power series. To simplify the work and the statement of theorems, the equations treated here will be restricted to those with polynomial coefficients. The difficulties to be encountered, the methods of attack, and the results accomplished all remain

essentially unchanged when the coefficients are permitted to be functions that have power series expansions valid about some point. (Such functions are called analytic functions.)

Consider the homogeneous linear equation of the second order,

bo(x)y" + bi (x)y’ + bo(x)y = 0,

(1)

with polynomial coefficients. If bo(x) does not vanish at x = O, then in some interval about x = O, staying away from the nearest point where bo(x) does vanish, it is safe to divide throughout by bo(x).

Thus we replace equation (1) by

y" + px)y’ +qx)y =0,

(2)

in which the coefficients p(x), g(x) are rational functions ofx with denominators that do not vanish at x = 0. We shall now show that it is reasonable to expect! a solution of (2) that is a

power series in x and that contains two arbitrary constants.

Let y = y(x) be a

solution of equation (2). We assign arbitrarily the values of y and y’ at x = 0;

y(0) = A, y'(0) = B. Equation (2) yields

y" (x) = —p(x)y'(x) — q(x) y(x),

(3)

| This is no proof. For proof, see, for example, E. D. Rainville, Intermediate Differential Equations, 2nd ed. (New York: Macmillan Publishing Company, 1964), pp. 67-71.

342

17.2.

Convergence of Power Series

343

so y”(0) may be computed directly, because p(x) and g(x) are well behaved at

x = 0. From equation (3) we get

y" (x) = —p(x)y"(x) — p’ a) y"(x) — g(x)’ () — g(x) 9),

(4)

so y”’(0) can be computed once y’(0) is known. The foregoing process can be continued as long as we wish; therefore, we can

determine successively y'” (0) for as many integral values of n as may be desired. Now by Maclaurin’s formula in calculus, oO

mr. yx) _= yO) + d YOR:

a

(5)

that is, the right member of (5) will converge to the value y(x) throughout some

interval about x = 0 if y(x) is sufficiently well behaved at and near x = 0. Thus we can determine the function y(x) and are led to a solution in power series form.

For actually obtaining the solutions for specific equations, we shall study another method, to be illustrated in examples, a technique far superior to the bruteforce method used above. What we have gained from the present discussion is the knowledge that it is reasonable to seek a power series solution. Once we know that, it remains only to develop good methods for finding the solution and theorems regarding the validity of the results found.

Convergence of Power Series From calculus we know that the power series co

)

aux”

n=0

converges either at x = 0 only, or for all finite x, or the series converges in an interval —R < x < R and diverges outside that interval. Unless the series

converges at only one point, it represents, where it does converge, a function f(x) in the sense that the series has at each value x the sum f(x). If oo

fa)

=

¥"

ane”,

—-R 2)

(6) Relation (6) may be used to obtain @, form are left arbitrary. We have

> 2 in terms of ap and a,, which

—4Aay 2

4

—4a|

AY

“B=

—4ar

5

3.9

—4a3

oe

—Aarp2

ok

—4ar,_|

DOK — 1)

OMS ORE DOK

In writing out these particular cases of equation (6), we have taken pains to keep the a’s with even or odd subscripts in separate columns. If we now multiply the

corresponding members of the equations of the first column, we obtain _

(je

| kak

a704°+++ ado, = “Ob,

0”

“**"G2k-2,

which simplifies to =| aap,

=

cp)

kak

(2k)!

dg

for k

=

1.

(7)

17.5

Solutions Near an Ordinary Point

349

A similar argument applied to the right column in the foregoing array gives us dy)

_ pi = (Ok+ i"!

fork ork

>= 1 1.

(8 )

We now wish to substitute the expressions for the a’s back into the assumed series for y, Co

(2)

p= Saye" n=0

we first rewrite (2) in the form

Since we have different forms for a2, and a2, y=agot+

y

aX

2k

aX

+ y

2k

aux’,

k=1

and then we use (7) and (8) to obtain

oo 1 = «| +2 y

Ak 2k (—1)*4*x —(Qk)!

oe t+),

+ ay

kak y2k+1 (—1)*4*x —_—_—__ Qk+)!——

9 (9)

|.

It is possible to rewrite equation (9) in the form (—1)* 0x7

(=1)*(2x)*

y = «| I a

io!oe

+ 4a

2 tea Qk+

1!

:

10 (10)

The two series in (10) are the Maclaurin series for the functions cos 2x and sin 2x, so that finally we may write y = day Cos 2x ++ 5a

sin 2.x.

Thus we have shown that the solution of equation (1) is a linear combination of cos2x and sin2x, a fact that could have been obtained immediately by the

methods of Chapter 7.

EXAMPLE

17.2

Solve the equation

(1 — x”)y" — 6xy’ —4y =0

(11)

near the ordinary point x = 0. The only singular points that this equation has in the finite plane are x = 1 and x = —1. Hence we know in advance that there is a solution %

y= > anx" n=0

valid in |x| < 1 and with ap and a, arbitrary.

(12)

350°

Chapter 17

Power Series Solutions

To determine the a,,

>

1, we substitute the y of equation (12) into the left

member of (11). We get oo

oO

Sinn

= Daa

oO

oO

— > n(n — l)a,x" — »

na=0

n=0

6na,x" — So 4anx"

n=0

=0,

n=0

or Le. @]

oO

Sinn

ge

— Yow

n=0

+5n+4ayx"

= 0,

(13)

n=0

in which we have combined series that contained the same powers of x. Next let us factor the coefficient in the second series in equation (13), writing oo

OO

Son Gi — Ta x**

Si(n +1)(n+4)a,x" = 0.

n=O

(14)

n=0

Relations for the determination of the a, will be obtained by using the fact that, for a power series to vanish identically over any interval, each coefficient in the series must be zero.

Therefore, we wish next to write the two series in equation

(14) in a form in which the exponents on x will be the same so that we can easily pick off the coefficient of each power of x. Let us shift the index in the second series, replacing n everywhere by (n — 2), Then the summation which started with the old = 0 will now start withn—2 = 0, or the new n = 2. Thus we obtain oo

Sonn

— jae

n=0

— Son

—1)(n + 2)a,_ox"~ = 0.

(15)

n=2

In equation (15) the coefficient of each separate power of x must be zero. For

n = (0 and n = 1, the second series has not yet started, so we get contributions from the first series only. In detail, we have n=

Q:

hol: n>2:

0-ay

=0,

0-a, =0, n(n — lay, — (a — 1)(n + 2)a,_2 = 0.

As we expected, ap and a are arbitrary. The relation form > 2 can be used to determine the other a’s in terms of ap and a). Since nin — 1) 40 for nm > 2, we can write

@y—2.

(16)

17.5

Solutions Near an Ordinary Point

351

Equation (16) is called a recurrence relation. It gives a, in terms of preceding a’s. In this particular case, each a is determined by the a with subscript two lower than its own and consequently, eventually, by either ap or a;, according to whether the original a had an even or an odd subscript. A recurrence relation is a special kind of difference equation. In difference equations the arguments of the unknown function (the subscripts in our relations) need not differ by integers. There are books and courses on difference equations and the calculus of finite differences paralleling the books and courses on differential equations and calculus.

It is convenient to arrange the iterated instances of the relation (16) in two

vertical columns [two columns because the subscripts in (16) differ by two], thus using successively n = 2,4,6,..., andn = 3,5,7,..., to obtain 4 ad. 2 = -a 5 0

=5 a3 a 30 _ 7

_ 6 d4

=

42

6

=

a

a5

=

ay

—

8 ag

6

4

53

9 aos

2k +2 Ik

aa, =

2k+3 eae

=

a2k+1

424-2

Next, obtain the product of corresponding members of the equations in the first column. The result, 4-6-8--+ (2k + 2) k

=

Le

dod4qdg:+

+: d2,

=

0024

2.4-6--- (Qk)

simplifies at once to k>l1:

ay,

=

(k +

Lao,

thus giving us each a with an even subscript in terms of ao. Similarly, from the right column in the array above we get 5-7-9.--(2k + 3)

og Ee k+l = 37507... (K+ 1).

k>1: or

2k+3 Rel:

A+)

=

i

a),

thus giving us each a with an odd subscript in terms of a).

ses A902;

352

Chapter 17

Power Series Solutions

We need next to substitute the expressions we have obtained for the a’s into

the assumed series for y, oO

y= aya,

(12)

n=0

The nature of our expressions for the a’s, depending on whether the subscript is odd or even, dictates that we should first split the series in (12) into two series, one containing all the terms with even subscripts and the other containing all the terms with odd subscripts. We write oO

oS

2S [a

| + [a

+ Seas

,

+ Ses

k=]

k=]

and then use our known results for a2, and a4) the form

to obtain the general solution in

= =, eS y=ao [ +S°kK+ oe | +a, E + ae ae). k=1

= k=1

(17)

3

These series converge at least for |x| < 1, as we know from the theory. That they

converge there and only there can be verified by applying elementary convergence tests. It happens in this example that the solution (17) may be written more simply as =

yy=a =a dX

k+1)x** ) +4

S

2

2k +3

——_ 3

rt

18 (18)

Indeed, the series can be expressed in terms of elementary functions,

do

Y=

ay(3x — x3)

7 BGP

Such simplifications may be important when they can be accomplished in a particular problem, but it must be realized that our goal was to obtain equation (17) and to know where it is a valid solution. Additional steps taken after that goal is reached are frequently irrelevant to the essential desire to find a computable solution of the differential equation. The length of the work in solving this particular equation is due largely to detailed steps, many of which will be taken mentally as we acquire more experience.

Solutions Near an Ordinary Point

17.5

353

EXAMPLE 17.3 Solve the equation y’+@—1)’y'

—4@

-

about the ordinary point x = 1. To solve an equation “about the point x = x”

Dy

=0

0)

Means to obtain solutions valid

of (x — xo). We in a region surrounding the point, solutions expressed in powers on (19) becomes first translate the axes, putting x — 1 = v. Then equati

d’y ,dy —4vy = 0. —+u—

20

(20)

,

dv

dv?

dy/dv, and so on. Always in a pure translation, x — x9 = v, we have dy/dx = As usual we put oo

(21)

av"

>

y=

n=0

and from (20) obtain oO

oO

Co

Sinn

a So 4anv"*

— 1)4,0"? + So nan

(22)

n=0

n=0

n=0

= 0,

Collecting like terms in (22) yields oC

co

n=0

n=0

Yo n(n — La,v"? + Si(n —4)a,v"*! =0, series, gives which, with a shift of index from n to (n — 3) in the second oo

oo

> n(n — Waqv"? + Yon —T)a,_30"~” = 0.

(23)

n=3

n=0

have Therefore, ao and a, are arbitrary, and for the remainder we n=2:

2a, =0,

n>3:

n(n— ay n =

la, + (n-7Tan-3 n—7 — n(n

— rr1)

= 0,

nrn

dp, from ay, and This time the a’s fall into three groups, those that come from from a2, We use three columns:

354

Chapter 17

Power Series Solutions

ay arbitrary =

a3

a, arbitrary

—4 —— 3. 940

da =0

9 4.3

a4

=———_da

ay

=

—1

I

as

=—

ag

=

—2 5. 4”

0

W665

2 ay = —~—=d5 9.3%

——

| gu

ajo tio = — 19.9”a

3k —7

a3, = ~3eGk

1)

3

0

—

=

0

0

a341=0,k > 2

a= in

— 2

7 as

= 0

0

a342 =0,k = 1.

With the usual multiplication scheme, the first column yields k>1:

“k=

(—1D*[(—4)(-1) - 2--- (3k — Tao

3.6-9-.-GON2-5-8--Gk

—-D]

For the a’s which are determined by a;, we see that a, = fay but that each of

the others is zero. Since a) = 0, all the a’s proportional to to it, ds, ag, and so on, are also zero.

For y we now have yea

4 5

So (= DAL(—4)(-1) - 2+ Bk — Ty]v 6-9.

GhIl2- 5-8.

Gk—]

a

+aj(u+

qv’).

Since v = x — I, the solution appears as

_ an fry!

(—1)*[(—4)(-1)- 2--- 8k — 1— 1)3* [3-6-9--- GR][2-5-8---Bk—D]

tal(e-1)+4+@—-1)*].

(24)

The original differential equation has no singular point in the finite plane, so the series in (24) is convergent for all finite x. In computations, of course, it is most

useful in the neighborhood of the point x = 1.

The coefficient of (x — 1)** is sufticiently simplify it. In the product 3 - 6 3. Thus we arrive at

complicated to warrant attempts to

-9-- + (3k), there are k factors, each a multiple of

3-6-9--- (3k) = 34(1-2-3---k) = 3*Kl. Furthermore, all but the first two factors inside the square brackets in the numerator also appear in the denominator,

With a little more argument, testing the terms

17.5

Solutions Near an Ordinary Point

355

k = 0, 1, 2 because the factors to be canceled do not appear until A > 2, it can be shown that

y=— 00)

Ss

fa — 1% 4D

DEE

GE

HE TAME

Da

,

¥

VD

(25) ba

In the exercises that follow, the equations are mostly homogeneous and of second order. Raising the order of the equation introduces nothing except additional labor, as can be seen by doing Exercise 16. A nonhomogeneous equation with right member having a power series expansion is theoretically no worse to handle than a homogeneous one; it is merely a matter of equating coefficients in the two power series. The treatment of equations leading to recurrence relations involving more than two different a’s is left for Chapter 18. i Exercises Unless requested otherwise, find the general solution valid near the origin. region of validity of the solution.

Always state the

2.

Solve the equation y” + y = 0 both by series and by elementary methods and compare your answers. Solve the equation y” — 9y = 0 by series and by elementary methods.

3. 4. 5.

y’+3xy' +3y =0. (1+4x2)y” —8y =0. (1—4x*)y"+ 8y =0.

14. 15. 16.

(1 —4x*)y” + Oxy’ — 4y = 0. (1+2x?)y" + 3xy’ —3y =0. y" +x2y" + S5xy’+3y =0.

6.

(L+x?)y” —4xy’ + 6y = 0. (4+x2)y"+10xy’+20y=0. (x? +4)y” + 2xy’— 12y =0. (x? —9)y” + 3xy’ — 3y =0.

17,

y’ +xy' +3y = x?, y"+2xy'+2y=0.

19. 20.

ll.

y” +2xy'+5y =0. (7? +4)y" + Oxy’ + 4y = 0.

21. 22.

y”+3xy'+7y =0. 2y’+9xy' — 36y =0. (x7 +4)y"+xy' —9y =0. (x?+4)y" + 3xy’ — By = 0.

12,

(1 +2x2)y"—Sxy'+3y=0.

23.

(1+9x7)y”— 18y = 0.

13.

y"+x?y =0.

24.

(1+3x*)y” + 13xy' + 7y =0.

25.

(1+2x?)y" + llxy’+ 9y =0.

26.

y" —2(x +3)y’ — 3y = 0. Solve about x = —3.

27.

y" +(x —2)y =0.

28.

(x? — 2x +2)y” —4(x — ly’ + 6y = 0. Solve aboutx = 1.

1.

7. 8. 9, 10.

18.

Solve about x = 2.

356

9 Chapter 17

Power Series Solutions

Computer Supplement Given the amount of algebraic manipulation used in series techniques, it is not surprising that Computer Algebra Systems can be very useful in solving differential equations requiring these techniques. There are two different approaches that we can employ, depending on the form of solution that we require. We will

illustrate both methods for the differential equation given in Example Section 17.5:

17.2 in

(1 — x*)y” — Oxy’ —4y = 0,

(1)

near the ordinary point x = 0. To simplify the calculations somewhat, we will add the initial conditions y(0) = 2 and y'(0) = 1.

The Maple command for entering this initial value problem is >Egqni:={ (1-x*2) *D(D(y) ) (x) -6*x*D(y) (x) -4*y(x)=0,D(y) (0)=1,y(0)=2}; If we employ the usual dsolve command we find that Maple produces a result,

but it is very complicated: >dsolve(Eqni,y(x));

:

y(x) =

(=1 $0?

5

2 Ge D/H 1? x?

3 (14a

et 1

of 1 ae x

+

(14+ 9)" @+ Dwi

pa

A minor modification to the dsolve command will produce as many terms of the series solution as we choose, in this case five. >Order:=5: >Soll:=dsolve(Eqni,y(x),series);

yx) = (2+

5

+4x° + 3x + 6x" + 0 (x°))

(2)

These are the first five terms of the power series solution and an error term of the

order of x°. Although this is an easy technique, for some applications what we really want is the recurrence relation for the coefficients. Maple can find this as well, albeit

with a little more work. We first change the format of the input for the equation: >Eqn2:=(1-x*2)

*di ff (y(x),x,x)-6*x*diff

(y(x),x)-4*y(x) =0;

Next, we create the representative terms a,_2x*~? + --- + agyox**? in a series

solution: >SeriesSol:=sum(a[n]

*x*n,n=k-2..k+2);

17.6

Computer Supplement

a7

We then substitute this solution into the differential equation and solve the resulting equation for a,, simplifying as we go. >simplify (simplify (subs (y(x)=SeriesSol,Egn2))); simplify (solve(coeff(lhs("),x*(k-2)),a[k]));

(kK + 2) ay_2 k This result agrees with equation (16) of Section 17.5. To check that this recurrence

relation agrees with our first result, we specify values for ag and a; based on the initial conditions, find the coefficients a2 ---as from the recurrence relation, and

form the polynomial with these coefficients. >a[O]:=2: sa GL): s=L: >for k from a(k]

:=(k+2)

2

to

5

do

*a[k-2]/k

od: >Sol2:=sum(a[j]*x*jJ,j=0..5);

5 3 TH 5 2txt4x?+ — +6x4 + which agrees with the first five terms of the solution that was found in equation (2) of this section.

BB Exercises

1. 2.

Use a computer to solve a variety of problems from the chapter. For the equation given in Example 17.1 in Section 17.5, add the initial conditions y(0) =

1, y’(0) = 2, and find the first m terms of the power series

solution form = 1---5. 3.

Have the computer plot the five functions from Exercise 2 on the same axes

along with the actual solution of the initial value problem.

MY Solutions Near Regular Singular Points

Regular Singular Points Suppose that the point x = xo is a singular point of the equation

bo(x)y” + bi(x)y' + ba(x)y = 0 with polynomial coefficients.

Then bo(x9)

(iy

= 0, so bo(x) has a factor (x — xo) to

some power. Let us put equation (1) into the form

y" + p(x)y’ + ¢(x)y = 0.

(2)

Because x = xg is a singular point and because p(x) and q(x) are rational functions of x, at least one (maybe both) of p(x) and g(x) has a denominator that contains the factor (x — xo). In what follows, we assume that both p(x) and g(x) have been reduced so that in each case the numerator and denominator contain no common factors. Ifx = xo is a singular point of equation (2), if the denominator of p(x) does not contain the factor (x — xg) to a higher power than one, and if the denominator of g(x) does not contain the factor (x — x9) to a power higher than two, then x = xg 1s called a regular singular point (R.S.P.) of equation (2). Ifx = xp is a singular point but is not a regular singular point, it is called an irregular singular point (1.S.P.).

EXAMPLE 18.1 Classify the singular points, in the finite plane, of the equation

ate

I

+ Dy

xPy! = ie? + Oe — Tp

For this equation X

PO)= CaaS D and

q(x)= 358

—(a? +3x = 1 x(x — 1)?2(x +.2)°

(3)

18.1

Regular Singular Points

359

The singular points in the finite plane are x = 0, 1, —2. Considerx = 0. The factor x is absent from the denominator of p(x) and it appears to the first power

in the denominator of g(x).

Hence x = 0 is a regular singular point of equa-

tion (3). Now consider x = 1. The factor (x — 1) appears to the second power in the denominator of p(x). That is a higher power than is permitted in the definition of a regular singular point. Hence it does not matter how (x — 1) appears in g(x); the point x = | is an irregular singular point. The factor (x + 2) appears to the first power in the denominator of p(x), just as high as is permitted, and to the first

power also in the denominator of g(x); therefore, x = —2 is a regular singular point.

In summary, equation (3) has in the finite plane the following singular points: regular singular points at x = 0, x = —2; irregular singular point atx = |. The methods of Section 18.10 will show that (3) also has an irregular singular point “at infinity.”

EXAMPLE

18.2

Classify the singular points in the finite plane for the equation

x*(x? + De — 1? y” +402

— Dy’ ++

Dy = 0.

Here p(x)

=

4

_

x 4+ De -1 x

“

-D@+)Q-D

and

q(x) =

x+]

x4(x + D(x — A(x — 1)?"

Therefore, the desired classification is R.S.P atx =1,

—i,

1;

LS.P atx = 0.

i Singular points of a linear equation of higher order are classified in much the

same way. For instance, the singular point x = x9 of the equation y+

py(x)y” + pa(x)y’ + p3(x)y =0

is called regular if the factor (x — x9) does not appear in the denominator of p; (x) to a power higher than one, of p2(x) to a power higher than two, of p3(x) toa power higher than three. If it is not regular, a singular point is irregular. This chapter is devoted to the solution of linear equations near regular singular

points. Solutions near irregular singular points present a great deal more difficulty and are not studied in this book.

360

9 Chapter 18

Solutions Near Regular Singular Points

Exercises For each equation, locate and classify all its singular points in the finite plane. 18.10 for the concept of a singular point “at infinity.”)

1

(See Section

x(e — Ly” +(x -— Dy’ +4xy =0.

2. x?(x? — 4)y” + 2x3y' + 3y = 0.

18.2)

3.

yt axy = 0.

4. 5.

x?y"+y=0. xty”"+y=0.

6.

G+ 1

—4Py"+@-4)y' +y =0.

7.

x?(x —2)y" + 3(x —2)y’ + y =0.

8.

x*(x —4)*y" + 3xy’ — (x —4)y = 0.

9.

x(x +2)y" + (x + 2)y’+4y = 0.

10. 11.

x(x+3)y"+y'-y=0. x3y”+4y =0.

12.

(«— 1) +2)y" + (x + 2)y’ +x7y =0.

13. 14. 15.

(L+4x*)y” + Oxy’ —9y = 0. (1+4x?)?y” + 6x(1 + 4x?)y’ — 9y = 0. (1+4x?)*y” + 6xy’ — 9y =0.

16.

(x—1)?(4 +4)?y" + (xe + 4)y’ + 7y = 0.

17.

(2x + 1)*y” + (2x + Dy’ — 8y = 0.

18.

xty’ + 2x3y’ +4y =0.

19.

Exercise 1, Section 17.3.

26.

Exercise 12, Section 17.3.

20.

Exercise 2, Section 17.3.

27,

Exercise 13, Section 17.3.

21.

Exercise 3, Section 17.3.

28,

Exercise 14, Section 17.3.

22.

Exercise 4, Section 17.3.

29.

Exercise 15, Section 17.3.

23.

Exercise 7, Section 17.3.

30.

Exercise 16, Section 17.3.

24.

Exercise 8, Section

17.3.

31.

Exercise 4, Section

25.

Exercise 9, Section 17.3.

32,

Exercise 7, Section 17.5.

17.5.

The Indicial Equation As in Chapter 17, whenever we wish to obtain solutions about a point other than x = 0, we first translate the origin to that point and then proceed with the usual technique. Hence we concentrate our attention on solutions valid about x = 0. We shall restrict our study to the interval x >

0, and if we then wish to find

18.2

The Indicial Equation

361

solutions of the same differential equation valid for x < 0, we can do so simply

by substituting x = —u and studying the resulting equation on the interval u > 0. Let x = 0 be a regular singular point of the equation

y’ + p(x)y’ +. q(x)y = 0,

(1)

where p and g are rational functions of x. Then p(x) cannot have in its denominator the factor x to a power higher than one. Therefore, p(x) =

ae

where r(x) is a rational function ofx and r(x) exists at x = 0. We know that

such a rational function, this r(x), has a power series expansion about x = 0. Then there exists the expansion ?

pe) = 2+

pi + pox

+ psx? ++

(2)

valid in some interval 0 < x < a.

By a similar argument we find that there exists an expansion go

ge)

=F

tant asx taux

+

(3)

t-~

valid in some interval 0 < x < pb. We shall see in a formal manner that it is reasonable to expect equation (1) to

have a solution of the form y=

00 y

a xe

=

fx®

+ ayx'te

+ aox?*

fee,

(4)

n=0

valid in an interval 0 < x < h, where fi is less than both a and hb.

If we put the series for y, p(x), and g(x) into equation (1) and consider only the first few terms, we get

c(e — lagx®* + (1 + e)cayx®! + (2+. c)(1 + c)agx® +++: Dp :

+ [= + pit pox +-: | [capx®"! + (1 + c)ayx® + (2 +e)agx!* +++] + |S

+ = +qo+

| apex’ aya?

gyn?

3 =] = 0.

Performing the indicated multiplications, we find that we have

e(c — Iagx®? + (1 + e)eayx® | + (2 +e) + e)aax® +++

+ pocagx’* + [po(l + c)ai + picag)x®| +--+ + qoagx°?

+ (goa)

+ gqiag|x*—!

+..-=0.

(5)

362

Chapter 18

Solutions Near Regular Singular Points From the fact that the coefficient of x°~- must vanish, we obtain

[c(c — 1) + poe + qolao = 0.

(6)

We may insist that aj 4 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

c* + (po — lie + qo = 9,

(7)

which is called the indicial equation (at x = 0). The po and gp are known constants; equation (7) is a quadratic equation giving us two roots, c = c, and c=>cC).

To distinguish between the roots of the indicial equation, we shall denote by c, the root whose real part is not smaller than the real part of the other root. Thus, if the roots are real, c; > c2; if the roots are imaginary, ‘i(c,) = (cz). For brevity we call c; the “larger” root.

Superficially, it appears that there should be two solutions of the form (4), one from each of these values of c. In each solution the ag should be arbitrary and

the succeeding a’s should be determined by equating to zero the coefficients of the higher powers of x (x°~!, x°, x'*°, and so on) in identity (5). This superficial conclusion is correct if the difference of the roots c; and c> is

not integral. If that difference is integral, however, a logarithmic term may enter the solution. The reasons for this strange behavior will be made clear when we develop a method for obtaining the solutions.

Form and Validity of the Solutions Near a Regular Singular Point Let x = 0 be a regular singular point of the equation

y" + p(x)y’ +. q(x)y = 0.

(1)

Then the functions xp(x) and xq (x) have Maclaurin series expansions that are

valid in some common interval 0 < x < b. It can be proved that equation (1) always has a general solution either of the form oO

oe =

A a

ae

+B

n=0

> by xt te

(2)

n=0

or of the form oO

y= (A+ Blnx) Yo an" n=O

oO

+ BY dyx™*®,

(3)

n=0

in which A and B are arbitrary constants. Furthermore, the infinite series that occur in the solutions above converge in the interval 0 < x < b.

184

18.4|

Indicial Equation with Difference of Roots Nonintegral

363

Indicial Equation with Difference of Roots Nonintegral The equation

2xy" + (1+x)y’—2y =0

(1)

has a regular singular point at x = 0 and no other singular points for finite x. Let

us assume that there is a solution of the form oO

y= >

ana,

ess 0)

(2)

n=O

Direct substitution of this y into (1) yields oO

> An te)(n te—

Dagx "+

oO

So(n

+ c)a,x"te-!

n=0

n=(0

co

oo A Ca, x"t?

fs Sin n=0

ane)

=i

a0"

n=0

or oo

Yon

oo

+c)(2n + 2c — 1)a,x"te-!

+ Sot

n=0

+e—2)a,x"**

= 0.

(3)

n=O

Having collected like terms, we next shift the index to bring all the exponents of x down to the smallest one present. This choice is used to get a recurrence relation for a, rather than one for a,,.; or some other a. In equation (3) we replace the index in the second summation by (7 — 1), thus getting oO

oo

n=0

n=l

Don + e)(2n + 2c — Vaux"! + Yn te

Baye"! = 0.

4)

Once more we reason that the total coefficient of each power of x in the left member of (4) must vanish. The second summation does not start its contribution

until 7 = 1. Hence the equations for the determination of c and the a’s are “= 0"

c(2c— lay = 0,

n>1:

(1 + c)(2n + 2e — la, + (n +e — 3)a,_)

= 0.

Since we may without loss of generality assume that ag 4 0, the indicial equation, that which determines c, is

c(2c — 1) = 0.

(5)

The indicial equation always comes from the n = Q term when the technique being used in this book is employed.

364

Chapter 18

Solutions Near Regular Singular Points

From (5) we see that cy ==e and c. = 0. The difference of the roots is S=c)-Q= i which is nonintegral. When s is not an integer, the method we are using always gives two linearly independent solutions of the form (2), one with each choice of c. Let us return to the recurrence relation using the value c = cy = 5 We have wei:

(a+ $)(2n + 1 — Da, 4: im $2 3 — 3)an-1

nel:

a,

=

—

= 9,

oe

2n(2n + 1)

As usual we use a vertical array and then form the product to get a formula for dy. We have ao arbitrary

(—3)ao

aqy=

2-3

(-lDay

(2n — 5)ay_| a,

=

—-

a,

G

2n(2n + 1)

so the product yields, forn > 1, 4,

=

"

(—1)"[(—3)(-D (1) --- (2n — 5) Jao

(6)

12-4-6-+--(Qn)J[3-5-7-+(nt 1)

The formula (6) may be simplified to form (—1)” - 3ao

Gn = Using a) =

2"n! (2n — 3)(Qn — 1I)Qn +1)

;

(7)

1, the a, from (7), and the pertinent value of c, c) = 5 we may

now write a particular solution. It is —

~*

yl/2

(—1y?3x24!/2

+

oa

oe

—DQn+1'

(8)

8

The notation y, is to emphasize that this particular solution corresponds to the root ¢, of the indicial equation. Our next task will be to get a particular solution y2 corresponding to the smaller root c7. Then the general solution, if it is desired, may be written at once as y = Ay; + By2 with A and B arbitrary constants.

18.4

Indicial Equation with Difference of Roots Nonintegral

365

In returning to the recurrence relation just above the indicial equation (5) with

the intention of using c = cy = 0, it is evident that the a’s will be different from those with c = c,. Hence it is wise to change notation. Let us use 6’s instead of a’s. With c = O, the recurrence relation becomes A=

1:

n(2n — 1)b, + (n — 3)b,_)

= 0.

The corresponding vertical array is

by arbitrary b=

— (—2)bo 1-1

b=

—

-

(-l)hy

2.5

by = _ O)by 3-5

h,

=

_f

—

ait

n(2n

—

1)

Then b, = 0 forn > 3 and, using bp = |, b; and b2 may be computed and found to have the values b} = 2 and by = ab = 7 Therefore, a second solution is

yp = 14 2x + 4x’,

(9)

Since the differential equation has no singular point, other than x = 0, in the finite plane, we conclude that the linearly independent solutions vy, of (8) and ys of (9) are valid at least for x > 0. The validity of (9) is evident in this particular example because the series terminates,

The student should associate with each solution the region of validity guaranteed by the general theorem quoted in Section 18.3, although from now on the printed answers to the exercises will omit the statement of the region of validity. i Exercises

Aion

eee

In Exercises | through 17, obtain two linearly independent solutions valid near the origin for x > 0. Always state the region of validity of each solution that you obtain.

2x(x+ Dy” +3(x + Dy’ -y =0. Ax?y"” + dxy' + (4x7 — Ly =0.

4x?y" + dxy’ — (4x7 + 1)y = 0. 4xy" + 3y'+3y =0.

2x7(1 —x)y” —x(1+7x)y +y =0. 2xy"+5(1 — 2x)y' -—Sy =0.

366

Chapter 18

Solutions Near Regular Singular Points

7.

8x2y"+ 10xy’-—(1+x)y =0.

8.

3xy"+(2-—x)y'-—2y =0.

9.

2x(x +3)y" —3(e4+ ly’ +2y

=0.

10.

2xy”+ (1 — 2x?)y’ —4xy =0.

ll.

x(4—x)y"4+

12. 13.

3x7y”+xy’-(1+-x)y =0. Qxy"+ (1+ 2x)y’ +4y = 0.

14. 15.

2xy”+ (14 2x)y’ — Sy =0. 2x?y” — 3x(1—x)y’ +2y = 0.

16.

2x?y"”+x(4x

17.

2xy” — (1+ 2x7)y’ —xy =0.

18.

The equation of Exercise 17 has a particular solution y. = exp (4x7) ob-

Q—x)y' +4y =0.

— ly’ +2Gx

—- l)y =0.

tained by the series method. Make a change of dependent vasiabile in the differential equation, using y = v exp (4 x7) (the device of Section 9.2), and thus obtain the general solution in “closed form.”

In Exercises 19 through 22, use the power series method to find solutions valid for x > 0. What is causing the recurrence relations to degenerate into one-term relations?

19.

=0. 2x?y”’+xy’-y

21.

9x?y"+2y =0.

20.

2x*y"” —3xy'+2y =0.

22.

2x?y"” + 5xy'’ —2y=0.

23.

Obtain dy/dx and d*y/dx? in terms of derivatives of y with respect to a

24.

new independent variable f related to x by ¢ = Inx for x > 0. Use the result of Exercise 23 to show that the change of independent variable from x to ¢, where t = In.x, transforms the equation! git arr

a, b, c constants, into a linear equation with constant coefficients. Solve Exercises 25 through 34 by the method implied by Exercise 24, that is, by changing the independent variable to f = Inx for x > 0.

25. Exercise

19.

26.

Exercise 20.

1 An equation such as the one of this exercise, which contains only terms of the kind ex' D‘y with c constant and k = 0, |, 2, 3, ..., is called an equation of Cauchy type, or of Euler type.

(8.5

Differentiation of a Product of Functions

27.

Exercise 21.

30.

x*y” + xy’—9y = 0.

28.

Exercise 22.

31.

x*y” —3xy'+4y =0.

29. 33.

x?y"+ xy’ — 12y =0. x?y"4+5xy’+5y =0.

32.

x7y"—5xy'+9y =0.

34.

367

(x° D? + 4x? D* — 8xD + 8)y = 0. You will need to extend the result of Exercise 23 to the third derivative, obtaining

Py ay

dy _ d°y dx3 dt

© dt?

dt’

Differentiation of a Product of Functions It will soon prove necessary for us to differentiate efficiently a product of a number of functions. Suppose that H = UjW2++ U3 Uy,

()

each of the u’s being a function of the parameter c. Let differentiation with respect to c be indicated by primes. Then from Inu = Inw,; + Inu. + Inw3+:-++Inu, it follows that

u ul | u2 u3 u —S=—F+HtH tee SS. “oo; og Uy i

f

f

t

f

Hence

owt

u Uy

t

uy Ws U3 f

u EE Uy

i

f

4g

(2)

Thus to differentiate a product, we may multiply the original product by a conversion factor (which converts the product into its derivative) consisting of the sum of the derivatives of the logarithms of the separate factors. When the factors involved are themselves powers of polynomials, there is a convenient way of forming the conversion factor mentally. That factor is the sum

of the conversion factors for the individual parts. The way most of us learned to differentiate a power of a quantity is to multiply the exponent, the derivative of the original quantity, and the quantity with its exponent lowered by one. Thus, if

y = (ac +b)*, then

dy =f = de °

ka

,

lac+b

the division by (ac + b) converting (ac + b)* into (ac + b)-!,

368

Chapter 18

Solutions Near Regular Singular Points EXAMPLE If

18.3

(e+)

“=

———_.———.. (4c — 1)3 (Fe + 2)6

then du —

dc

=u

2 1 4- + — c

12 - —

ctl

-

4-1

4?

.

Te+2

Note that the denominator factors in the function wu are thought of as numerator factors with negative exponents.

el EXAMPLE If

18.4 _

c+n

y= er DeE+2+-(e+n—1’ then dy

oc

ctn

dc

1

|

|

ctl

ct2

c+tn—-l1

1

I

_

a EXAMPLE If

18.5 27 3

w=

[(e + 2(e+3)-(+n + DP

;

then dw —_

dc

3 =

w

3

|

n

|

4

-—

c

i ewe

c+2

e¢+3

1 ———___——_

.

ctatl

@

Indicial Equation with Equal Roots When the indicial equation has equal roots, the method of Section 18.4 cannot

yield two linearly independent solutions. The work with one value of c would be a pure repetition of that with the other value of c. A new attack is needed.

18.6

Indicial Equation with Equal Roots

369

Consider the problem of solving the equation

x’y” + 3xy'+ (1 —2x)y =0

(1)

for x > 0. The roots of the indicial equation are equal, a fact that can be determined by setting up the indicial equation as developed in the theory, Section 18.2. Here 3

p(Qxy=-,

q(x)=

1— 2x

——,

x”

“we

so py = 3 and qo = |. The indicial equation is c+2c¢+1=0, with roots ¢; = co = —1. Any attempt to obtain solutions by putting

w y=

a

(2)

n=0

into equation (1) is certain to force us to choose c = —1

and thus get only one

solution. We know that we must not choose c yet if we are to get two solutions. Hence let us put the y of equation (2) into the left member of equation (1) and try to come as close as we can to making that left member zero without choosing c. It is convenient to have a notation for the left member of equation (1); let us

use

L(y) = x?y" + 3xy' + (1 — 2x)y.

(3)

For the y of equation (2) we find that

L(y) =

itn

t+c)(n te — la, x"t

+ \> 3(n + cayx"

n=0

n=0

co

oo

+ Y. DiggVF n=0

> 2a, enretl n=0

from which ow

oO

L(y) = Sohn +c)? +2(n +c) +

1a,x7* —

n=0

> Da, etter! n=0

The usual simplification leads to 0°

LO = Son n=0

oO

+eo+1Pa,x7 — » Day 1x", n=l

(4)

370

Chapter 18

Solutions Near Regular Singular Points

Recalling that the indicial equation comes from setting the coefficient in the n = Oterm equal to zero, we purposely avoid trying to make that term vanish yet. But by choosing the a’s and leaving c as a parameter, we can make every term but that first one in L(y) vanish. Therefore, we set equal to zero each coefficient,

except for the n = 0 term, of the various powers of x on the right side of equation (4); thus ne

1a,

(n+e+

Le

— 2a,-)

= 0.

©)

The successive application of the recurrence relation (5) will determine each

dn, n > 1, in terms of ap and c. Indeed, from the array _ dy|

2a

~ (e+2/ 2a

a=

(c+ 3)

24-1

ad, = ————

"

(e+n+ 1?

it follows by the usual multiplication device that 2" ay

n=l:

=

[ic +2)(e +3)

(c+n4+)P

To arrive at a specific solution, let us choose ap = 1. Using the a’s determined above, we write a y that is dependent upon both x and c, namely, oO

y(x,e) =x

x > 0,

+ Soan(ox"™,

(6)

n=l in which

Qn n>l:

ate

=

[c+ 2)(e + 3)---(c+nt+ DP

(7)

The y of equation (6) has been so determined that for that y, the right member of

equation (4) must reduce to a single term, the n = 0 term. That is, for the y(x, c) of equation (6), we have

Liy(x, c)] = (e+ 1)?x°.

(8)

A solution of the original differential equation is a function y for which L(y) = 0. Now we see why the choice c = —1 yields a solution; it makes the right member of equation (8) zero.

18.6

Indicial Equation with Equal Roots

371

But the factor (c+ 1) occurs squared in equation (8), an automatic consequence

of the equality of the roots of the indicial equation. We know from elementary calculus that if a function contains a power of a certain factor dependent upon c,

then the derivative with respect to c of that function contains the same factor to a power one lower than in the original. For equation (8), in particular, differentiation of each member with respect to

c yields ©

dc

yee. e]=L

a) dc

= 2(c + 1)x° + (c +1)?x* Inx,

(9)

the right member containing the factor (c + 1) to the first power, as we knew it

must from the theorem quoted above. In (9), the order of differentiations with respect to x and c was interchanged.

It is best to avoid the need for justifying such steps by verifying the solutions, (13) and (14) below, directly. The verification in this instance is straightforward but a bit lengthy and is omitted here. From equations (8) and (9) it can be seen that two solutions of the equation L(y) = Oare

vy =([[email protected]]_, =x, -) and _ | dy, y2 =

c)

ae

a

because c = —1 makes the right member of each of equations (8) and (9) vanish. That y; and yz are linearly independent will be evident later. We have oO

y(x,c) = xo+ s° a, (e)yx"t*

(6)

n=l

and we need dy(x, c)/dc. From (6) it follows that dy= (x, c)

_

gt l

= nx + D> ay(e)x

n+e

l

nx+) =

a) a()xer

_

n=!

n=1

which simplifies at once to the form

dy(x, c) dc

= y(x,c)Inx + S“al(e)x"™**, n=]

(10)

372

Chapter 18

Solutions Near Regular Singular Points

The solutions y; and y2 will be obtained by putting c = —1 in equations (6) and (10), that is, co

yi =a + So an(-Dx""|,

(11)

n=l CO

yo =yiInx+ So ah(-1)x" 1.

(12)

n=l

Therefore, we need to evaluate a,,(c) and a/,(c) at c = —1. We know that on

ay(c) =

[(e+2)(e+3).-(c+n4+ DP’

from which, by the method of Section 18.5, we obtain immediately

/

|

ay(e) = —2an(0)

1

5 c+t2

+

I

I c+3

bot}. c+tnt+l

We now use c = —1 to obtain Oe

ay(—1)

=

(n!)2

and

Anhons =”

aye

+t 2° b 3ste

te nf’

A frequently used notation for a partial sum of the harmonic series is useful here. It is

iGo

2

Se g st3 tactea nm gk

We can now write a’ (—1) more simply as a (-l)=-—

ntl ey aor

Finally, the desired solutions can be written in the form

yexte

yO ly

a7

(13)

and y2 = yi Inx -

Seer yet == (nl)? n=]

(14)

18.6

Indicial Equation with Equal Roots

373

The general solution, valid for x > 0, is

y= Ay, + Byo, with A and B arbitrary constants. The linear independence of y; and y2 should be evident because of the presence of In x in y3. In detail, xy, 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 the method is due to the fact that the n = 0 term in L(y) contains a square factor. @ Exercises Obtain two linearly independent solutions valid for x > 0 unless otherwise instructed,

1

x?y”—x(l+x)y’+y =0.

2.

4x?y” + (1—2x)y =0.

3, 4. 5.

x7?y”+x(x —3)y' +4y =0. x?y"4 3xy'+ (1+ 4x7)y = 0. x +x)y’+(1+5x)y +3y=0.

6.

x?y”—x(1+3x)y’ +

7.

x?y"+x(x— Dy’ +(1—x)y =0.

8.

x(x —2)y" +2(« — l)y’ — 2y =0.

9.

Solve the equation of Exercise 8 about the point x = 2.

— 6x)y =0.

10.

Solve aboutx = 4: 4(x — 4)?y” + (x —4)(x — 8)y’ +.xy = 0.

ll.

xy” + y’+xy

= 0.

This is known as Bessel’s equation of index zero.

It is widely encountered in both pure and applied mathematics. Sections 19.5 and 19.6.) 12,

xy" +(1—x7)y’—xy

13.

Show that

(See also

=0.

ae

P+atgte

|

+s

_=

l

Mae — i

and apply the result to simplification of the formula for y2 in the answer to Exercise 12.

14.

x*y”4+x(3+2x)y’ + (1 + 3x)y = 0. In simplifying y2, use the formula given in Exercise 13.

15.

4x7y" + 8x(x + 1l)y’+y =0.

16.

x?y”+ 3x(1 +x)y’+ (1 —3x)y = 0.

374

Chapter 18

Solutions Near Regular Singular Points

17. 18.

xy"+(—x)y

-y Refer to Exercise 17. change of dependent differential equation

=0. There one solution was found to be y; = e*. Use the variable y = ve*, to obtain the general solution of the in the form

y= ke [

Be? dp. ky

18.7|

Indicial Equation with Equal Roots: An Alternative In Section 18.6 we saw that when the indicial equation has equal roots, cz = cy,

two linearly independent solutions always appear of the form YI

=

xe!

+

oO an™,

(1)

n=] oo

y2 = yylnx + D0 by x"*",

(2)

n=1

where c1, d,, Db, are dependent upon the coefficients in the particular equation being solved.

It is possible to avoid some of the computational difficulties encountered in Section 18.6 in computing the b, of (2) by first determining c; and yj, then substituting the y2 of (2) directly into the differential equation and finding a recurrence relation that must be satisfied by b,. The resulting recurrence relation may well be difficult to solve in closed form, but at least we can successively produce as many of the }, as we choose.

EXAMPLE 18.6 For the differential equation of Section 18.6,

L(y) =x?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 4

oo

gn ynol

n=!

(n!)?

(4)

yx .

We know that a logarithmic solution of the form oO

yo = yi Inx + 0 bpx”| n=1

(5)

18.7

Indicial Equation with Equal Roots: An Alternative

375

exists and we shall determine the b,, by forcing this yz to be a solution of (3). We have oO

yy = yylnx tary + So(a = Ddpx”™, n=! fe. 2)

yy = yf nx + 2x7!y) — x?yy + So(n

~ Din — Dax,

n=]

so that

L(y2) = L(y) Inx + 2y, + 2xy, + SOG = I)

= 2)dy x"!

n=1 oO

oo

+ S230 = Wdyx”| + n=|

co

> by x”! — 2° Dax", n=1

n=!

from which oO

L(yo) = 2y + 2xy} +1 + > (07By — 2dp—1)x". The logarithmic term vanishes because L(y,) = 0. Substituting from (4) for y, yields

L(y2) (y2) ==2x7! 2x +2

oO

»

Qi ynnl

ae

— 2x7'+2

oO

2" (n

-

x,

iat

(n!)? co

+ by

+ Yo (nb,

=

2b,

a",

n=2

or

a

net!

L(x) = by +44 2, E by — 2By—1 + aa

| get

'

If y2 is to be a solution of equation (3), then b} = —4 and b, must satisfy the recurrence relation gn+l

nb, —2bpit——=0, (n!)?

n2=2.

(6)

A simple calculation yields b, = —3 and b; = —22/27, but a closed form for b,, is difficult to obtain from (6). In Section 18.6 we found the values of b, to be —2ntl H,,

Pn =

(n!)2

(7)

376

Chapter 18

Solutions Near Regular Singular Points

It is not difficult to show that this expression satisfies the recurrence relation (6), but it is difficult to obtain the form (7) from (6). Even so, for computational

purposes the alternative form for y2 given by the series (5) and the recurrence relation (6) is useful.

EXAMPLE 18.7 Solve the differential equation

xy" —x(1+x)y’+y=0

(8)

of Exercise 1 of Section 18.6. The two roots of the indicial equation are found to be c; = cy =

| and the

nonlogarithmic solution is oo

M=

ltl

(9)

;° 0)

ne

We seek a second solution of the form 00 y2 =

V1 Inx

+ Maa, n=]

so that co

yy = y,Inx +x

ly, + Sia

+ 1)b,x",

n=]

ya = y, nx 43x! yy — x7 a

+ Snir

$ 1b.

n=1

Substitution of these expressions into (8) gives oo

oO

2xy, —2y, — xy + yh

i Son

n=l

+ 1)b,x"*? =0

n=]

or CO

2xy, —(2+x)y1

+ b,x? + Sb,

—nb,_\)x"*!

= 0.

n=2

Using the series (9) for y; gives

py

eet

er

x

a

co

Bi -2F

yttl

3

oe

n=| oo

=

xt?

> “2a n=l

+ b,x?

+

S n=2

(nb,

—

nb,

part!

==—(.

18.8

Indicial Equation with Difference of Rootsa Positive Integer: Nonlogarithmic Case

SUT

which may be written yltl

(1 + b))x? + Ya maa

DI +Sin

§

by, — nby—)x"t!

= 0.

n=2

It follows that b} = —1 and nb, —nb,_;

I

+ ———

(n — 1)!

=0,

n>

2.

(10)

If this problem is solved by the method of Section 18.6, we obtain

=

—HA,

It is easy to show that this expression satisfies the recurrence relation (10).

eS Exercises For Exercises 2 through 8 of Section 18.6, find the logarithmic solution by finding a recurrence relation for the b, of equation (5) of this section.

Indicial Equation with Difference of Roots a Positive

Integer: Nonlogarithmic Case Consider the equation

xy” —(4+x)y’+2y =0.

(1)

As usual, let L(y) stand for the left member of (1) and put oO

p= yan.

(2)

n=0

At once we find that for the y of equation (2), the left member of equation (1)

takes the form

L(y) = Yin toe)(n Fe = 1) — An Fc) anx"*°! — Yn + = Dax", n=0

n=0

or

a

oo

L(y) = Di + e)(n $e = Spanx — YN te = 3a 10, n=0

nel

378

Chapter 18

Solutions Near Regular Singular Points

The indicial equation is c(c — 5) = 0, so Gy

Hoy

co = 0,

SSC

= 5.

We reason that we may hope for two power series solutions, one starting with an

x° term and the other with an x° term. If we use the large root c = 5 and try a series 00 y

ge

n=0

it is evident that we can get at most one solution; the x° term would never enter. On the other hand, if we use the smaller root c = O, then a trial solution of the form Co

y

ayx?

n=0

has a chance of picking up both solutions because the n = 5 (n = s) term does contain x°. If s is a positive integer, we try a series of the form (2) using the smaller root ce). If ap and a, both turn out to be arbitrary, we obtain the general solution by this method. Otherwise, the relation that should determine a, will be impossible

(with our usual assumption that @ 4 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 section. Let us return to the numerical problem.

Using the smaller root c = 0, we now

know that for

ym

(3)

Sa n=0

we get oO

oO

n=0

n=l

L(y) = )on@ = 5)anx"7! — SO(n = 3)ayx"|. Therefore, to make L(y) = 0, we must have n=0:

0+ ao = 0

(ao arbitrary),

n>1:

n(n — S)ay — (n — 3)dy_) = O.

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 Equation with Difference of Roots a Positive Integer: Nonlogarithmic Case

w= A

1s

379

— 4a, + 2a = 0,

Oh

— 6a, +a,

= 0,

n=3:

— 6a;

n=4:

— 4a, — a, = 0,

+ 0- ay

= 0,

n=5;

0- as —2a4 = 0,

my 2205

a, = ant n(n — 5)

It follows from these relations that

$0 ds is arbitrary. Each a,,n is found that

,

a4

y

0+ as

1

I]

CO

sw J aqy= 7a.

oOo

a3

do,

So

a2

ae

Ne

a

> 5, will be obtained from as. In the usual way it 3 ag

=

6-1

oat L_

7, qt

(n — 3) an

= SOOT

rs

n(m — 5)

from which 3-4-5---(n—3)

ay

=

3-4-5

[6-7-8--nla—5)! >

@—aa@—baw—S)!

Therefore, with ag and as arbitrary, the general solution may be written

=

y=

==

ao(l + 3+ x") +as | -b

L

aL

2

60x"

|x" + DY i = 5) lent — Dr 2) n=6 5

Se

a

The infinite series may also be written, with a shift of index, in the form

=

60x09 nl(nt+5)(n+4)(n + 3)

.

380

Chapter 18

Solutions Near Regular Singular Points

Before proceeding to the exercises, let us examine an equation for which the fortunate circumstance ag and a, both arbitrary does not occur. For the equation

xy” + x(1—x)y’ —(1+3x)y =0

(4)

the trial oo y=

agurre n=0

leads to

++

L(y) = Soa +e4+])(n+e — Dayx"h — Sia n=0

Daye".

n=]

Since c; = | and c. = —1, we use c = —1 and find the relations ne

Ll:

n(n — 2)ay, — (1 + l)ay-)

= 0,

with ao arbitrary. Let us write down the separate relations out to the critical one, need

;

— a; — 2ap = 0,

n=2: n

>

3

0-a — 3a, = 0, :

&,

1+

Da,

= (a+ Dan-1

n(n — 2) It follows that a, = 0. aj>=

—2a0, 3a)

=

—6a.

These relations cannot be satisfied except by choosing ag = 0. But if that is done, a2 will be the only arbitrary constant, and the only solution coming out of

the work will be that corresponding to the large value of c, c =

1. This is an

instance where a logarithmic solution is indicated and the equation will be solved in the next section. A good way to waste time is to use aj = 0, a; = O, a arbitrary, and to determine a,,, 1 > 3, from the recurrence relation above. In that way extra work can be done to get a solution that will be reobtained automatically when solving

the equation by the method of the next section. Exercises Obtain the general solution near « = 0 except when instructed otherwise. validity of each solution.

lL.

x?y" + 2x(x — 2)y’ + 2(2 — 3x)y =0.

State the region of

US Bo

18.9

Indicial Equation with Difference of Roots a Positive Integer: Logarithmic Case

381

x*(1 + 2x)y” + 2x(1 + 6x)y’ —2y =0. x?y" +x(2+3x)y’ —2y = 0.

GO! gl

ON Gr

xy” —(3+x)y'+2y =0.

9.

x(L+x)y" +(x + 5)y’ —4y =0. Solve the equation of Exercise 5 about the point x = —1. x?y" + x%y' —2y =0. x(1—x)y"” —3y'+2y =0. Solve the equation of Exercise 8 about the point x = 1.

10.

3y =0. xy” + (44+ 3x)y’+

12.

xy”’+(34 2x)y’+4y =0.

Ll.

xy” —2(4+2)y'+4y

=0.

13.

x(x +3)y" —9y'-—6y = 0.

14.

x(1 —2x)y"—2(2+x)y’

15.

xy” + (x3 — ly’ +x7y

16.

x2(4x — ly” + x(5x + ly’ +3y = 0.

+ 8y = 0. =0.

Indicial Equation with Difference of Roots a Positive Integer: Logarithmic Case In the preceding section we examined the equation

x?y"+x(l—x)y’-U4+3x)y=0

x>0

and found that its indicial equation has roots c; = 1, cp = —1.

(1) Since there is no

power series solution starting with x, we suspect the presence of a logarithmic term and start to treat the equation in the manner of the previous logarithmic case, that of equal roots. From the assumed form _ y

oo y

—_

n+e AnX

n=0

we easily determine the left member of equation (1) to be oOo oo

L(y) = Yon

+e+Dint+e—

Vax"

- Sit

n=0 co

= Sia

Keb Sa

n=0 oo

tot+l)(nte—

layx"

— So(n 4e+t Da,ix™.

n=0

n=1

As usual, each term after the first one in the series for L(y) can be made zero by choosing the a,, 2 > 1, without choosing c. Let us put n>1:

Gn

_

(n

+e

t+

2)ay-1

(antec+1I(nte—1)’

382

Chapter 18

Solutions Near Regular Singular Points

from which it follows at once that n>:

a=

(c+ 3)(c +4)---(c

L@-+ 2-3):

(e+ a

+n + 2)ag

Tle)

way

=. (e-+-a — I

or

n>:

ay

_

(c+n+2)ag

~ (e+ 2)fele+ le (e+n—1)]

From the a,, obtained above, all terms after the first one in the power series for L(y) have been made to vanish, so with

(c +n + 2)anx"t¢ Y=_ ax +2,e Gepee+)-e4ao Dl n=|

”

it must follow that

(3)

L(y) = (ce + 1)(e — lagx’. From

the larger root, c =

1, only one solution can be obtained.

From

the

smaller root, c = —1, two solutions would be available, following the technique of using y(x, ¢) and dy(x, c)/dec, as in the case of equal roots if the right member of (3) contained the factor (c + 1)? instead of just (¢ + 1) to the first power. But ao is still arbitrary, so we take

ag = (c+ 1) to get the desired square factor on the right in equation (3). Another way of seeing that it is desirable to choose ay = (c + 1) is as follows.

We know that eventually it is going to be necessary to use c = —1 in equation (2). But within the series, the denominator contains the factor (c + 1) for all terms after the n = | term. As equation (2) stands now, the terms n > 2 would not exist with c = —1. Therefore, we remove the troublesome factor (¢ + 1) from the denominator by choosing ay = (c + 1). With ag = (c + 1) we have y(x,e) = (c+ 1)xo+

| n=l

(e+ l(c +n +2)x"¢ (c+ 2fe(e+1)---(c+tn—-1)]

(4)

for which

Liy(x, o)] = (e+ 1° - Dx*.

(5)

The same argument as the one used when the indicial equation had equal roots

shows that the two linearly independent solutions being sought may be obtained as yi = yx, -1)

(6)

18.9

Indicial Equation with Difference of Roots a Positive Integer: Logarithmic Case

383

and

Ae ( a "ee

|

dc

(7)

c=-1

Naturally, it is wise to cancel the factor (c + 1) from the numerator and denominator in the terms of the series in (4). But the factor (c + 1) does not enter the denominator until the term n = 2. Therefore, it seems best to write out the

terms that far separately. We rewrite equation (4) as y(x,c) = (e+ 1)x°+

(e+ 1(e+3)xit©

(c+ 4)x7t

(c+ 2)c

(c+2)e

(c+n+2)x"*

oO

.

+ 2& (c+ 2el(c+2)(e+ 3)---(C+n—- DI .

(8)

Differentiation with respect to c of the members of equation (8) yields

dy(x, y(x,c) = 5, a dc

1

e+1

c+3

c+2

(c + 4)x2t¢

l

|

|

(e+2)c

|[ct+4

ct+2

c

(c+ 2)c

pSn=3 beta

|

1

£

1

(e+ Iye+ 3)xit

c¢

tet sta tese [ads “o (e+ 2)cel(c +2)(c +3)---(e+n—1)]

n+e

|

(9)

Allthat remains to be done is to get y, and y2 by using c = —1 inthe expressions above for y(x, c) and dy(x, c)/dc. In the third term on the right in equation (9), we first (mentally) insert the factor (c + 1) throughout the quantity in the curly brackets.

The desired solutions are thus found to be _—.

yy =O-x

(n

yo!

Sy

+0-x

.

t+)

he 1)x?7!

oo ee

a

oe

ih

oho

n=3

ee

and

yp = yy Inx +x7! —2x° —3xf{f-14+ 1}

wo nt I)x +2, 1

n=3

n-l

|

{——

1

+ (145+ (-Dfl -2---@—2)] 2 if

[xf

7

a

wie

|

+.)

Solutions Near Regular Singular Points

These results can be written more compactly as ¥y=

—3xx

-—

= 8 (n+ 1)x" ole ee ! dFl aD

10 (10)

and

1) Hp-2]x"7!

[1 — (2+

5 Ee (n we pilns-tat—2—-¢- n=3 — 2)!

(11)

It is also possible to absorb one more term into the summation and to improve the appearance of these results. The student can show that (n + 3)x"1 ie ae co

!

n=0

es

and =

n+l

l—-

yee yo = yi Inx+x7'-2- n=O

i. |

as long as the common conventions (definitions) Hp = 0 and 0! = 1 are used.

The general solution of the original differential equation is

y = Ay + Bya, and is valid for all x > QO, since the differential equation has no other singular

points in the finite plane. The step taken in passing from equation (4) to equation (8) should be used

regularly in this type of solution. Without its use, indeterminate forms that may cause confusion will be encountered, An essential point in this method is the choice aj = (c¢ — cz), where c2 is the smaller root of the indicial equation.

Exercises

be os

Find two linearly independent solutions, valid for x > O, unless otherwise instructed.

xy"+y=0. x?y” — 3xy'+ 3+4x)y

=0.

2xy" + 6y +y = 0. oy Pe ge

Chapter 18

Ax?y” 4+ Ix(2—x)y’ — (1 4+3x)y =0. x?y"” —x(6+x)y’ + 10y = 0. ky" +3 -+2e)y' +89 — 0.

a

84

x(l—x)y"” +2(1 —x)y’ +2y =0.

Solution for Large x

18.10

385

Show that the answers to Exercise 7 may be replaced by xt!

co

y3 = 1—- x;

ya = ysInx — tx!

+2

- 5° n=3

(n — 1)(n — 2)

Solve the equation of Exercise 7 near the point x = 1. x2y" + xy’ + (x? — 1)y = 0. This is Bessel’s equation of index one. (See

9. 10.

also Sections 19.5 and 19.6.)

11. 12.

x?y" — 5xy' + (8+ 5x)y = 0.

13. 14.

9x?y" — LSxy' + 7(1 +x)y = 0.

xy’ + (3—x)y’ —Sy =0.

x2y" + x(1—2x)y’ —(x+ Dy =90.

Solution for Large x The power series solutions that have been studied up to this stage converge in regions surrounding some point x = xo, usually the origin. Such solutions, although they may converge for large values of x, are apt to do so with discouraging slowness. inves

For this reason, and for others of a more theoretical nature, we shall

tigate the problem of obtaining solutions particularly useful for large values

of x.

Consider the equation

bo(x)y" + bi@x)y’ + ba(x)y = 0

(1)

with polynomial coefficients. Let us put

w=. x

(2)

Then

dy dwdy

dx

1 dy

dxdw

_

, dy

x?dw

dw

and

d*y dw d OF ae dx? dx dw —

2

ey?

> dy dw

2

d°y

° ( "dw? coat

ic

4 d’y 4 tur, 3 dy =yt— 0 dwt a

Pe dw

2

dy

aE

we)

386

Chapter 18

Solutions Near Regular Singular Points

Thus equation (1) is transformed into the following equation in y and w:

1 2 pe (2) wt. 4 | eats w

dw

1 (2) — we,

t\] FL

w

4

w)|

dw

1 tm (—)y=0. w

8)

Since bo, b;, and bz are polynomials, equation (3) is readily converted into an

equation with polynomial coefficients. If the point w = 0 is an ordinary point or a regular singular point of equation (3), then our previous method of attack will yield solutions valid for small w. But

w = 1/x, so small w means large x. As a matter of terminology, whatever is true about equation (3) at w = Ois said to be true of equation (1) “at the point at infinity.” For instance, if the transformed

equation has w = 0 as an ordinary point, then we say that equation (1) has an ordinary point at infinity. (See Exercises 1 through 6.) EXAMPLE 18.8 Obtain solutions valid for large x for the equation

xy" + 3x —ly'+y=0.

(4)

This equation has an irregular singular point at the origin and has no other singular points in the finite plane. To investigate the nature of equation (4) for large x, put x = 1/w. We have already found that dy dx

d

= yrdw

(5)

and

d?y

—

=

4d’y:

Pe

dw?

® dw

dxt

;, dy

6

(8)

With the aid of (5) and (6) we see that equation (4) becomes

| 4a’y w (» dw?

—

, dy ew dud

3 \w

2w? —

3 dy dw)?

——1){-—w*—

=,

or dy

dy

°— dw- — w(l-w)— + y =0, dw

(7)

an equation that we wish to solve about w = 0. Since w = 0 is a regular singular point of equation (7), the point at infinity is a regular singular point of equation (4). From the assumed form oo

y=

» n=0

dn wre,

18.10

Solution for Large x

387

it follows from equation (7) by our usual methods that oO

&

te— Wranw" + Site = Vapi", L(y) = Do n=0 n=] in which L(y) now represents the left member of equation (7). The indicial equation has roots c = 1, 1. Let us set up y(w, c) as usual, From the recurrence relation mek

en n+ce-1

it follows that

(—1)"ao

~ cle (e+n= + De 1) Hence we choose \

°

cS

(—1)"w"t?

D’

Sa

an

FE

1

y(w,c) =w and then find that

Ry,

ak dc

zg

1

a

hor

"

y(w,c)Inw — o

1

+te

ele +1)---(c+n—1)

n=l

Employing the root c = 1, we arrive at the solutions a

navy

n+l

or

and _

y2 = yy nw

—

wt!

evr

.

Therefore, the original differential equation has the two linearly independent solutions co

(1x

n=

and 2

(_

"Fy,

ee |

n=1

ne

—n—l

(9)

188

Chapter 18

Solutions Near Regular Singular Points These solutions are valid for all x > 0.

|_| Exercises In Exercises | through 6, the singular points in the finite plane have already been located and classified. For each equation, determine whether the point at infinity is an ordinary point (O.P.), a regular singular point point (R.S.P.), or an irregular singular point (1.S.P.). Do not solve the problems.

lL.

x3(x — Dy” + ( — 1)y’ + 4ay = 0. (Exercise 1, Section 18.1.)

2.

x7(x? —4)y"” + 2x3y’ + 3y = 0. (Exercise 2, Section 18.1.)

3. 4.

y"”+xy =0. (Exercise 3, Section 18.1.) x?y"+ y =0. (Exercise 4, Section 18.1.)

5.

xty” + y =0. (Exercise 5, Section 18.1.)

6.

xty” +2x3y' + 4y = 0. (Exercise 18, Section 18.1.)

In Exercises 7 through 19, find solutions valid for large positive x unless otherwise instructed.

7. 8.

xty” +x(1 + 2x?)y’ + 5y = 0. 2x3y"” —x(2—5x)y’+y =0.

9.

x(l—x)y"” —3y' + 2y = 0, the equation of Exercise 8, Section 18.8.

10.

xty"” + x(2—3x)y’ — (5 — 4x)y = 0. See Exercise 13, Section 18.6.

11.

2x*(x — 1)y” +.x(5x —3)y’ +xe+

Dy =0.

12.

Solve the equation of Exercise 11 about the point x = 0.

13.

2x7(1 —x)y” —S5x(1+x)y'+(5—x)y

14.

Solve the equation of Exercise 13 about the point x = 0.

15.

x(+x)y"+(1+5x)y’+3y

=0.

= 0, the equation of Exercise 5, Section 18.6.

16.

x*(44x7)y" +2x(44+x7)y’'+y =0.

17.

x(1—x)y"

+ (1 — 4x)y’ — 2y

=

0, the equation of Exercise

18 in the

Miscellaneous Exercises at the end of this chapter. 18. 19.

x + 4x)y" + (1 + 8x)y’ + y = 0, the equation of Exercise 49 in the Miscellaneous Exercises at the end of this chapter. The equation of Exercise 6.

Many-Term Recurrence Relations In solving an equation near a regular singular point, it will sometimes happen that a many-term recurrence relation is encountered. In nonlogarithmic cases, the

18.11

Many-Term Recurrence Relations

389

methods developed in Chapter 18 are easily applied and no complications result except that usually no explicit formula will be obtained for the coefficients. In logarithmic cases, the methods introduced in Chapter 18, constructing y(x,c)

can become awkward

and dy(x,c)/dc,

when

a many-term

recurrence

relation is present. There is another attack which has its good points, Consider the problem of solving the equation

(1)

t+ txtx)y =0 Ly) =xy"+xB4+xny for x > 0. From co

(2)

y= Daa n=0

it is easily shown that oo

L(y) = Sota

co

e+

oO

+ c)ay— x"

4 Yi

1) a,x"

n=0

n=]

ae

+

(3)

n=2

Therefore, the indicial equation is (c + 17 =0.

Since the roots of the indicial equation are equal, c = —1, —1, it follows that there exist the solutions y=

So anx"

|,

(4)

n=0

oO

yp = ypInx

+ So dyx"|,

(5)

n=]

valid for x > 0, The region of validity is obtained from the differential equation; the form of the solutions can be seen by the reasoning in Section 18.6. We shall determine the a,, n > 0, by requiring that L(y,) = 0. Then the b,,

n > 1, will be determined in terms of the a, by requiring that L(y2) = 0. From L(y)) = 0 it follows that oO

oo

CO

n=0

n=l

n=2

Soran” | + Yo = Danis” | + DY anax"™| = 0. Let us choose ay = 1. Then the rest of the a’s are determined by n=l:

a, +0-a) = 0,

a Sls

nan + (n — 1)ag—1 + Gn-2 = 0.

Therefore, one solution of the differential equation (1) is oO

i= > mee, n=0

(6)

390

Chapter 18

Solutions Near Regular Singular Points in which ag = 1, a; = 0,

n>=2:

(a =a

ay = —

1)an—4

+

5

ay—2

ai

Next we wish to require that L(y2) = 0. From

y2 = yp Inx +

yx"!

(5)

n=1

it follows that

yy = yj Inx +x! y, + \0@ = Ddyx”? n=1

and oo

ys = yi nx + 2x7!y) — xy

+ Sin

— yn — Db,a*".

n=l

Now direct computation of L(y.) yields co

L(y) = L(y) nx + 2xy, — yy tay +3y + >

7b

n=] 00

+ D1 = 1x

CO

+ Dna n=3

n=2

Since L(y,) = 0, the requirement L(y2) = 0 leads to the equation oo

ye

co

bat

oO

i Sia

n=]

_

1)b,

x"!

i

n=2

S-

bax!

n=3

= —2xy, — 2y, — xy oo __

oo

S 2nayx"

|

_

n=]

saa,

(7)

n=l

in which the right member has been simplified by using equation (6). From the identity (7), relations for the determination of the b, from the a, follow.

are: =;

b;

=

n=2:

Abo

n>3:

n’ by

—2a, + b)

=

+ (a —

— ap, —4ax

— ay,

1)bn-1

+ Ppa

= —2nay

— Gy-1-

They

18.11

Many-Term Recurrence Relations

391

Therefore, the original differential equation has the two linearly independent solutions given by the y, of equation (6) and by co

w=

Inx

+

yee,

(8)

n=]

in which b} = —1, bz = 3, n>

3%

jp

Se

If the indicial equation has roots that differ by a positive integer and if a logarithmic solution exists, then the two solutions will have the form co

n=

yaa, n=0 oO

¥2

=

BI

Inx ae

;

bax".

n=0

where c, is the larger and c2 the smaller root of the indicial equation. The a, and

b, can still be determined by the procedure used in this section. i Exercises

Solve each equation for x > 0 unless otherwise instructed.

1.

x2y"4+3xy' + (l+x+x)y

2.

Qx(l—x)y’ +

3.

xy” ty +xC14+x)y =0.

4.

— (1 — 3x + 6x7)y = 0. x2y"+x(1+x)y’

5.

Show that the series in the answer to Exercise 4 start out as follows: y = ao(x7'

6.

=0.

— 2x)y+Q24+x)y =0.

+2+44x?

r +...) — 3x3 4+ Bart - Bar

— Lett BP bo). bane fat + Be equation has roots c =

xy” +xy’ + (1+x+)y = 0. Here the indicial

0,

c = 1, and an attempt to get a complete solution without Inx fails. Then we put oo y=

)

a,"

n=0 oo yo

=

1

Inx

+

bax”: n=0

392

Chapter 18

Solutions Near Regular Singular Points

The coefficient b;(s = 1) turns out to be arbitrary and we choose it to be zero. Show that the indicated y,; and y2 are solutions if |

|

|

ao9=1,a)=—-1l,a.= 5,43 = —-5%, ag = 5, mes:

a,

(nh +

=

l)da-1

+ ans

n(a+

1)

oS

and if the b’s are given by by = —1, b) = O0(so chosen), bo = 1, by = —3, by =H, n>

5. b,

=

re

+

bys

_

(2n

n(n— 1) 7.

~~

Vay

+

Gy-2

n(n— 1)

For the y, in Exercise 6, prove that the a, alternate in sign. Also compute the terms of y; out to the x’ term.

8.

x(¢ —2)2y" — 2x — .

2y’ + 2y = 0.

Solve the equation of Exercise 8 for x > 2.

10.

2xy"+

1 —x)y'-U4+x)y

11.

Show that the answers to Exercise 10 are also given by y=

e*,

=0.

y=

e|

exp (—3e

)dBp.

0

18.12}

Summary Confronted by a linear equation

L(y) =9,

(1)

we first determine the location and nature of the singular points of the equation. In practice, the use to which the results will be put will dictate that solutions are desired near a certain point or points. In seeking solutions valid about the point

xX = Xo, always first translate the origin, putting x — xq = v. Solutions valid near an ordinary point x = 0 of equation (1) take the form CO

y= Daa"

(2)

n=0

with a and a, arbitrary, if equation (1) is of second order. If x = 0 is a regular singular point of equation (1) and we wish to get solutions valid for x > 0, we first put

y= Dax,

(3)

18/2

For the y of (3), we obtain the series for L(y) by substitution.

Summary

393

From the n = 0

term of that series, the indicial equation may be written. When the difference of the roots of the indicial equation is not an integer, or if the roots are equal, the technique is straightforward following the method of Section 18.4 or 18.6. When the roots differ by a nonzero integer, the solution may, or may not, involve Inx, The recurrence relation for n = s, where s is the difference of the roots, is the critical one.

We

must then determine

whether the relations for

n= 1, 2, ..., s leave ag and a, both arbitrary. If they do, two power series of the form (3) will be solutions of the differential equation. If aj and a, are not

both arbitrary, the case is logarithmic. Then the device of Section 18.9 may be used. The technique can be varied, if desired, by always choosing the a, in terms of c so that the power series for L(y) reduces to a single term. Thus a series of the

form y(x,c)

= do c -[- »

fox

(4)

|

n=|

will be determined for which

Liy(x, )] = aole — er) (e — 2) x,

(5)

where k = 0 or | for the equations being treated here and c, and c» are the roots of the indicial equation. Then it can be determined from the actual coefficients in (4) whether the use of ¢ = c, andc = ¢) will result in two solutions of the differential

equation. If c, = co, the results would be identical and the use of dy(x, c)/dc is indicated. The other logarithmic case will be identified by the fact that some one or more of the coefficients f,(c) will not exist when ¢ = root. dg

Then

again the differentiation process

is needed,

cz, the smaller

after the introduction

of

=C—C.

The method sketched above has a disadvantage in that it seems to tempt the user into automatic application of rules, always a dangerous procedure in mathematics.

When a student thoroughly understands what is happening in each of the four possible cases, this method may safely be used and saves some labor. Extension of the methods of this and the preceding chapter to linear equations of higher order is direct. As an example, a fourth-order equation whose indicial equation has roots c = 2, 2, 2, 4 would be treated as follows, A series would be determined for y(x, c),

yx,

c) =a9

[»

4

S° th ow n=!

for which the left member of the original equation reduces to one term, such as Lly(x, c)] = (e - 2) (2¢ — 1)aox°.

Chapter 18

Solutions Near Regular Singular Points

Then four linearly independent solutions could be obtained: yy = y(x, 2);

y=

dy(x,c

a

dc

dy(x,c)]

m= laa |!

ae

Miscellaneous Exercises In each exercise, obtain solutions valid for x > 0.

xy” —(2+x)y’-y =0. x?y" +2x7y' —2y =0.

2x? y" —x(QQx+7)y +24 5)y =0. x*(1 + x2)y" + 2x(3 + x2)y' + Gy = 0.

(1 —x2)y"— 10xy’ — 18y =0 2xy" + (1+2x)y' —3y =0.

y" + 2xy’ — 8y = 0. x(1 —x*)y" — (7427) y' +4xy = 0.

x?y"

Ax? y"” —x7y'+y =0.

+ x(x? — 3)y’ + 4y = 0.

(1 +x?)y” —2y =0. Qx?2y" — x(1 + 2x)y’+ (14+ 3x)y

= 0.

y" +xy = 0.

Ax?y" + 3x7y'

+ (1+ 3x)y =0

NY

x7y"+x3+x)y’ + (1+ 2x)y=

NY

xy” + (1 —x?)y’ + 2xy = 0. Ax*y" + 2x7y' — (x + 3)y = 0. x(1 — x?)y" + 5(1 — x?)y’ ae = 0).

x*y” + xy’ — (x? +4)y =0.

NNO

fey

NN

Pe

NY

YK

YK

Ee

Ke

HK

xy” —x(1+x*)y’ +(1—x7)y =0 axy"+y'+y=0.

Rr

2x7y” —x(1+2x)y’' + (14+ 4x)y = 0. 4x?y" —Ix(2+x)y'+34x)y =0.

RK

eer Petar

SS

au pwn © CON

xy” —(2+x)y' -—2y =0.

x(1 — 2x)y" —2(2+x)y'+ xy" + (2-—x)y-—y=0.

Av

394

18y = 0.

I

4

i c=2

18.12

27.

x?y"” —3xy’+4(1+x)y

28.

y” — 2xy' + 6y = 0.

29: 30. 31.

Ax*y” + Ix(x — 4)y’ + (5 — 3x)y =0.

32. 33.

4x2y" + 2x(x + 2)y’ + (5x — Dy = 0.

34, 35: 36.

Ax?y"” + (3x +1l)y =0.

=0.

x*y" — x3 +2x)y'+G—x)y

= 0.

x(1—x)y"” — (44+ x)y' +4y =0. Solve the equation of Exercise 31 about the point x = 1. x(_—x)y"+

(1 —4x)y’ —2y

= 0.

Show that the solutions of Exercise 35 may be written in the form

y=

(1 -x)’,

yz = (1 —x)77(Inx — x).

30s

xy" +(1—x)y' + 3y = 0.

38. 39.

x?y"” + xx — Dy’ + Gxt ly =0.

40.

xy" + x(4x — 3)y' + (8x + 3)y =0. x?y" —3x(1+x)y'+4(1—x)y =0.

41. 42.

2x(1—x)y" + (1 — 2x)y’ + 8y = 0.

2(1-+ x7)y” + Txy’ + 2y = 0. 2xy" + (3—x)y’ -—3y =0.

43. 44,

xy” — (1+ 3x)y' -—4y =0.

45,

xy" +3y’-—y=0.

46, AT, Ag. 49,

xy" —26 4+ 2x)y' + 4—z)y = 0.

50. 51. 52.

Summary

3xy" +2(1 —x)y’ —2y =0. 2x(1—x)y”+ y+4y =0.

xy” + (3 —2x)y’+4y =0. x(1 +4x)y" + (14+ 8x)y’ + y =0. 2x7y” + 3xy’ -(1+x)y

=0.

xy" + x(x —3)y' + 4e+3)y =0.

395

a

~

Equations

of Hypergeometric Type +:

Equations to Be Treated in This Chapter With the methods studied in Chapters 17 and 18, we are able to solve many equations that appear frequently in physics and engineering as well as in pure mathematics. We shall consider briefly the hypergeometric equation, Bessel’s equation, and the equations that lead to the study of Laguerre, Legendre,

and

Hermite polynomials. There are, in mathematical literature, thousands of research papers devoted entirely or in part to the study of the functions that are solutions of the equations to be studied in this chapter. Here we do no more than call to the attention of the student the existence of these special functions, which are of such great value to theoretical physicists, engineers, and many

mathematicians.

An

introduction to the properties of these and other special functions can be found

in Rainville.!

The Factorial Function It will be convenient for us to employ a notation that is widely encountered in advanced mathematics. We define the factorial function (a), for n equal to zero or a positive integer by

(a), =a(a+)D(@t2)+++@tna-)

(a)o = 1

Thus the symbol

fora # 0.

forn> 1;

(1 )

(a),, denotes a product of n factors starting with the factor a,

each factor being one larger than the factor before it. For instance, (7)4=7-8-9- 10,

(—5)3 = (—5)(—4)(-3),

(—3)3| = (—3) (3) G3). The factorial function is a generalization of the ordinary factorial. Indeed, (1), =1-2:3+-:n=al. | B,D. Rainville, Special Functions (New York: Macmillan Publishing Company, 1960).

396

(2)

19.3

The Hypergeometric Function

397

In our study of the gamma function in Section 14.9, we derived the functional relation C(xt+

1) =x

(x)

fora

0:

(3)

By repeated use of the relation (3), we find that if 7 is an integer, T(a+n)=(at+n—1)C(at+n-—1) =(at+n—l\(a+n—-2)P(a+n

—2)

=(a+tn—I1\(at+n—2)---(a)l (a)

= (a),T (a). Therefore, the factorial function and the gamma function are related by r

(Q)_ = Tatn)

n integer, n > 0, anda > 0.

P(a)

(4)

Actually, (4) can be shown to be valid for any complex a except zero or a negative integer.

The Hypergeometric Function Let us now consider any second-order linear differential equation that has only three singular points (one could be at infinity). Suppose that each of these singularities is regular. It can be shown? that such an equation can be transformed by

change of variables into the hypergeometric equation x(1—x)y" + [e — (a+ b+

1)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 indicial equation has roots zero and (1 — c). We put y=

oo )

Ae"

n=0

in equation (1) and thus arrive, after the usual simplifications, at oO

Sinn

oo

te

De x

n=0

— So(n t+ta)(n+b)e,x" =0.

(2)

n=0

Shift the index in (2) to get CO

oO

n=0

n=l

Yin(a te = Vex"! — Si(nt+a—VYtb— Denix"! =0. 2 See, for example, E. D. Rainville, Intermediate Differential Equations, 2nd ed. Macmillan Publishing Company, 1964), Chapter 6.

(3)

(New York:

398

Chapter 19

Equations of Hypergeometric Type

We thus find that eo is arbitrary and for n > 1,

(nta—W(n+b-1) eg = oom

C=1

n(n+e-1)

(4)

The recurrence relation (4) may be solved by our customary device. The result is, forn > 1,

1)(b+2)---(b+n—

a(a+1)\(a+2):--(a+n—1)-b(b+

nic(e + 1)(e+2)---(ec+n—-1)

ao

leo

,

(5) But (5) is greatly simplified by use of the factorial function. We rewrite (5) as

(a)n(B)n

=

6

n(n

A

°)

Let us choose eg = | and write our first solution of the hypergeometric equation as

r=

> (An (B)nx" .

]

*

(7)

(c),n!

n=]

The particular solution y, in (7) is called the hypergeometric function and a common symbol for itis F(a, b; c; x). That is,

Fa, bs ce; x) =14+ 0

co (Q)n

n=1

(b),x"

(c),n!

and y; = F(a, b; c; x) is a solution of equation (1).

The other root of the indicial equation is (1 — c). We may put oo y=

5

a

n=0

into equation (1), determine /;, in the usual manner, and arrive at a second solution n=

wee

-

co

(a+ 1—e)y(b+1—c)yx"*' n

n

l—e

Ds n=|

:

(2—c),n!

(8)

In the hypergeometric notation this second solution (8) may be written yy =x'“F(atl—c,b+1-—c;

which means

0 4, (—3), = 0, for

(—3)4 = (—3)(—2)(—1)() = 0.

19.4

Laguerre Polynomials

399

If ¢ is an integer but a and } are 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 may or may not involve a logarithm. To save space, we omit the logarithmic solutions of the hypergeometric equation.

Laguerre Polynomials The equation

xy” +(1—x)y'+ny =0 is called Laguerre’s equation. equation (1) is a polynomial.

If m is a nonnegative

(1) integer,

one solution of

Consider the solution of (1) about the regular singular pointx = 0. The indicial equation has equal roots c = 0, 0. Hence one solution will involve a logarithm. We seek the nonlogarithmic solution. Let us put

into (1) and obtain, in the usual way, oO

oo

> ax!

— Sok —1—n)a,_)x*-' =0.

k=0

(2)

k=]

From (2) we find that eS

1:

a= Saint _

(—n)(—n + 1)---(—n +k — lao

7

(k!)? (—n),

~ (ep? If n is a nonnegative

integer,

(—n),

=

0 fork

>

n. Therefore,

with aj chosen

equal to unity, one solution of equation (1) is a

yi =>! k=0

ial ee

(3)

(k1)?

The right member of (3) is called the Laguerre polynomial and is usually denoted by L,, (x):

(—n)ex OY (—Dkntxk La (x) *) = » (kK!) oh, zs (k)2(n — k)! A

4)

400

Chapter 19

Equations of Hypergeometric Type

The student should prove the equivalence of the two summations in (4) by showing that

(—n)_ =

(—1)‘n! (n—k)!

One solution of (1) is y; = L,(x). The associated logarithmic solution may,

after considerable simplification, be put in the form yo = L(x)

2 Inx + >

(—n); (Hy — Hy —_ “ee

k=]

— 2H)

x* =

.

3 (—Ly"al(k — tack

Tk + mip

2.

*

The solution (3) is valid for all finite x; the solution (5) is valid for x > 0.

Bessel's Equation with Index Not an Integer The equation

x*y" + xy’ + (x? —n*)y = 0

(1)

is called Bessel’s equation of index n. Equation (1) has a regular singular point at x = 0, but no other singular points in the finite plane. At x = 0, the roots of the indicial equation are c; = 1, c. = —n. 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 m is not an integer, two linearly independent solutions of (1) are oO

(— Lee"

_, * =) d. 2K *amy, cd

”)2

(—Tyiete*

“e dK

0. The function l

¥3

co

(—I)kx**4

~ Pd +n)°)~ » kek (k+n+1)

also a solution of equation (1), is called J,,(x), the Bessel function of the first kind and of index n. Thus , 00

y3

(x)

C.(

2. DRI

En

1)

"

is a solution of (1) and the general solution of (1) may be written y=

AJ, (x) + BU_n(@),

n %

an integer.

(5)

19.6

Bessel’s Equation with Index an Integer

401

That J_, (x) is a solution of the differential equation (1) should be evident from the fact that the parameter 7 enters (1) only in the term n’, It is also true that

I

Tay

©) = ord—m =

Se

De

Bessel's Equation with Index an Integer In Bessel’s equation

x*y" + xy! + (x? —n?)y =0,

(1)

let n be zero Or a positive integer. Then (—1)*xk

oo

= J,(x)=

5

=

y.2k-+n

-2

ED

WO = 2 Teen

w=

is one solution of equation (1). Any solution linearly independent of (2) must contain Inx. We have already solved (1) form = O in Exercise 11 of Section 18.6,

and for

= | in Exercise 10 of Section 18.9.

For n an integer > 2, put 3

;—

y=)

yite

j=0

proceed with the technique of Section

ajxe

ns,

18.9, determine y(x,

a

c) and 3c

- y(x, c),

and then use c = —n to obtain two solutions: 0°

— a

(—Lyhettnn

pee, dG —n),-\(k —n)lk! k=n

3 @)

and nal =

In

4

“seas

—"

4 1)k24-" ee

ee

+ 2 oR — mel 274-11

k=n

rn

4 Ay — Hy) x2#>"

(—1)*+! (Hy

: x

— n)p_i(k

— nn)! k!

,

A shift of index in (3) from k to (k + 7) yields

(—1 yk +1 2k

_ .

ee But forn > 2, (1 —n),_1

4

Qtr (1 — maki

(kK +A)

= (-1)""!(n — 1), so y=

VH(n—D!

JAK):

We can therefore replace solution (3) with yi

=

Jy (x).

(3)

Chapter 19

Equations of Hypergeometric Type

By similar manipulations, we replace solution (4) with n—-l

yg = J, (x) Inx +

k+1 (-1)*Ti(n — I)!x 2k—n ee Q2krI—nkl (1 —n)y

) ih

oO

+5

)

(—1)**"2 _ASER (Ae ET+ Seka Aen)x

SORA

k=0

For n an integer >

2k-+n

(6)

DENK (k +n)!

1, equations (5) and (6) can be used as the fundamental

pair of linearly independent solutions of Bessel’s equation (1) for x > 0. NS

402

Hermite Polynomials The equation y” —2xy' + 2ny = 0

(1)

is called Hermite’s equation. Since equation (1) has no singularities in the finite plane, x = 0 is an ordinary point of the equation. We put co

y=

)

ajx!

j=0 and employ the methods of Chapter 17 to obtain the general solution

09 Ohh (et 2) om (2K y=do

fred

= a]

(2k)!

k=1

+a,|x+

OF — all — wD)il ae

a,

|

(2k +1)!

i

&

valid for all finite x and with ap) and a, arbitrary.

Interest in equation (1) is greatest when n

is a positive integer or zero. If n is

an even integer, the coefficient of ag in (2) terminates, each term for k > $(n +2)

being zero. If n is an odd integer, the coefficient of a; in (2) terminates, each term for k => 4(n +1) being zero. Thus Hermite’s equation always has a polynomial of degree n, for n zero or a positive integer. It is elementary but tedious solution, to obtain from (2) a single expression for this polynomial solution. The result is k H,(x) = [5"] 5 (-—1)*n! (2x) n—2k

i

ki (n — 2k)!

'

(3)

in which [37] stands for the greatest integer < 3n.

The polynomial H,,(x) of (3) is the Hermite polynomial; y = solution of equation (1).

H,(x) is a

19.8

Legendre Polynomials

403

Legendre Polynomials The equation

(1—x?)y” —2xy’+n(n+1l)y is called Legendre’s equation.

=0

(1)

Let us solve (1) about the regular singular point

x = 1. We put x — 1 = v and obtain the transformed equation v(u+2

d’y

Ly = 0.

I?

(2)

At v = 0, equation (2) has, as roots of its indicial equation, c = O, 0. Hence one solution is logarithmic. We are interested here only in the nonlogarithmic solution.

Following the methods of Chapter 18, we put

y= >

oo

k=0

qv

into equation (2) and thus arrive at the results: ag is arbitrary and ee

Ll?

—(k-n—

a

l)(k + n)az_ L

2

(3)

Solve the recurrence relation (3) and thus obtain

a

(—1)*(—n)e(1 +: 1) a0

ee

2k (kI)2

with the factorial notation of Section 19.2. We may now write one solution of equation (1) in the form

yh =14+90§ (=Di (eanTaN+ Dae = DE

(4)

Since k! = (1), we may put _ in the form

y=

y=

1 =e

(—n)(n +1),

oeCoit

(1—x\"

( ; )

x

(5) 5

The right member of equation (5) is an example of the hypergeometric function that we met in Section 19.3. In fact, =F

(=n,

n+1;

1;

l-—x 5 ).

(6)

If 2 is a positive integer or zero, the series in (4), (5), or (6) terminates. It is then called the Legendre polynomial and designated P,,(~). We write our nonlogarithmic solution of Legendre’s equation as

y=

Po)

=F

(=m nth

1;

l-—x

2

).

(7)

W

Partial Differential

Equations

20.1)

Remarks on Partial Differential Equations A partial differential equation is one that contains one or more partial derivatives. Such equations occur frequently in applications of mathematics. The subject of partial differential equations offers sufficient ramifications and difficulties to be of interest for its own sake.

In this book we devote the space allotted to partial

differential equations almost entirely to a kind of boundary value problem that enters applied mathematics at every turn. Partial differential equations can have solutions involving arbitrary functions and solutions involving an unlimited number of arbitrary constants. The general

equation of order n may involve n arbitrary partial differential equation is almost never exceptions) of any practical use in solving with that equation.

solution of a linear partial differential functions. The general solution of a (the wave equation is one of the few boundary value problems associated

Some Partial Differential Equations of Applied Mathematics Certain partial differential equations enter applied mathematics so frequently and in so many connections that their study is remarkably remunerative. A sufficiently thorough study of these equations would lead the student eventually into every phase of classical mathematics

and,

in particular, would

lead almost at once

to contact with special functions that are widely used in quantum theory and elsewhere in theoretical physics and engineering. The derivation of the differential equations to be listed here is beyond the scope of this book. Some of the ways in which these equations are useful will appear in the detailed applications in Chapters 23 and 25. Let.x, y, z, be rectangular coordinates in ordinary space. Then the equation eV 4.2

ax

404

+

eV

BV

ey)

52 az

aye

=0

(1)

20.2

Some Partial Differential Equations

405

is called Laplace’s equation. It enters problems in steady-state temperature, elec-

trostatic potential, fluid flow of the steady-state variety, involving equation (1) is such that a physical object in cylinder, then it is possible that cylindrical coordinates the problem. We shall encounter such a problem later.

and so on. If a problem the problem is a circular will facilitate solution of It is possible to change

equation (1) into an equation in which the independent variables are cylindrical coordinates r, @, z, related to the x, y, z of equation (1) by the equations

HF

COS O;

y=rsing,

2 = Zz.

The resulting equation, Laplace’s equation in cylindrical coordinates, is

ave

iav

1a°Vv

of por

av

Og?

= (0).

age

(2)

Note that the use of z in both coordinate systems above is safe in making the change of variables only because z is not involved in the equations with other variables. That is, in a change of independent variables such as X=S=XTM+T

A,

y=Exi-N,

Z= 21,

or its equivalent

xy=gaty—-z),

wm=ge-y-2,

u=%

incorrect conclusions would often result from any attempt to drop the subscript on the z;, even though z = z). For instance, from the change of variables under

discussion, it follows that av

lav

lav

Be 29q

av

dy” Der

Hence

av

av

az

az

even though z = z). Let us return to Laplace’s equation. tox, y, z by the equations x = psingcosé,

In spherical coordinates p, @, #, related

y= psingsinég,

Z= pcos,

Laplace’s equation is av

dp=

et

2aVv

=

pp dp

13°V

cotoedaV_—

ee

pv age

p-

ap

ee

csc? pa?V

do

pe

5

og

5

=O.

(3)

With an additional independent variable ¢ representing time, and with a con-

stant denoted by a, we can write the wave equation in rectangular coordinates,

PV ar?

= 2

2(®Y ax?

ev Pv —+— dy*

az?

}.

(4)

406

Chapter 20

Partial Differential Equations

Equation (4) occurs in problems involving wave motions.

We shall meet it later

in the problem of the vibrating string. Whenever the physical problem suggests a different choice for a coordinate system, usually by the shape of the objects involved, the pertinent partial differ-

ential 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 point with rectangular coordinates x, y, z and at time t. The origin of coordinates and the initial time ¢ = 0 may be assigned at our convenience. If there are no heat sources present, the temperature u must satisfy the heat equation

au —

or

,(eu =h*

| —

(os

+

au

dru

—

aw

Is

dy) ae

©)

5

in which h? is a physical constant called thermal diffusivity. Equation (5) is derived under the assumption that the density, specific heat, and thermal conductivity are constant 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 equation 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 stress function ¢, which must satisfy the partial differential equation

Hy 5 ax4

—, dx2dy2

ad dy

0.

(6)

6

Numerous other partial differential equations occur in applications, though not with the dominating insistency of equations (1), (4), and (5).

In this book two methods for solving boundary value problems in partial differential equations will be examined. The Laplace transform, which was studied in Chapters 14 and 15, is a useful tool for certain kinds of boundary value problems. The transform technique will be developed further in Chapter 24 and used in Chapter 25. A second method, the classical one of separation of variables, will 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 problems involving partial differential equations.

Method of Separation of Variables Before attacking an actual boundary value problem in partial differential equa-

tions, it is wise to become somewhat proficient in getting solutions of the differential equations. When we have acquired some facility in obtaining solutions, 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 du

au

—=/—

dt

1

Ox?

a

with A constant. A solution of equation (1) will in general be a function of the two independent variables ¢ and x and of the parameter h. Let us seek a solution that is a product of a function of ¢ alone by a function of x alone. We put u=

f(t)v(x)

in equation (1) and arrive at

f' (ux) = h? fv" (x),

(2)

where primes denote derivatives with respect to the indicated argument. Dividing each member of (2) by the product f(t)u(x), we get

f(t)

_ h?v" (x)

ft)

(3)

va)

Now equation (3) is said to have its variables (the independent variables) sepa-

rated; that is, the left member of equation (3) is a function of ¢ alone and the right member of equation (3) is a function of x alone. Since x and ¢ are independent variables, the only way in which a function of x

alone can equal a function of ¢ alone is for each function to be a constant. Thus from (3) it follows at once that

f'() f(t)

=k;

4

h2v"(x) _

7

(5)

in which & is arbitrary. Another way of obtaining equations (4) and (5) is this: Differentiate each member of equation (3) with respect to f (either independent variable could be used) and thus get

4fO_y dt f(t)

since the right member of equation (3) is independent of ¢.

First we obtain

equation (4) by integration and then (5) follows from (4) and (3).

7

408

Chapter 20

Partial Differential Equations

Equation (4) may be rewritten

df

—=hkf,

dt

I

from which its general solution

f=ce™ follows immediately.

Before going further into the solution of our problem, we attempt to choose a convenient form for the arbitrary constant introduced in equations (4) and (5). Equation (5) suggests that & be taken as a multiple of h?. Let us then return to equations (4) and (5) and put k = h?B?, so we have i f'(t)

Had — 72p?

6

fn?

(6)

and Dei

iow

u(x)

gig

(7)

Using real 6 and the choice k = h?*, we are implying that the constant k is positive. Later we shall obtain solutions corresponding to the choice of a negative constant. From equations (6) and (7) we find at once that

f (0) = cr exp (h? Bt)

(8)

u(x) = co cosh Bx + ¢3 sinh Bx.

(9)

and

Since u = f(f)v(x), we are led to the result that the partial differential equation

.

2

has solutions

u = exp (h’ B*t)[a cosh Bx + b sinh Bx], in which f, a, and b are arbitrary constants.

(10)

The a and b of equation (10) are

respectively given by a = c\cz and b = cj c3 in terms of the constants of equations

(8) and (9). If we return to equations (4) and (5) with the choice k = —h’a*, so k is taken to be a negative constant, we find that the partial differential equation (1) has

solutions

u = exp(—h7a’t)[A cosax + B sinax], in which a,

A, and B are arbitrary constants.

(11)

20.3

Finally, let the constant k be zero.

Method of Separation of Variables

409

It is straightforward to determine that the

corresponding solutions of the differential equation (1) are w= C, + Cox, in which C, and C2 are constants. Direct verification that equations (10), (1) is simple. Since the partial differential solutions by forming linear combinations (12) with varying choices ofa, 6, A, B, solutions of (1) as we wish.

(12)

(11), and (12) are actually solutions of equation (1) is linear, we may construct of solutions. Thus from (10), (11), and a, b, Cj, C2, we can construct as many

The distinction between equations (10) and (11) is dependent upon the parameters and variables remaining real. Our aim is to develop tools for solving physical problems; hence we do intend to keep things real. Wi Exercises Except where other instructions are given, use the method of separation of variables to obtain solutions in real form for each differential equation.

l

a°u = are

2,

a 452av dx? ay

3.

a1

=a

5 ou ox?

=0.

au

a7u

= t + 2b— =a “. at? ar ax?

Show that this equation has the solutions

u = g(t)(B, coskx + Bo sinkx),

where g(t) can assume any one of the forms

e"(Aje” + Are"),

y? =b* —a?k?

ifb? —a7k’ > 0,

or

e~"'(A3cosét + Aqsindt),

b& =a’k?—b?

ifa’k? —b? > 0,

or

e'(As+ Act)

if a’k? —b* =0,

and the solutions

u = (Aq + Age 7") (B3 + Byx). Find also solutions containing e** and e~**, aw

ow

Oy

“Ox!

410

Chapter 20

Partial Differential Equations 5

ow -

6.

ow

ax

x—

=

ay

WwW

a.

Subject the partial differential equation of Exercise 5 to the change of de-

pendent variable w = v/y and show that the resultant equation for v is

dv

dv

A=

=

ox

7.

SS

ay

Show that the method of separation of variables does not succeed, without modifications, for the equation

au

au

au

x2 of axdt + oa

8.

=O,

For the equation of Exercise 7, seck a solution of the form

u =e" f(x) and thus obtain the solutions u = exp [k(t — 2x)[A, coskx + Ao sinkx], where k, Aj, A2 are arbitrary. Show also that the equation of Exercise 7 has the solutions u = Ax + B andu = Ct+ D, with A, B, C, D arbitrary.

9.

For the equation

a°u

du

a?u

— +4x— + — =0, ax? r vax ay? put wu = f(x)g(y) and thus obtain solutions

u = [Aye

+ Are

Bi file) + Bo faa],

in which f(x) and f>(x) are any two linearly independent solutions of the

equation

f+ 4xf +41 f = 0. Obtain also the solutions that correspond to k = 0 in the above. For k # 0,

the functions f,; and f2 should be obtained by the method of Chapter 17. They may be found in the form

2 Lo fi@y=1+ De x

=

Alx)=x+>0

+4) +++ (kK? + 2m — 2)x2™ . (Ay (k2)(k2 + 2)(K2 nae (-4)" (k* +

1)(k?

+ 3)(k?

om

+ 5) ico (k? +2m

BT

m=!

Find similar solutions involving cos 2ky and sin 2ky.

—

ii

.

20.4.

10.

A Problem on the Conduction of Heat in a Slab

411

Use the change of variable u = e”' g(x) to find solutions of the equation

aru ‘ 11.

au se au

ax? dxdt dr? Show by direct computation that if f\(y) and f2(y) are any functions with continuous second derivatives, f/’(y) and f;’(y), then

u= fi(x —at) + fox + at) satisfies the simple wave equation (Exercise 1) ‘

vs

anu

2 0°u

—_7,

=a

ar?

ax?

m

12.

Show that v = f(xy) is a solution of the equation of Exercise 6.

13.

Show that w = f(2x + y7) is a solution of the equation of Exercise 4. Show that u = f\(t — x) + xfo(t — x) is a solution of the equation of Exercise 10.

14.

A Problem on the Conduction of Heat in a Slab Among the equations of applied mathematics already stated is the heat equation in rectangular coordinates, Ou

at

= he (

au

ax?

au

dru

day?

az

dag F

)

(1)

in which x, y, Z = rectangular space coordinates, t = time coordinate,

h> = thermal diffusivity, u = temperature. The constant /? and the variables x, y, z, f, wu, may be in any consistent set of units.

For instance, we may measure x, y, z in feet, ¢ in hours, u in degrees

Fahrenheit, and h? in square feet per hour. The thermal diffusivity (assumed to be constant in our work) can be defined by 2K = ay in terms of quantities of elementary physics, K

= thermal conductivity,

o = specific heat, 6 = density,

all pertaining to the material composing the solid whose temperature we seek.

412

Chapter 20

Partial Differential Equations

For our first boundary value problem in partial differential 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 problem, w will be a function of only two independent variables (x and 1), 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 ho-

mogeneity in texture, Let the thickness of the slab be ¢ units of length. Choose the origin 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 function of x alone,

and let the surfaces x = 0, x = c be kept at zero temperature for all t > 0. 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 w at any time ¢ and position x is determined by the boundary value problem

a a? SW ot Ox-

for0

t

cosna

2cosnz

2

ni

near

n373

Hence the Fourier sine series, over 0 < x
)|

(nx)3

1

nex

sinnx.

(5)

The series on the right in (5) converges to the function exhibited in Figure 22.6,

fx)

Figure 22.6

440

Chapter 22

Fourier Series

that function being called the odd periodic extension, with period 2, of the function

O 0+, =

(1)

forO

coshx =

“=

— oO

arctanx

= )* oh

.,

all x;

(3)

(4)

x2

(2n)!

»

sinhx =

1)!

all x;

;

all x;

(5)

ntl

———., a+ 1)! —|)"

all x;

(6)

|x| < 1;

(7)

2n+1

= 2n + 1

467

468

Chapter 24

Additional Properties of the Laplace Transform

1

——

= 1+

(l—x)"

min

+ 1)---(m +n — 1x" n!

2

In(1 n(1+x) =)

oo

X

[fey

In

(=

————_, Hi oO

=

l-x

|

;

Ix|
0, and let L~'{ f(s)} = F(t). Prove that

3

s* sinh (3s) |

=

9.

1 ||

“a

|

At)

"ma

=2)

=

(kt (kt) }

and compute F(12).

Compute #(10).

_

_

_ _ 1)" F(t — 2nc — c)a(t — 2ne — ce).

_4y\n

Lete > 0,5 > 0,and let L~'{ f(s)} = F(t). Prove that oO

ze

{ f(s) tanh (cs)} = F(t)+2 YD"

Fe — 2nc)a(t — 2nc).

n=l

10.

Let 0 0.

k>0,

= — exp (—2k,/s),

We shall use (23) in the form LT!

{:

exp (-24y5|

=

erfc

(=)

‘i

k>

0,

s>

0.

(24)

In Chapter 25 it will be important to combine the use of equation (24) and the series methods of Section 24.1. Consider the problem of obtaining pe

{ sinh (x./s)

s sinh /s

|.

Dee

J,

om 0:

(25)

If x were greater than unity, the inverse in (25) would not exist because of the

behavior of sinh (x./s)/ sinh ,/s as s > 00. Because we know (24), it is wise to turn to exponentials. We write

sinh (x/s) _ exp (x4/s) — exp (—x./s)

sinh./s

exp (,/s) — exp(—4/s)

©

(26)

As in Section 24.1, we seek a series involving exponentials of negative argument. We therefore multiply numerator and denominator on the right in (26) by

exp (—./s) and find that

sinh (x./s) _= exp [—d — x)./s] — exp[-—C + x) /5] . 1 — exp (—2,/s) sinh ./s

an

27

Now |

mep Ca

oO

=

=

Dep

—2na/s). ni/s)

(28) 28

Therefore,

sinh (x./s) al Ssinh Js = >_< exp [—(. — x + 2n)J/s] — exp[-(. +x + 2n)4/s |}. n=0

476

Chapter 24

Additional Properties of the Laplace Transform

For 0 < x < | the exponentials have negative arguments and we may use (24) to conclude that

Lt

feces Se

—

> ert

Geen wa =) — erfc Ce oS _ ) » (29)

n=0

Exercises Show that for all real x,

jerfx| < 1.

2.

Show that erf x is an odd function of x.

3.

Show that lim a x70

4.

=

a

Xx

fe

Use integration by parts to show that x

1

erfy dy = x erf x — —=[1 — exp (—x?)].

5.

Obtain the result of equation (11).

6.

Start with the power series for erf x, equation (5), and show that

Ll’ 7.

2

1

VTSS

vs

erf (/t)} = ——arctan—,

s>0.

Use the fact that 1

_l-vilt+s

_

1+/T+s 1-(1+s)

Eg

5

METS

Ly

ss

si +s

and equation (3) to show that 1

8.

—t

Use equation (3) to conclude that

apg] ne to and therefore that ]

|e 9.

Evaluate L~!

l

(wa |

.

ae ete WD.

I+s

=

t

24.2

1

10.

valuate L~! Evaluate

{|

11.

Define the function #(t) by

The Error Function

477

:

o(t) = L! {ert *| ‘ Ss

Prove that

12.

Show that for x > 0,

i

13;

sty

|

x“

i

2Jt ——— ; =2 2 —-1 )" erfefc | ———— ;

|

|.

Show that for x > 0,

cc

25> erfe a. n=0

14.

Derive the result

_ v8 exp (—2k./s), A(s) = L {1-2 exp (-“)|-~

k>0,s>0

directly from the definition of a transform. In the integral

A(s) = [

exp (—st — k?t7!)17-7?? dt

put B = 4/7 to get

A(s) =2 | © p>? exp (~s6? — KB)aB or

°

oo

A(s) = exp (-2kV5) |p exp l—(Bs — kB“) 148 Show that

afds __2 | 0 ” exp (—s8? — 2B) dB = —2exp (245) |

exp [—(B/s — kB-')’] dp.

Thus arrive at the differential equation

478

Chapter 24

Additional Properties of the Laplace Transform

A aff & —kA = —2,/m exp (—2k,/s) and from it obtain the desired function A(s).

Bessel Functions The Bessel function oo

)

Dn

=

kel

(=1)"(32)

d

y2k+n

Pr&+n 4D)

vi

a

1

of the first kind and of index n, appeared in Sections 19.5 and 19.6. We meet J,(z) in a simple application of the series technique of Section 24.1. If we can expand a given function of s in negative powers of s, surely we can get the inverse transform term by term. A simple example is the following:

1

(-*)

=i

a

=

[5 ex (

=

(1h

=

k'stok+l

_ oo

)

”

(—1)*x*t*

2

kik!

°)

When nr = 0 in (1) we get, since P(k + 1) = K!,

2, (1G)

Joe) =}

—

(3)

k=0

By comparing (2) with (3), we get

I Lo! {exp (-*)| =IQV/xt); §

x>0,5>0.

(4)

S

From

I gttl

x\

ee (kx!

exp (==)

~ a

kl gk+n+!

we get

rt

ft oe (-2)) x

|

|

exp

(

=)

= SR oO

aU

(—

1)'e***

are+n

FD

wag RITA +n+

1)

24.3

Bessel Functions

479

Therefore, at least for n > 0, nf2 =

| =

exp

(-*)|

s>

Jn (2/xt),

= (

0.

(5)

With more knowledge of the gamma function, we could use series methods to

obtain the transform of J, (xt) for general n. Here we restrict ourselves ton = 0

for simplicity. From (1) we obtain ryickanye*

Jo(xt) = soe. k=0

Then

(—1)* (4.x)? (2k)! L{ Jo(xt)} = 2arr ae But (2k)! = 2*k! [1-3-5--- (2k — 1)]. Hence

L{ Jo(xt)} = a +

(-1)*[1 -3- a ae — Tie Gl

or

1

|

wy?

LiJjoah} { Jo(xt)} =-ll+—75a =( 2

;

Therefore, Li} Jo(xt)} {

0

}

] = ——_.. 2

+

(6)

x2

From (1) it is easy to conclude that d — de J o(Z) = —J)(z). 1(Z) Then d

apo)

=

—xJ\(xt)

and we obtain d

L{-xJ;@t)} = L {0.

(7)

The conditions (4) and (5) become x-O0',

wo

2

=) e

(8)

lim w(x, 5) exists.

(9)

X00

We now solve the new problem, (7), (8), and (9), for w(x, s) and then obtain y(x, t) as the inverse transform of w. Let us rewrite (7) in the form

d?w —~ l6s*w = 16 (10) dx? and keep in mind that x is the independent variable and s is a parameter. When we get the general solution of (10), the arbitrary constants in it may well be functions of s; they must not involve x. The general solution of (10) should be found by inspection. It is w=->Z

I

Ss

+ c)(s) exp (—45x) + c2(s) exp (45x),

x>O0,s>0.

(11)

25.1

Boundary Value Problems

483

Because of (9), the w of (11) is to approach a limit as x > oo. The first two terms on the right in (11) approach limits as x —> 00, but the term with the positive exponent, exp (4s), will not do so unless co(s) = 0.

(12)

That is, (9) forces (12) upon us. The w of (11) then becomes w=->sfe

1

(13)

5 > 0.

x>0,

+ €;(s) exp (—4sx),

Application of condition (8) to the w of (13) yields ] a

=c¢i(s}

—

Py

2 cis)

= aa

] +

a

Thus we find that 1 2 1 w(x,s)=—->St+|atrsa se so gt

exp (—4sx),

x SO,

© SiO)

(14)

We already know that if

L'{f)} = FO, Ee

(15)

f(s)} = FE — edad — 0).

Therefore, the application of the operator L~! throughout (14) gives us

yx, t) = —t + [(t — 4x)? + (¢ — 4x) ]ae(t — 4x),

x>0,t>0.

(16)

It is our contention that the y of (16) satisfies the boundary value problem (1) through (5). Let us now verify the solution in detail. From (16) it follows at once that oy

Go = 1+ 2

= 4x) + Hele — 4),

x>0,t>0,144x.

(17)

Note the discontinuity in the derivative for ¢ = 4x. This is forcing us to the admission that we obtain a solution of the problem only on each side of the line t = 4x in the first quadrant of the x7 plane. Our y will not satisfy the differential equation along that line because the second derivative cannot exist there. This is a reflection of the fact that (1) is a “hyperbolic differential equation.” Whether the “solution” does or does not satisfy the differential equation along what are called characteristic lines of the equation depends upon the specific boundary conditions. We shall treat each problem individually with no attempt to examine the general situation. From (17) we obtain dey

ay

aE

— 4),

x>0,t>0,t44x.

(18)

484

Chapter 25

Partial Differential Equations:

Transform Methods

Equation (16) also yields

; ee =[-8(t 4x) —4Ja(t—4x),

2x>0,1>0, 144k,

(19)

xX

and ay 9x2 7 = 32a(t — 4x),

2

x>0,t>0,

¢ 44x.

(20)

Equations (18) and (20) combine to show that the y of (16) is a solution of the differential equation (1) in the xr region desired, except along the line t = 4x,

where the second derivatives do not exist. Next we verify that our y satisfies the boundary conditions. To see whether y satisfies condition (2), we must hold x fixed, but positive, and then let r approach zero through positive values. As

t+ 0*, y>0+4[(—4x)? + (—4x)Jo(—4x) =0

= forx > 0.

Thus (2) is satisfied. Note that a(—4x) would not have been zero for negative x. From (17), with x fixed and positive, it follows that as

a t > 0*, a > —1+[2(—4x)+

e(-4x) =—-1

forx > 0.

Thus (3) is satisfied. Once more the fact that x is positive plays an important role in the verification. Consider condition (4). In it we must hold ¢ fixed and positive. Then, by (16),

as

x27 0t, yo

tt (P 4+ He() = t+? 41=2

fort > 0.

Then (4) is satisfied. Finally, the y of (16) satisfies condition (5), since lim y(x,f)

=-f+0=-t

fort > 0,

XO

because for sufficiently large x and fixed ¢, (ft — 4x) is negative and therefore a(t —4x)

= 0. This completes the verification of the solution (16).

@ Exercises In each exercise, solve the problem and verify your solution completely. a

ay

Ox

or

Le ee= Be

die bow O, ee

t>0t,y>0

for x > 0;

x

fort > 0.

0t, y > 27

Dh

25.2

a0 taea =a

i

toO0t,y>0 3.

for x > 0;

>0t,y> 2

fort >0.

Solve Exercise 1 with the condition as f > 0+ replaced by t > 0*, y > x, Solve Exercise 2 with the condition ast > 0* replaced byt > OF, y > 2x. ay

5.

485

fort > 0, x > 0:

%

x

The Wave Equation

a2

a°y

7

a

fort

t>O0T,y>0 a

tot,

x30,

>

0, x

> 0;

for x > 0;

% + -2

for x > 0;

yor

for t > 0;

lim y(x, t)exists

fort > 0.

xX CO

a2

6.

oy

a2

_ 4?

fort > 0, x > 0;

ar? ax2 t>0T,y30 0

fom Ot, 25

for x > 0;

eB

for x > 0;

x > 0+, y > sint

fort > 0;

lim y(x, t) exists

fort > 0.

x7 00

The Wave Equation The transverse displacement y of an elastic string satisfies the one-dimensional wave equation

Pyat? _ tyax?

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 ¢ = 0, y=0

and

dy

—=0 for x > 0. at We assume the string long enough that the assumption that it extends from x = 0 to co 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 — Oas x — ox, but that at the y-axis end the string is moved up and down according to some prescribed law, y +

F(t) asx

—

O*, with F(t) known.

Figure 25.1 shows the position of the string at some t > 0.

486

Chapter 25

Partial Differential Equations:

Transform Methods

yy, ty)

P(t)

O

x=aty

x

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: a

ar

eo

t>0*T,

a

fort > 0,

yo0 a

t>0*,

=

x—20*,

yo

> 0 F(t)

lim y(x,t)=0

x > 0;

(1)

for x > 0;

(2)

for x > 0;

(3)

fort > 0;

(4)

for allt > 0.

(5)

X00

The prescribed function F(t) must vanish 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(@,o}=u@,s),

L{FO})= fo).

Note that F(t) must be continuous because of its physical meaning here.

(6) The

operator L converts the problem (1) through (5) into the new problem d*u

seu = ara

forx > 0;

(7)

x2 0t, u> f(s);

)

lim u(x, s) = 0.

(9)

25.2.

The Wave Equation

487

From (7) we write at once the general solution $x u(x, s) = c)(s) exp i=—) With s > 0,

oe + ca(s) exp (=).

(10)

x > 0, the condition (9) requires that

ols) = 0.

(11)

Thus (10) becomes u(x, Ss) = e¢,(s) exp (-=),

(12)

a

and (8) requires that

f(s) = e,(s). We therefore have .

ox

u(x, s) = f(s)exp (-=), a

x>0O0,

s>0.

(13)

Equation (13) yields the desired solution, x x =F(t—=)a(e-=),

ya.

a

x>0,

t>0,

(14)

ad

in which we assume that F(t) is defined in some manner for negative argument so that Theorem 15.3 of Section 15.4, can be used. Verification of the solution (14) is a simple matter. Note that

er (—Se(-2). Babe (-Del-

and

vy,

x

x

ge = F'(t- Ze (0-3).

ay a?

1, ae

x

x

(1-Z)a(s

a)

We are forced to assume the existence of two derivatives of the prescribed function F(t). It is particularly convenient to choose F'(r) so that F'(O) and F”(0) vanish

along with F(0), so that the continuity of y and its derivatives are not interrupted along the line x = at. Completion of the verification of the solution is left to the student. In Section 23.5 we studied the transverse displacement of a string of finite length held fixed at both ends. Fourier series methods seem superior to Laplace transform techniques for such problems. Try, for instance, transform methods on Exercise | of Section 23.5.

488

Chapter 25

Partial Differential Equations: Transform Methods

Exercises |.

Interpret and solve the problem:

vy _ ax? dy

Oe

P30, 1+ 0t,

fort>0,0 0. We may visualize, for example, a huge flat slab of concrete with an initial temperature distribution dependent only upon the distance from the plane surface of the slab. If the temperature at that surface is thereafter (t > 0) maintained at some specified function of f, or if the surface is

insulated, the problem of finding the temperature for all positive x and ¢ is one involving the simple heat equation (2) of Section 23.1.

EXAMPLE 25.2 Consider a semi-infinite slab x > Q, initially at a fixed temperature u = A and thereafter subjected to a surface temperature (x — O*) which is u = B for Q fp. Find the temperature within the solid for x =O; t> 0. The boundary value problem to be solved is

a

+=

a?

5

forx > 0, t > 0;

(1)

t>0t,ucA

for x > 0;

(2)

x70t,u>B

for 0 0t,u>0

fort > to;

lim w(x, f) exists xXxOoO

< fo,

for each fixed t > 0.

(3) (4)

25.3

Diffusion in a Semi-Infinite Solid

use the a function to reword

We

B, and h? are constants.

In this problem A,

489

the boundary condition (3) in the form x—>

07,

u—

B[l —al(t — f)]

fort > 0.

(3)

Note also that the physical problem dictates that the 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 suggest the use of the

Laplace transform. Let

L{u(x,t)}=w,s),

«>0,5>0.

(6)

The equation (1) with condition (2) is transformed into

sw-Az=

aa

pe™ dx?

j

0 0,

x

or

dws ie

A

ee

x

> 0.

(7)

Conditions (4) and (5) become lim w(x, s) exists

for fixed s > 0

(8)

—[l — exp (—‘95)].

(9)

x7 00

and x—>

+

0,

w—>

B s

The differential equation (7) has the general solution : + cxe(*)

Ww = Cc} exp (-*4)

A +4

x>0,s5s>0,

(10)

in which c; and c2 may be functions of s, but not of x. As x — oo, the w of (10) will approach a limit if, and only if, c2 = 0. Hence condition (8) yields the result co =0

(11)

and the w of (10) becomes w = c) exp (-*

A

=) +e. h 5

By letting x — O* and using (9), we obtain B Su

—

exp

(—ios)]

(12)

A =

er

+

—-

(13)

Therefore, the solution of the problem (7) through (9) is

ves

-

— exp (-)]

h

+ = exp (-*)n

— exp (—fos)].

(14)

490

Chapter 25

Partial Differential Equations: We know

Transform Methods

that

i {5e(-* 5

h

*)|

= erfe (=)

x>0.

2h/t)

(15)

Hence we may write

i

|

{: exp (-*)

exp (-105)} serie (sai)

a(t — to),

(16)

where absolute value signs have been inserted to permit ¢ to be used in the range 0 to fp, 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), Forx > Oandr u(x,t)=A

> 0, X

! — erfc (5) +B

ert

x

(7)

x

— erfc Gren.

a(t — 0)

»

(7)

or

Xx

u(x,

f)=A ert( =) Ke

+B

ert

( 0°, u>A-04 ast —>0t,

Bll —a(t —f)] = Bll — a(t —t)]

u>A-1+B0-0)=A

asx-7-o,u->A-1+B-0=A asx >oo,

u>A-14+B(0-O0=A

fort > 0; for x > 0; for0 < ft fo.

25.4

Canonical Variables

491

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:

a a? = fl ot dx?

fori >'0,. O
0

fort > 0;

(3)

x>ec,u

fort > 0.

(4)

26

0

& < @:

(1)

A consistent set of units for the measure of the various constants (parameters) and variables in this problem is

u = temperature (°F), t = time (hr), x = space coordinate (ft),

h* = thermal diffusivity (ft7/hr), c = length (ft), A = initial temperature (°F).

We seek dimensionless new variables ¢, t, y, proportional to the physical variables x, t, u. For the moment let

x=pl,

t=yt,

u=dy,

(5)

in which f, y, 4 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

baw

yor

hb ay

B

ae2

fort

> 0,0 < Bo 0T, SW OA

for0 < Bt 0,

(11)

for y > 0,

(12)

v 30

for t > 0,

(13)

0

fort > 0.

(14)

for fixed positive f,

(15)

wo

1

condition (2) will be satisfied. From condition (3) we get

x>0T, and from (4),

yoOF,w

Conditions (5) and (6) will be satisfied if lim v(x, f) exists X00

498

Chapter 25

Partial Differential Equations:

Transform Methods

and

for fixed positive f.

lim w(y, ft) exists yro0

(16)

We must now find v from a

=o

fort > 0, x > 0;

1

0T, vo

to x

a°

>0t, lim

v30

u(x, f) exists

(9)

for x > 0;

(11)

for tf > 0;

(13)

for fixed positivef.

(15)

X—>> OO

The function w must satisfy (10), (12), (14), and (16); it is therefore the same

function as v except that y replaces x. To obtain v we use the Laplace transform. Let Liv(x, tp} = 24,5) = |

0

e

v(x, t) dt.

(17)

Then (9) and (11) yield sesg —|

d’g

Tx’ —

=

(18)

for which the general solution is easily written by inspection because of our experience in handling equations with constant coefficients. We thus get

| gS a

exp (—x4/s) + c2(s) exp.x4/s). ci(s)

(19)

The function g must, because of (13) and (15), satisfy the conditions

g— 0,

x2 0t,

lim g(x, 8) exists.

(20) (21)

x00

Because of (21), c2(s) = 0. Because of (20),

I

). 0=-+e,(s s Therefore,

cyn=

|

=

l

eRe (—xa/s).

(22)

25.6

Diffusion in a Quarter-Infinite Solid

499

The function u(x, f) is an inverse transform of g(x, 5):

v(x,f) = 1 — erfc (=).

(23)

But | — erfez = erf z. Hence

v(x, t) = erf (=).

(24)

2/t

Therefore, the solution of our original problem (1) through (6) is u = erf (=)

erf (2)

‘

(25)

The student should verify that the uw of (25) satisfies all the conditions of the boundary value problem (1) through (6) introduced at the beginning of this section. i Exercises 1.

Show that for theu of (25), 0 < u < 1, forall x, y,t > 0.

2.

Letthe point with coordinates (x, y, f) bein the first octant of the rectangular x, y, t space. Let that point approach the origin along the curve x” = 4a"t, y* = 4a°t,

in which a is positive but otherwise arbitrary. Show that as x, y, f > OF in the manner described above, u may be made to approach any desired number between zero and unity.

Answers to Odd-numbered Exercises

Chapter 1 Section 1.2 1. ordinary, linear in x, order 2 3.

ordinary, nonlinear, order 1

5.

ordinary, linear in y, order 3

7.

partial, linear in u, order 2

11, 13: 15.

ordinary, ordinary, ordinary, ordinary,

linear in x or y, order 2 linear in y, order | nonlinear, order 3 linear in y, order 2

Section 1.3

lL.

y= txt+xPte.

y = 3e* +3.

3

y = 2sin6x +c.

y = 3e*,

5.

y =arctan (x/2)+c.

y = —2cos 2x.

Chapter2 Section 2.1 1. r=roexp(—2r7).

3. 5. 7. 9, ll.

y= 5V 10x? — 4, y= GD. y=In2—In[1l + exp (—x’)]. r?In(r/a) =r? cos@ — a’. yin|e(l—x)| =1.

Ls, 19, pill 23: 25. 2: 29, 3,

= c.

(x + 1)? + y?+ 2In|e(x — 1)| = 0. e(x —1) = Qy + 1)/@y?) +e. x(Qyvt1)=(U+ exe’.

4In|sec y+tan y| = 2x +sin2x +c. 2x +sin2x=e+(14+ry. caB = exp (—3a — B).

13.

e* +y?%=c.

is.

#? Sceys.

33.

y—c=—WVa? — x2, the lower half of the circle x? + (y —c)’ =a’,

35,

psa

37.

In(@x?+1) =y? —2y4+4 In |e(y + DI.

a

500

PY

r=c(1—bcos@).

Answers to Odd-numbered Exercises

501

Section 2.2 All functions are homogeneous except those of Exercises 2, 5, 6, and 19.

Section 2.3

1 3.

x8 =c(9x? + y?). xt = c7 (4x? 4+ y’).

9. 11.

5.

x*(y + 2x) =c(y +2).

13.

xv? =c(x +2v).

7.

x(y +x)? =c(y — 2x).

15.

4xIn|x/e| — 2y + x sin (2y/x) = 0.

27. 29, 31. 33.

x-—y=5(y + 4x) Inx. y4(3x? + 4y?) = 4. 3x3 -x?y —2y? =0. y+3x = (y + 4x) In(y + 4x).

17.

r*+2r(sin@ —cos@) =c.

17.

2yaretan (y/x) = x In[e*(x? + y?)/x4].

19. 21. 23. 25. 35.

5? = —2r7 In fest]. (y—x)(y +3x) = cx}, 22x+3y)+(ety)in@+y)=0. x? =2y4+1. x? = 2y*(y + 1).

x? +4y? =c(x+y). x? (x24 2y?) = ct.

Section 2.4

1. 3.

x?4+2xy—y? =c. se.

19.

rsind —r*cos*6 =c.

5.

x7 4+2y? =4xy +e.

wy-x+iy

21.

yx? +y?)=c.

9.

xy? —x?y 4+ 3x2 -2y =e.

23.

xy? +2xy—x2 =.

Il.

5sin2y + xcos2y —x3y?

13. 15.

Ix+y*1+x)?=c. x*?y—xtany=c.

=c.

25.

27.

xy? — y3+5xy

—3x

=5.

x*y+y? +2x? +y exp (—x7) =e.

Section 2.6

1 3.

2y=x +x. 20x =4y —1l+e(v+ 1774.

5:

xu =ce™

—u— 3,

+x?)(e +x

7. 9, ll,

13.

y=(

15.)

y=c3e™'*

17.

x?y = Ga? +18 +e(a? +1).

y=csinx —cosx. y(secx +tanx) =c+x—cosx. xysinx

=c + sinx — x cos x.

— arctanx).

+ e3e"™*,

where c3 = cy /(m)

— m2).

19.

(«-Dy=(*4+ D(ce+x -—2In|x 4+ 1).

21. 23.

3ycos'x =c+3sinx — sin’ x. y = (x? +a?) [e(x? + a?) — I].

25.

Ifn=0, y=bx+c—ablIn|x+al. Ifn =—-l,y=ab+c(x +a) + bx +a) In|x +a].

27.

2y = (2x +3)!/? In (2x 4 3).

31.

y=2x-1.

99,

jak h ep (-*)|.

33.

s=(1+7)[3 —exp(—?)].

R

L

Miscellaneous Exercises

1.

2c? =e*

3.

y=2etltt+ee~t)™.

+c.

5.

yix+y)=ex.

7,

2y=x7—1+4exp(1

9,

x?y = c(2x + 3y).

11.

xy

+1

= yi(e+3x

—x?). —3aretanx).

502

Answers to Odd-numbered Exercises

13.

4ln|secx +tanx|=2t+sin2r+c.

21,

x?y3=3(e+y—xy).

15.

x =yln|exy|.

23.

y=b/a+ce™.

17. 19. 29.

x? +4xy + y* =e. ky4+4xy3 4x4 =c. arcsinx + arcsiny =

30.

arcsinx + arcsiny

25. xsiny+ycosx=c. 27. y=2sin’x sin? tx. c, ora part of the ellipse x* + 2cyxy + y? +c} — 1 =0;

where cy = cose.

=

47, or that arc of the ellipse x* + xy + y? = 4 that is

indicated by the upper solid line in Figure 2.4.

31.

—4z7, or that arc of the ellipse x? + xy + y? = + that is

=

arcsinx + arcsiny

indicated by the lower solid line in Figure 2.4.

35. 37. 39.

xy=cy?-1. 2y =sinx +(x +c)secx. y=x—x3 +e(1—x7)!.

4].

y=sinx +ccosx.

Sl.

@w@+Dy=@-—D_e+14+2In@

53.

xty=2x-1.

55.

2x2(x —2)y =x? —5.

43. 45. 47,

xycosx =c+cosx+-xsinx. x= y[l+yln(-y)]. y?+2xy+1=2Iny.

49,

y=-3(x+1).

— 1).

Chapter 3 Section 3.2 1. The results are found in Table 1. The correct solution is computed from the solution y=2e* —x—-1. TABLE

1

x

y

x+y

dy

Correct

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

1.00 1.10 122 #136 1.53 1.72 194 2.19 248 2.81 3.18

1.00 1.20 142 1.66 1.93 2.22 2.54 2.89 3.28 3.71

0.10 0.12 40.14 0.17 0.19 0.22 0.25 0.29 0.33 0.37

1.00 1.11 1.24 1.40 1.58 1.80 2.04 2.33 2.65 3.02 3.44

Section 3.4

1.

ye) = 14-24 4x75 yo(x) = L+xtx? + 32%; ys(x) = 14x txP?4 xh 4 txt

3,

yx) = 142-1)

4+ 5-1);

?+ yo(x) = 1+4+2(%-1I) + 3(x —1)

h(x — 1)3;

— @ — 1) + 53-1? +3@ — DP + ye 1+2) ys=(x

Answers to Odd-numbered Exercises Section 3.5

+ bot 1. ys(x) = 1+ tx? + det

3,

+2@—D?

ys(x) = 142-1)

Section 3.6

1.

+h@-

+4@-17

+ Lxt + dx5,

lL.

ys(x) = 1 txtx24tx3

7.

+ Bao t Tex" ye) Lex + fe? + bat + fa

+ ay — DP. 3. ys(x) = 1+ 20 = I + 3 — DP ++ 4 3 — — 28D++ HegD" —2)* + gyle — 2). 5. y5(x) = —1+ (x —2) + @ — 2)? Section 3.7 The results are found in Table 2. 1.

TABLE2 0.20 0.30 1.25 142 1.52 1.93 0.25 0.35

0.40 1.64 2.53 0.45

y+ 3hK, K>

133 1.71

152 2.19

1.77 2.93

y+4hK, Ky

1.34 1.73

1.53 2.22

1.79 3.00

xth y+hK3; K, K

0.30 142 1.93 1.72

0.40 1.64 2.53 2.21

0.50 1.94 3.51 2.98

x y K, x+35h

2.

0.50 1.94

The results are found in Table 3.

TABLE 3 %

0.00

0.10

0.20

0.30

0.40

y Ky

1.00 1.00

Ll 1.21

1.24 144

140 1.70

1.58 1.98

x+th | 0.05

0.15

0.25

0.35

y+$hK, | 1.05

1.17

131

149

1.68

K> y+ 5hK,] K;

110 1.06 111

1.32 1.18 1.33

1.56 1.32 1.57

1.84 149 1.84

2.13 1.69 2.14

0.10 | 111 1.21 1.J1

0.20 1.24 144 1.33

0.30 140 1.70 157

040 1.58 1.98 1.84

0.50 1.79 2.29 2.14

xt+h ythK; K4 K

0.45

503

504 = Answers to Odd-numbered Exercises

Section 3.8

yf?= 1.93; y = 1.94; yi? = 2.36; yo? = 2.37. yl? = 1.58; yp? = 1.58; ys?= 1.80; y = 1.80.

1, 2,

Chapter4 Section 4.3 1. 1.5 miles/sec.

3.

u=70—52exp(—0.29r);

whent = 5, u = 58.

5.

56°F.

19.

v=b+ (vy —b) exp (—gt/b).

7.

At 2:05 P.M.

21.

1.9 sec and 10.5 ft/sec. Zaye

a t=RS2./x0/k.

23, x=aln2 7 V2y _ Jat,

Ll.

13. 15.

Ifb>a,x >a; ifbb.

25.

(a)s = 160 lb; (b) tf = 64 min.

126 hr.

27.

(a) $5.13; (b) 5.13%.

Section 4.4 3.

t=_

“ing

-1Ing

o 2.7 hr.

-kT

yoe (d) [=e 7

7.

P(t)=

-

=

rhe

aE

5

where O = /k?(d +b)? +B?,a

= arceos ©

— and c) = Py — a

k - ae

Chapter5 Section 5.1 I. x(xy +1) =cy.

15.

yIn|ex| = x(1 — x7).

3.

x3y3 = —3In [cx].

17.

x*?y+xty=cxy’.

5.

y(3x4+y?) = cx?.

19.

x*+exy+y?

7.

x(y? +1) =cy.

21.

xy +aretan (y/x) =c.

9.

x*y* = 2In|cx/y].

23,

2x22 + y? = ex’.

Ll.

x?y? = 2Inlex"/y"|,

25.

3(x? — y?)? = 4(x9 +4).

13,

x(Qxty*)=cy.

27.

x3y3 4+ 4x? —Txy + 2y =0

29.

Ifn #1, n—1)(xy)""!(0? + y? —c) = 2a. Ifn = 1, x? + y? —c = —2a In |xy].

= 1.

Section 5.2

lL

x? -y?+xy—-l=cx.

9.

y(2x — y) =cexp(—x?).

3.

2x*+xy+2yln|y| =cy.

Ll.

x*(y*+2xy —y42)=c.

5.

y(Qx + y) =ce*.

13.

x*(y?+xy—y+2x)

7.

xPy(Qx —3y) =.

17.

x* =4y? In|y/el.

19.

wu? = 2v* In|ev?/ul.

Sc.

Answers to Odd-numbered Exercises Section 5.4 lL. S(x+y+ce) =2In|i5x—10y+11|.

3.

5. 7. 9. 21.

5Q5

11.

2k arctan (ue~?”) = In|e(u? + e*”)].

3tan (6x +c) = 2(9x +4y +1).

13.

+ 6y —7| = 0. 3x-—3y+c+4l1n|3x

xtc=tan(x+y)—sec(xt+y). x3(siny —x)? =csin’ y. (u—v—3)exp(4u) =c(u—v+1). Ifm=landk £0, x* =kIn|cy/x\.

15. 17. 19

(k+n—Dy!l" = (l-n)x*+ex!, (x+2y—1f% =2yte. y(e—x) =x.

Ifn = 1andk =0, y =cx?.

Ifn £lbutk+a=1,

y'" =(1—a)x! Injex|.

23.

4arctan(3x + y) = 8x +7.

25, 27,

x? = yr (x +2). 2y? = x?(3x - 1).

Section 5.5 lL x—3=(2-—y)In|c(y —2)|.

3.

(x+y—3) =c(2x +y —4)*.

5. 7.

x+2y+e=3ln[x+y4+2]. 3 oe In{(x — 1)° + 9(y — 1°) sane

9

y-—1=3(y

—3x—1)In|[ceGx

ll.

(-y-43 =ceet+y—2).

13. 15.

3x? 4+ 4xy — y? 4+ 14x +7 =c. yto=—In|2x-—y+4l.

i Bee 19.

—y+

Pre

= —20r — In —— 3(y-2)

x-l i

|

DI.

21.

y— 5x48 =2%y—x)In2—.

23. 27.

(x —2y —1)? =c(x —3y —2). (2y—x+3)?=c(y—x 42).

31. Uy +1) = -(+2y) In oce + 2y)I. 33.

+3)? =c(y —x +2). Qy—x

Section 5.6 l. In |cy| = x — 5/7 erf x.

3.

y =exp(—x*) [c+ fp exp (B") dB].

5.

y=e® fi Fads.

7.

2y = exp (x?) — x + 54/7 exp (x”) erf x.

Miscellaneous Exercises 1 y?—3y-xt+l=ce™, 3.

2(y—1)= (+ y—3)In|e(e + y — 3).

5. 7. 9,

Ix24+2xy—3y?7-8xt+24yec. 15xty=4y5 +e. x2 =y*(1+cy’).

19 21. 23,

x+yte= 8ln|4x +3y +25). 2(y—3x +c) =5ln|2x —2y —3|. (xty—5)?(x—2y4]1) =c.

ll.

xeOot+e)y

25.

Injx+2y—-—1]=y—2x +c.

13. 15.

(w+ty—8)t=c@—-2y41). y(S+xy4) = ex.

27. 29,

y(Qx—y)=ce™. b*y =cye’* —abx —a —cb.

17.

y*(2x2y —3) = ex?.

31.

xv2(Sx —l =H.

=e.

506

Answers to Odd-numbered Exercises

33.

yrs (xty)*+2in[x+y|t+e.

37.

y*(5 —3x) = x7(54 3x).

35.

x°-+x—3xy—y*?+4y=c.

39.

Injx-y+1l)/=c—x.

Chapter6 Section 6.2 l x >lorx 3.

0.

5.

y=3e*+e™.

7.

y=

3,

W=2e"*,

Te* —3xe*.

Section 6.4 lL

W=O!1!2!.--@—TD!.

Section 6.8 l. 4D?-—7D—-2.

ll.

(D-—1)(D-4)(D +5).

3.

D+D+10.

13.

(D+2)3(2D— 1).

5.

(D+2)2D-—1).

15.

(D—2)(D+3)(D?

7. 9.

(D—1)(D+2)(D — 3). D?(D—2)(D +2).

17. 19.

D?+1—x?. xD?.

21.

+4).

x?D?+2xD—2.

Section 6.9

lL.

y= (ey teax +.03x7)e7*.

3.

y = (C1 +c2x) exp (5x).

Chapter7 Section 7.2 lL. y =cye* +c,e7**.

3. 9.

5.

y=cye™ +c,e7*. 7 y=cre* +c. exp (4x) + ¢3 exp (—3x).

ll.

x =c; +e2e! +c3e7*".

13.

y=cje*

+c

yc;

+oeoe* +c3e7**.

ysowe

+co.e" +c3e".

exp (4x) +¢3 exp (3x).

15.

y=ce7*

17.

y =cye* + cne* +3 exp (—4x) + cq exp (3x).

+e. exp[(2+ V2)x] + ¢3 exp [(2 — /2)x].

19.

y =cje™* + c9e7** + ¢3 exp (4x) + cq exp (3x).

21. 23.

y=ce™ + ce, yoe*—e*,

29.

Whenx = 1, y= —19.8.

25. 27.

Whenx = 1, y =e? 4 3e7! = 21.2, Whenx = 1, y = 20.4.

Section 7.3

lL

y=(e)+eox)e*.

3.

y =e)

5. 7.

poe tox +(e3+cyxje™. y= eye + (cz +¢5x) exp (4x).

17.

19,

= ce

y=

9.

+ (co +¢3x) exp (— 5x).

+ (c2 + e3x)e™*

(ce; teox)e*

y= (ce, tox

+c4x*)e™.

Ll.

ye,

13. 15.

y = (cy +eox)e*/? + (c3 + c4x)e™. y =cye** + (cp 03x + c4x7)e.

+ (c4 + esxje*!?,

+ cye(ltv3)x + cgel-VDx

beox +e3x7 + cge* +c5e7.

Answers to Odd-numbered Exercises

pals

23. 29.

507

—e*,

y=(14+x)e.

25.

y=2-e*

y= 2e* + Bx —2)e*. Whenx =2, y=e-°.

27.

Whenx = 2, y =4e.

Section 7.6 3. y=c e* cosx + cpe* sinx. 5. y=c,cosh3x + cz sinh 3x.

7.

y =cye* cos V3x + cpe™ sin 3x.

9,

y=c)

tex

+03e* cos3x + cye™ sin 3x,

ll.

y = (ce) + ¢2x) cos 3x + (c3 + cqx) sin 3x,

13.

y =c,cosx

15. 17.

y= yocoshx. y =e7>* sin 2x.

19.

x = (vp/k) sinkt.

21.

x =(vo/aje~™ sinat; where a = Vk? — b?.

+ce2 sinx + (c3 + c4x) cos 2x + (c5 + Cox) sin 2x.

Miscellaneous Exercises

5. 7.

lL 3. 9 Ll.

y=e; tee. y=cye* + ce. ye (cy terx +.63x7). y =cye* +2 exp (5x) +.c3 exp

13. 15.

y= (ce) +¢2X) exp (4x) + c3 exp (—5x). y =e*(cy + cox) +c; cos 2x + cy sin 2x,

—$x).

17.

y=c,cosx

19. 2.

y= (ce, teox +03x7)e* + cge™. yse (cy tewx) + c3e¥* + cye7Y*,

23.

y=ae*+

25.

yoe*(c)

27.

+c; sinx. +c. cosx* y =cye

+c2sinx

y=cye* +e (co +034). y =cye* + (cy +. €3x) exp (5x).

+ ¢3 cos 2x + c4 sin 2x.

ce7** + ¢3 exp (—}x)

+ c4exp 31.

+e.x) + ce",

+e3e* +ege*.

—$x). y= cye7* + (c2 + e3x) exp (4x).

* + ere y=cye

y =e, teax te3x? +c4e" +cese73*.

29.

y=cye™ +e.

37,

FS

39. Al, 43,

y=se*(c, + e2x) + c3e"* + c4cos x + c5 Sinx, y =e (cy + cox +03x7) +e (cq + €5x). y =cye™® +e (cy + rx +.€3x").

45,

ysoe*(cy +e2x) + eye" + (e4 + ¢5x) exp (—4x).

cye7* + (c2 cos x + c3 sin x) exp (—4x).

47.

y=l—-e7?.

49,

y=cye*+

51.

y=e

e*

+3e7*

+7

(cq + 5X).

+ c3e* + c4cosx + cs sinx. (cy + cox)

Chapter 8 Section 8.1

lL

(D-2)(D+l)y =0.

(c3 + c4x). +77

33. 35.

3.

D(D—-4)y =0.

508

Answers to Odd-numbered Exercises

5. 7. 9. ll.

(D?—2D+10)y =0. (D?-6D+10)y =0. (D?4+2D+5)*y =0. (D?+k)y =0.

19. 21. 23. 25.

m=O, 0, 2, 2. m=1,1. m= +2), m=2i, 2i, —2i, —2i.

13. 15.

(D?—1)y =0, n= 2) 2;

27. 29.)

m=2i, 2i, —21, —2i. m= ti, +37.

17,

m=—-1t4i.

31.

m=0,

33.

sbi chi, bi.

Section 8.3 lL y=ecy+ee*

+ 5 COs x _ i sinx,

0, 0,

-—1, —1,

7

yoce™

3.

y =cye*

+epe7™* 46x27 -18x421.

9

yscye* + (ce. + pxde® — 4,

5.

y=c)cos3x+e,sin3x+5e"—18x.

11.

+eaxy)+7—

+ ce-*

—l+i.

—5e*.

y=cye* +e.e% +2cosx —4sinx.

13,

y=e (ey

12.cos 2x — 9 sin 2x.

15.

y=c,

17.

y =cye* +ceoe*

19,

y=cy +e2e* +c3e7* — $x?.

21.

y=cye™ +e,e7* + (c3 — xe +x 44.

23.

y=qetla-

25.

y =c,cosx

27.

y=cye™

+oe%

—4x +3 +4008 2x +3 sin 2x,

29,

y=(ey+

5xje

+ ee * +e,

31.

y=

33,

p=cjye*

35.

y= (ce) — 2x)e* + me* + c3e7% — 2x,

37,

ye — Se 42x — 3.

41.

x = (1+e7) sint-—(1—e7) cost.

39,

y=e*(13sinx

43.

y= 2e7% — e* cos2x — e*.

45.

Atx=2,y=e%, yy =1.

47. 49. Sl.

Atx = 2, y = —0.7635, y’ = +0.3012. Atx=4,y=8-—e7!-e?-e7, y=(ce+x)sinx.

53.

The point is (1, 0); the solution is y = x7 + 1 — 2exp (x — 1).

cosx +c) sinx + $x sinx. — 2e™ sinx.

ixje™* +3

(ce) —2x)e-* + ee

Section 8.4

cosx + c4sinx.

+c2sinx + 6 — 2cos 2x.

+ (ex texxye*

—3x+

3,

+ 3 + 4cos 2x — 2 sin 2x.

—cosx)+5e-3*,

a, 5.

yes, y= 2.

19. 21,

y= 2x+4-3e*. ym Je.

71

y= —$

23.

yx

9

y= 3x.

25.

y=—4cos2x.

y= 3m, y=—3x?,

27.

y=

— ih

29.

y= —2x — 5 cos 2x.

y=2sinx.

31.

y= —3 sinde.

1h, 13.

15.

17.

y

te, se

+ 3x. i

3 a

Answers to Odd-numbered Exercises

y= 3e*. y=3e™.

33. 35.

I

37. 39. 41,

y=—qe". y= —4sinx, y= 2e* —5.

43.

= —te%

45,

= fe™.

47.

= * sin 2x 1

49,

509

= 5 c0s 2x.

Chapter 9 Section 9.2 l yooe’+toe*

—x+1.

3.

y= (ce; teax)e*™ +e".

5.

y =e) sinx +c¢2cosx +x sinx +cos x In| cosx|. y=eyx+ 13. yse*(ey teox —In|l —e™)).

7.

=c31sinx +c2cosx cscx. +5 tos y =e)

ll.

Is.

cyx

y=—2-+aln

2,

1 ** l-x

Section 9.4

1. 3.

y =cye* +e9e* + 4xe" — fe" — 1. y=ye—xcosx +sinxIn| sins],

9.

y=y,.—cosxIn|secx

5: 7.

Y= Yet }secx. + tanx|. y=ye—cosxIn|secx

+ tanx|—sinxIn|cscx + cotx|.

ll.

yoyte*In(l

+e").

15.

y=y, —sin(e*) — e* cos (e™).

17.

y=ye- 5x? -1.

13.

y= ye — e** cos (e™*).

19.

y=y—e“*(x—4).

25.

y= yocos(x — x9) + yg sin (x — x) + i f (B) sin (x — B) dB.

Miscellaneous Exercises Ai, y=ye—xe* — te* In(l+e™).

3.

y=yet+4tanx + 5 cosx In|secx + tan x].

5.

y=y,—sinxIn|csex

7.

y=y.e+xsinx +cosx In| cos x].

+cotx|.

9, Il. 13.

y=Ye— xe + $(e* — e*) In|1 — e~* J. y=ye— 4 —coshx arctane™*. y= ye —e7>* sin (e*) — e~* cos (e*).

15.

y=y.+e*

17.

y=ye+4sinx

In|sece*|. tanx — x sinx —cosx In| cos x|.

Chapter 10 Section 10.3

1.

x = 4(cos 8r — 3 sin 87).

3. 5.

x = }cos16r+ $ sin 160. x = 4(t —2)sin 8.

9,

t=7/8, 2/4, 1, 37/8 (sec) andx = —0.15, +0.05, +0.03, +0.04 (ft), respectively.

ll.

Att = 7/8 (sec), x = —§ (ft), v = —8 (ft/sec).

510 = Answers to Odd-numbered Exercises

13. 13% 21.

x =0.33cos11.3¢—-O0.71sin11.3t. 35 in.

23)

—

t=7/8, ‘,

17. 19.

Att = approximately 0.4 sec. x =0.50cos 9.8¢ — 0.82 sin 9.8r.

w/4, 37/8 (sec).

i w/16, 1/8, 37/16 (sec) and x = +0.11, +0.14, —0.54, +0.93 (ft), respec-

tively. Section 10.4

I.

13(1/sec), x = Ste,

(a)

y =

(b) (c)

x = 1.6e~!* sin St. x =0.77(e78 — @7 19-24),

3.

x = fe

(2sin7¢ — cos 71).

5. 7.

x= —je~* sin 8. x= (5 +4t)e* — i cos8r.

9.

t =0.3 sec, x) = —12 in; tg = 0.8 sec, x» = +6 in.; ts = 1.3 sec, x3 = —4 in.

ll.

14.4 sec.

13.

x = 0.30e-** cos 6.4¢ + 0.22e~4* sin 6.4t — 0.05 cos 8¢ (ft).

15.

x = $(8t+ De® — 4 cos 8; fort > 1, x =—t cos 8.

17.

x =exp (—32)(0.30 cos 8.0¢ — 0.22 sin 8.0t) — 0.05 cos 4t + 0.49 sin 4r.

19,

x=

sat.

21.

v* = up? + 2gx.

23.

x = 4gt? + vot — zat?(3v0 + gt) + Harr (4up + gt).

25.

(a)

x =—3e-* sin4r.

(b)

1=0.23+44nz,n=0, 1, 2,3, ....

(c)

0.095.

Section 10.5 1. 38 times.

3.

9.

3.

Te

Orage = OF + Bag).

0.8 (radians/sec) at 0.2 sec.

Chapter 11 Section 11.2

lL. 3. 5.

y =u,w =6u-—8y+x42. yw =u,u’ =—pu—qyt f(x). yY=u,u=v,v =w.w'=y.

7.

vi =2w+2x —-3,w’ = —2u+

9

You

=wi

(2) . (6), ee

6

‘

-—2

5e* —6x

=-v+2w4+1,w'

+9.

=—-y+4v—2u.

s (1 3). 2 (La ei

1

3

Answers to Odd-numbered Exercises

21, (5)x =«( —2“

17. (5)x =«(2).1 es yi Zz

19.

23.

=

1

-l

2

1

1 “) 1

2

25.

X’=AX,whereA= | 1

27,

X’=AX,whereA=|t?

zt 2

1 € y|=b|-l] 0 Zz

—1

t 1

and B= |

4d

0 1}.

+c

sint

¢

O)}

|

¢

1).

andB=

0

Section 11.5

_

1\

3

at

d+ea({)e

1.

X=c

(;

3.

X=c;

(3)

3

e* +e

(3)

1

eth,

1

5.

X=c

(j)+a

7.

X=¢

(3)

a

et +e

3

(a 5

(3) e

es [( power (0) sna] +0[(9) 2+ (2) se] s xeae|(2) nr (9)sn] eae (9) s+)sor]

Section 11.6

x=aet|(! 4) 0821 — (_ 9) sin2r] + ex [ (2) eos r+ ("§) sin

5,

2 7.

0 1} 0

tee

X=c,|—-3)e 2

cos 2t —

0

+ ce! (

) cos 2t + (' —l

9,

ia}. 0

y=e “(cj cosx +c) s8inx).

ll,

y =c,cos2x + c2sin2x.

15.

W(X,(0), X2(0)) = bq £0.

1 xma (Let ree{(2)e4(°)] » xea(eral( Jade MN

Section 11.7

0 O | sin 2r —1

$11

= Answers to Odd-numbered Exercises

I 7

X=

|e;

0

OO}

+e

1

-1 9

ez]

1]

—3

l 1lpe%+] 1

X=c,]

| e'+

ec

3 O} —2

e%

| | t+e3]2] 0

|e

Chapter 12

. X=re(2)+¢()+(°),

Section 12.1

_ 1 (-207 — 181 — 6

>. xp=4( 5.

24? + 14¢ )

X, = —3te! @

— 3e! (3).

Section 12.2

1.

x(t) =4e + 3e7" + 1, y(t) = 8e7' — 3e-” 4 2. Stable arms race.

3.

x(t) =e

4 6e* — 2, y(t) = —e~™ + Ge — 3. Runaway arms race.

Pe Se

Section 12.4

I = ER"[1 —exp(—RtL~')], I= EZ~[-oL cost + Rsinet + wL exp(—RtL~!)]. 1(O) = 1.25(amp), 7 (1) = 1.25 exp (—250r)(amp). Imax = 3e~'(amp).

8

512

1=EZ~(Rsinwt—y coswt)+ EB Ze“ [By cos Bt—a(y+207!C-!) sin Bt].

13.

fy = 2(1—eP™), Ip = 33

Jy = 2 + e300",

Chapter 13 Section 13.2

x2

1

gok

3.

e*,

Chapter 14 Section 14.3 5,

6 2 4 —~-—+—,s>0.

st

3

2

15.

2s + 10

ag p(s—4)(s+2)°~ e ols

(s + a)(s +b) I

Th

2k

> 4 4k) oe 52

b-a ————_.,5 a

oe OF

a

s =

e775)

s> 0.

19.

=e a

1°?

11.

gah

Section 14.10

2k(3s? — k*) (@aep

9

l

L{F(t)} =s7'(1 —e-**), s > 0.

13.

a Ge

cs

s>0.

he 0,

—a,

hs

—b).

2

Answers to Odd-numbered Exercises

—1 s—l

15.

7

1-—expe(l—s) sol. 1—exp(—es) ’

"

o

1

5242

1—exp(—sz/w)

513

Chapter 15 Section 15.1

5,

1.

4e7' sin 3t.

3.

e

9.

e*(2r — 32°).

te,

e7' (2. cos 2t + 3 sin 2r).

(3cos3t — 2sin 31).

Section

135.2

Ll.

“(1

e")

3.

De! — en,

5.

3e% —3— 21.

7,

£~De

Lert,

cos at — cos bt

9.

moa

Section 15.3 lL y=.

11.

x(t) = e% (20? — 2t — 1).

y= fe*—ter™.

13.

y(t) =2cosht

5.

y =cosat.

15.

x(t) =14 4¢— 3 sin2r.

e

y=

9.

ys=e*+2r—-1.

17. 19.

x(t) =sint — cost + 2e'. y(x) = 4e* + cos 3x — 2 sin 3x.

3.

hel

—e%

+

56%.

WU,

x(t) =20 -—6t+7-—8e* +e,

27.

y=e*(c

43,

x(t)=5(1—1)e”.

cosx +c) sinx) + 10x —8 + 4e 3x

Section 15.4 4 9.

= s

2

+e

l

(5 g

ay

ii

+ sinht — 2cost.

‘=

=

13.

*). s

{2

2

2

s

ss?

33

ig,

3c 5

1 — exp (—2s — 2) s+]

Oe”

—tSs

s27+9

4(t — 4)? exp[-20 — 4) ]o(t — 4).

17.

19.

FG) =F(1)=1, F3)=—-1,

23.

x(t) =3—2cost +2[¢ —4—sin(t — 4) a(t — 4).

25.

x(t)= ¢[1 —a(¢ — 2x) ](2 sine — sin 21).

a7.

x(l)=2+e7!, x(4) =143e7!+4e%

Section 15.5

|

t td-—e*) 31

a Fad’ a

6 3.

9.

Tafe

1

st(s

—1) 1

t

yo= sf

H(t — B)sinhkdg.B

7

11.

3(sint —f cost).

f

F(t — B) dB. xy =e™[a+ w43ay]+ | pe? 0

514

= Answers to Odd-numbered Exercises Section 15.6

1. 3. 5.

F(t) = 1422. F(t)=ett $e. F@ =P + 50.

13.

FQ) =443P4+

7 9. Il.

Fa) =? - 34. H(t) = Se + 4e-' — 6te". g(x) =e*(1—x)*.

$1.

Section 15.8

1.

Ely(x) = Bwoe?x? — Awocx? +

3.

Ely(x) = Awoc?x? — ja wocx? + 3 wo[x* — (x — c)ta(x — &)

lex" =x? +(x —cPa(x — 0)].

Section 15.9

l.

x(t) =2— je! — fe* —e, y(t) = 1h +5 —e! + Be,

3. 5.

x(t) = (1+ 2ne! + 2e*, y(t) = (1 — ne! — e*. x(t) =—t — 3sine + 4sin2z, yt) = 14 $ cost — 2 cos 2¢.

7.

x(t) =—lt+e'", y(t) = —te’, z(t) = 1—e' — te’.

9.

x(t) =¢; cosht + cp sinht + j[cosh BF(t — B) + sinh BG(t — B)] ap,

11.

y(t) = c, sinht + 2 cosh + [cosh BG(t — 8) + sinh BF(t — B)] ap. x(t) =cos2r + fj[cos 2B F(t — B) + sin2BG(t — B)] dB, y(t) = —sin2t + fj [cos 2BG(t — B) — sin2B F(t — B)] dp.

Chapter 16 Section 16.2 l.

y=cx,

3. 5. 7.

yse,x?, y = cox}, y =cy exp (5x7), y = —In|eox|. y=In|e;x|, x = In[eyI.

17.

xy = ¢p.

9.

Ll. 13. 15.

x=ylnleyyl, y2x + y) =e.

ye? +¢)) = -2, x3 =3Inleoyl. x =—yln|e,y|, y(2x — y) =e. y—x? =cly, yGx? + y?) =e.

y?(y? + 2x?) = c1, y? = 2x? Inez].

25.

y= (3—2x)!:

27.

y= (3—2x)! forx !

2n |

518

Answers to Odd-numbered Exercises

78

pk y2k-+3/2

y =x’

19.)

yp = x5 yo =

21.

yy

93,

ae a T115Gk 3)

+ os

17.

=x7+

1+)

oy a oe

12), exp (52°)

axT!?,

yy

BH

y=

=

x!/3,

d*y _ oom d*y _ dy

dx

dt’ dx?

dt

dt |

29. 31.

yp =x8syo Sat, y= x2(cy +2 In x).

33.

y =x *[c; cos (Inx) + cp sin (Inx)].

Section 18.6 co 1

yatl

y=) n=0

— ny

= xe";

y

Ms

(—1)"(n

3.

=

+

co

H,x"t!

n=l

ths

=y:Inx-

1

Ixtt?

Let!

De

a=0

5.

yy

ening

ati(n+

OO

.

n=l

yw =l+3 0D"

1A,

es

tt

OE

=

+ In +22";

n=!

y2 = yInx —3(y1 — 1) +3 $e 1)"(2n + 3)x", n=l (-1}*x")

7,

y=X;

9,

=1 .

ya

wins

YO

— 2); yo = y, In(x ane PE 0 ky 2k

eb yo BRK

ll

y=

IS.

= yp=x rns

oo

+

ne? yo = yj Inx “2 —

(-1)"|C-1)[

r=

yt

Te? ne

] "/2 + yas gules a5 (Qn — 3} |x"

3-++(2n — 3)](2Hon—2 — Ay—1 — 2Hy — 2)x"-'/? oO

nalts

=e

w=

H,

x"

yinx—

n=!

a=

5

Section 18.7

~0.n>2,

ae

3. by = Band nd, + (n+ Wby1 + he

5. 7.

.

(n 1)?

n=] Vi.

(=1)"(n + 1) (x —— 2)" EG SRG 2 . 2?n(n — 1)

— 2) — 2(x —2)— 2, 32 Ax 2k (—1)* kK py,

i

by = Zand nb, + (2 + 2)bn—-) + (-1)"(n + 1) = 0, n = 2. b) =—landn?h, + (n — 1)b,_) =0,n > 2.

Section 18.8

1.

y =ag(x — 2x? + 2x7) +4; [

n-3

ry sore n=4

itl

|

Answers to Odd-numbered Exercises

519

2(- 1y"- 3gn— ayn 2 =

-2

xo!

y = a(x ~ —

+35 pal

|

7

ery n=4

Yaga

Sa” 7) + ag(1+ $x + $x).

y = ay (x i-h+ay i

(-1)"*!(n

ee n!

n=3

Dxh= I

y =ao[(x — 1)? +400 — W'] +a[i+ 3-1)

+i(¢—- 1’).

60 - DP"

y=agl+xt+

bas

oo

— 5) In(n — 1)(n — 2)

nas S

cree] ly" (a+

[35

y=ao(l—3xtix—dx) tay E + (— 1)kx 3k

LS:

oo

ofr +d Se] a

Sy gu-4

—l)*x 3k+-2

a|s +) cots z:en

Section 18.9 oo

1.

oO

(—1)"x" —nye yagi +e OYE Dk » He Bear a =) Leiba oo

(Ay + Ay)" (—1)"

niin —1)!

(a1jtly?

Lyn Kn — 2)!" _ y= yilnx ty

nae een

Ly

2

n4+-2 (n + 1)x"*

n=3

2—3)!

oo

1

(—1)" (Ay + Ay, -2)x" ae

yo = winx + hy) +27 —¥ Last

—9

~

2'n\(n — 2)!

t =

Fe

oO

_

45

n+]

—

[1

ee

ie

ant

H,

ee

2(n — 3)! Dyn!

1 os 1-5 + 2x; yo =y= yp Inx+x7 Yi =-242

—_ bea + Lm (—1)"*!(n

») =I

Vise

nine—- hei -se- 45 n=2

n—I

(—1)"*!5"x -n+2

LL

n= _ w=

ni(n oa — 2)! Vy oo

13.

|

nx (ory

ypesx yg

4

z 3 = +3% +2

(—5)" (An n@

+

Ry 2a =)!

tls

Y=) 2, gaan ol 3%-Tn Mn — 2)! yo = yy Inx +: 3x'9 + tx

S 4 y~$

(-7)"( "(Ay A, ++ Ay 2Ay—2 —-1

+

1)x .n+1/3

32"-In Wn — 2)!

I(x

_

1)"

$20

Answers to Odd-numbered Exercises

Section 18.10 1. RSP. 3.

LSP

5.

R-S.P

=

=

—15x—% OO

7

,

9.

y=ag(x? +2x+3) + jas

co

By Ba

Hl

RIOR -DOk= DOES

«0D

ge

be

34.

gt]

nt 4yxr

n=0 60

a

ll.

=

eS

xl:

xno le

Laon 1

=~

co

oo.

13,

42) nt Dx? y= 7g D+ Ant3)(2n+5)x| yo = § VD

15.

y=

n=0

n=0

Spain

_

Lx"!

n=2 oo

G? a

+ yep

yo = yiIn(i/x) +x! +x

— Dx7*,

n=2 oo

oo

+o

yo = yp In(1/x)

se

17,

y=

19.

y,; =cos(2x7'); yo = sin(2x7').

n=l

n=0

Section 18.11 co

1.

yi

baa

in which ap =

1, a)

=

5 ;yw=y

l

—1l, a) = f,

n=0

iS>

3

:

ay

=

_ an- 1+ ay- 3.

=

—

net

Dobe

,

in whichby = 2, by = —3, bs = Bn > 4: by = — et Ps _ Pan n?

a

co

3h

JY

=

age,

in

which

ap

= 1,

ay

=0,

a,

=—4,

n=0

n>3:a,=

Se

+

eed n

y2 = yj Inx + Sax

w =

7

Dy-2 + bys en

Rone:

ypsx-x? tix

x4 + Gx? “2,

= (x ——2 2); yp = yyy, InIn(x

9.

—2

t

+i n=4

2a)

ue wae? as.

“yeea —2y

ie n=]

Miscellaneous Exercises

in which b; = 0, by = 4,

n=]

.

Answers to Odd-numbered Exercises 3 — 2H),-3)x

STR

Ja syilnx +1 fot fat fats yes

-1

-lyws=-x

2

—3 — 2)x" nl (—2)"?(n

eee

+6

oo

(-1)"x"

n=]

oo

=1+

yee

=x° 53 oat

ll?

YI

oo

ntl/2

2n

(n — 2)(n — 3)x

d

.

2

—

+ Sins

a?

pp el

——— (2n — 1)(2n — 3)n!

y= 143)

Les

n!

»

— 3x7}; yp = x? — Ge

yi =

v1

521

x

xnt3/2

3/2

sere

+

n=l

13: 15.

ma

os

aaj —

n

-te

er 3.

vi a *

xl/2

19,

=

1/2

_

If

yee

SSDEB 8.7

a 2

Bio

ne

CO

DA.

=

5

Bt

feo

2

1

3

f_ayn (—3)"3-5---(2n+ 1x

(—3)"3.-5

n+1/2

]

mal

/

++ (Qn + xt?

Sag ne

yy extemal? ney

71

1

Toad

30

— Ay |.

(ere 2-l(n + 2))

(—1)"x"" |

(—1)"+',

x"

_

1

OO" Be

y= yinx¢

YI = n=0

.

n=1

.

oo

25.

2 2)

= — ant DL Gz a Dal n=0

eye

+d,

nine +

oO

23)

Dette

232(n 1)?

3/2 7 wee

Co

y=

“Das 5..-(n+ In!

1-3-+: (2a — Wxrtl? : i "(nF

n=1

aiid

1x yn tl/2

a

se

jain 17.

1/2

yg

yy =24+9x + 24x? +4009; yp = DO" -— H+

It 2x",

a=5 oo

24,

nan+2

yine 2)mai CO yey nao [(n + 1) !] y= =

4)" H

n+2

[(n+1)!]

5/2

29.

ye — Vo oiG@eat eb yp =x a2 1/2x pole m=)

31.

yattat tet y= Din 4-3 + De" n=5

33.

yi =e — 144-40 —

3 $5 — 1)

yp = 14 F@—D +5

-D?.

§22

Answers to Odd-numbered Exercises co

35.

oO

= 04

Ix"; y2 = yi Inx — Dn”.

a=0

37.

n=1

yp = 1—3x+ dx? - 2x3; oo

SE=yInx eT+ 7x — Bx? + bx? -6 )> ° ——_______ Raya oo

39.

=1-8

a

41.

yp=x

8x2; yp =

oop]

ae

ae

Qn

axle

+ 3)x"t1/2

SS

+3)

On = aGr

Dal’

10-13---(3n + 7)x"t2 age

10-13---Gn+7)x"??

ee=y1 nae

[3

ey,

et I(n +2)x"

3

| oe

(n!?

10°

ol

7>! (2n +

--(Qn+1)’

G5 47.

yeh

a

yy == 2"? exp 3x; yo =

49.

ey ey

x" Bee

yeeetie

my

Qn + 3x00?

=

ed

ae On

1

— 2H, |.

347

a

=

2,

2"n!

G — Ay — Ay42)x" ee

niin +2)!

ye n

hogST. = “Grad

= —24 + 32x — 8x?; =

Inx +27? + 8x7! git yn—2

Tet 51.

26

—

70

ie x?

gee i

yet + ss

oo

— 24

aE xr!

Ses a

= [i ae

if

ntl

gael ee ts 7. (n+3)n!

Chapter 20 Section 20.3 l. u=(A,cosaBt + A sinaBt)(B, cos Bx + Bz sin Bx), u = (A3 + Aqgt)( Bs + Byx), and u = (As coshaft + Ag sinhaBrt)(Bs cosh Bx + Bg sinh Bx), in which f and the A’s

and B’s are arbitrary constants.

5.

w= Axkyk-!. & and A arbitrary.

Chapter 22 Section 22.3

1,

A rhe 5 1 -— (—1)"} (-1)"}cos cos — tn f@)~ ~ at

3.

Fe)

5.

ic

c?

ae 5

AITX

(—1)" cos (nwx/c)

ee

2 Asin [2k + )xx/c] FG) ha odes s+ a »k=ih oe 2k +1

sinig, — | ATTX

Answers to Odd-numbered Exercises

523

2. sin 2kx

f(x) ~2n -2 Ri f(x~ 11;

a + —7

;

Lap =|

x

-(- 1)"} cos 5inmx +nx(—-1)""! sin jn

(—1)" cosnx

a=l

13. 15; 17.

2

fQ~

#

r

nen? —24+2(-1)"] sinna

n

48

(-1)'*! cosnx

ote 2» On —DOn +’

fle) ~ sinhe >> 2(—1)" sinh c[c cos (nsx /c) — nx sin(nxx/o)) ct +n? eog —— 5 [sin Lae

ag ti

f(x)

eet

4

‘

19;

1)"*!

5 roe “Tt

fay~ L425° Coen

+ (cos nm — cos + nor) sin

c

@

I]

fx~

; + =

f~

i+

+ (cosa — cos $nz) sin =|.

hs 7 [sin 5 ar cos“

NIX

E

7

21,

laa

1)" } cos =

[ka

Cc. —

sin

Section 22.4

1.

sin [(2k + l)rx/c]

&

4

ie)~ est

| sinamex——. ee wae so | bt AED")

F(x) ~ 2e¢ yj 205

1

eae

Onn

n=]

n[(2k + 1)rx/c] (2k + 1)3 ,

fix) =

fo

ihr

11.

2

251

1 nitty \ . nmt — { 1 — cos —— } sin —.

co

_yyrtl

15.

!

fee), | n=l

13,

vi) in

n=1

nie

_

‘

ee

dunn:

n? sr?

ase (2k+ 1) sin [(2k + 1)x]

POO~ FL

Ok — DQK+9)

2nx[1 — (-1)"e~*] sin (nex /) ioe

>

c2 + nr

n=l

Li

2. Inx[1+(-1)"*! coshke]

fa~)n=!

(key? + (nn?

| c

. nax

aa

AX “2.

I

524

= Answers to Odd-numbered Exercises 19,

Fie)

2ct

~

=

i » \ 2

21.

—pytl io

=

fw~az -

12

an

I

2_.2

—[n?x n

n=1

I

24

24

—_ |

7nd i m+n?

—2+2(-1)

n)

t rn

nIEX

sin

——.,

|e

ot

|sinnzx.

Section 22.5 |

1

f)~5+- le = (2 cos

4c

2°

&

faye

8 aa 7

9.

f(%) ~ cos2x.

»

13. 3,

f(x) é

;

(2k +1)? | 2

y

c ie 3

11.

ATX

1)" [cos “=.

cos [ (2k + l)xx/c}

moe

te

ji

1

(cosmz

— cos =)

COS AITX.

“1

nICX

zal nov sin 55nm — 2(1 — cos $pn) |cos —.

nh ke

.

2ke(—1)" nex Eig (ke)? + (na? cos ;

+ sinh ke y

ke

n=!

15.

o

f(x)

i+eh[S

os

1 —(-1)"

5 SE

Ico

nITX

~s

Chapter 23 Section 23.1

| 3.

u=Alc eye.

5.

u=Al(e—x)/e+ Sob

fore

h

exp -

where So bn sin Bee é

2

(=)

n=l

"

c

ce

2k-+1

or

sin (2k + l)mx

—h* 7? (2k + 1)7t

I

_ dug ?=

t | sin alam

e

f(x) —A(e—x)/e.

n=l

7

11.

w=

30+

180 1

Sy

S1-

n=l

3am

/4

eT n

0.

3.

By EE s—

5.

F(t)=4) (t—2n)’a(t — 2n); F(5) = 17.5.

7.

344.

ag oR anil

n=0

ll.

:

J)

E

=.

t

= x0 (- a)

2E

ae

R

) a=l

(

—])"

"exp (

_

t—ne RC

ace

—

),

ne)

p dp.

526

= Answers to Odd-numbered Exercises Section 24.2 I —— ae—e’

erfc

f

(./t)(./f).

Chapter 25 Section 25.1 lL. yx, t) = 1? +3(¢

— 4x? a(t — 4x).

3.

yx, Qa=x- jt-P + [30 — 4x)? + F(t — 4x) Jo(t — 4x).

5.

y= 3

—4x)a(t — 4x) — 20.

Section 25.2

oo

Lo ysxnx?-P +5 \(-1)"[(t-n ae (tnx) +¢—n 1 +2) n=0

Section 25.5 lL

u=l1—-)

—_

2,

(-1"

ent

(

2n +x

Ji

In+2—x

Di

) + erfc (74 =)|

(tn 1+4)].

_ Index

Abel’s formula, 163 Alpha function, 286 Analytic functions, 342

Damping, critical, 173

Damping, overdamped, 173 Degree of homogeneous function, 24 Dependence, linear, 102

Beams bending moment, 307 boundary conditions, 307 deflection of, 307-310 shearing force, 307

Bernoulli’s Equation, 86-89 Bessel’s equation of index one, 385 of index zero, 373 with index an integer, 401-402 with index not an integer, 400-401 c-discriminant equation, 325-326 Canonical variables, 491-493 Catenary, 338-341 Cauchy type equation, 366 Change of variable equation of order one, 26, 84-86 in Cauchy type equation, 366 linear equation of order two, 152-156

Characteristic equation, 198 Characteristic polynomial, 198 Chemical conversion, 65-69

Clairaut’s equation, 330-334 Class A functions, 261-263 Complementary function, 108 Compound interest, 69 Concrete dams, temperature of, 453-454 Constant coefficients, linear equations with, 117

Continuing methods, 58-61 Convergence, improvement in rapidity of, Convolution theorem, 294-297

Cooling, Newton’s law of, 64 Critical damping, 173 Damped vibrations, 172-177

Damping factor, 172

Dependent variable defined, 2 missing in equation, 334-335 Difference equation, 351 Differential operators, 109-111 exponential shift, 114 laws of operation, 111-113

products of, 109 Differentiation of a product, 367-368 Diffusivity, thermal, 411 Dimensionless variables, 491

Drugs, dissipation of, 72 Eccentricity of a conic, 182 Eccentricity of an orbit, 183 Eigenvalues of a matrix, 196

Eigenvectors of a matrix, 196 Electric circuits, 232-235 Electric networks, 235-241

Eliminating the dependent variable, 328-330 Envelope, 325 Epidemics, 72 Error function application, 488-490

complementary, 472 defined, 471

Escape, velocity of, 62-64 Euler type equation, 366 Euler’s method, 45-48

modified, 48-49 Euler’s theorem on homogeneous functions, 25 Exact equations, 29-34 Existence of solutions, 14—15, 243-246 Exponential function with imaginary argument,

123-125 Exponential order, functions of, 258-261 Exponential shift, 114

527

528

Index Factorial function, 396-397 Families of curves as solutions, 5—10 geometric interpretation, 10-13 Flux of heat, 448 Fourier series at discontinuities, 430 convergence theorem, 430 cosine series, 441-443 defined, 430 improvement in rapidity of convergence, 444-445

numerical analysis, 443-444 numerical examples, 431-438 periodicity, 431 sine series, 438-441 Gamma function, 267-269

General solution homogeneous linear equation, 106-107 nonhomogeneous linear equation, 107-109 Gravitation, Newton’s law of, 178 Half-wave rectification, 270 Harmonic series, 372 Heat conduction quarter infinite solid, 496-499 semi-infinite solid, 488-490

slab, 411-417, 448, 455 slab of finite width, 493-496 sphere, 457-458

Heat equation cartesian coordinates, 411

spherical coordinates, 457-458 validity of, 453-454 Hermite polynomials defined, 402 orthogonality of, 424 Homogeneous coefficients, equations with,

25-29, 83 Homogeneous functions, 24-25 Euler’s theorem on, 25

Homogeneous linear equation with constant coefficients, 117 with variable coefficients, 106-107 Hooke’s law, 165

Hyperbolic functions, 127-133 Hypergeometric function, 397-399 Impedance, steady-state, 239 Impressed force, 166 Inclined plane, motion on a, 68 Independent variable, 2 Independent variable missing, 335-338

Indicial equation defined, 362

difference of roots integral logarithmic case, 381-385 nonlogarithmic case, 377-38|

difference of roots nonintegral, 363-367 equal roots, 368-374 alternate method, 374-377 Infinity

point at, 386 solutions near, 386 Initial value problem, 245 Insulation, 448 Integral equations, special, 298-303

Integral transforms, 252 Integrating factor for equations with homogeneous coefficients, 83 found by inspection, 75-79 linear first order equation, 35 Irregular singular point, 358 Isocline, 13-15 Kepler’s first law, 180-182 Kepler’s second law, 179, 180 Kepler’s third law, 182-184 Kernel of an integral transform, 252 Kirchhoff’s laws, 233

Laguerre polynomials defined, 399-400 orthogonality of, 424 Laplace operator, 253 Laplace transform a convolution theorem, 294-297 defined, 253 deflection of beams, 307-310 derivatives of, 266-267 inverse of, 274-277 linearity of, 253, 275

obtained by power series, 467-471 of of of of of of of

a step function, 286-293 derivatives, 263-266 elementary functions, 253-257 functions of Class A, 261-263 functions of exponential order, 258-261 periodic functions, 269-273 sectionally continuous functions, 257-258

table of, 318 vibration of springs, 303-307 Laplace’s equation in three dimensions cartesian coordinates, 405 in two dimensions, 461-463

Legendre polynomials defined, 403 orthogonality of, 422-423

Linear coefficients, equations with, 89-94 Linear combination of functions, 99-100 Linear dependence, 102 Linear equation change of variable in, 89-94 defined, 4 homogeneous with constant coefficients, 117

Index homogeneous with variable coefficients,

106-107 irregular singular point of, 358 nonhomogeneous with constant coefficients,

134 nonhomogeneous with variable coefficients,

107-109

validity of, 347 Orthogonality

of Hermite polynomials, 424 of Laguerre polynomials, 424 defined, 418 of Legendre polynomials, 422-423

of polynomials, 419-420 zeros of, 421

order one, 35-43 ordinary point of, 345 solutions near, 347

Overdamped motion, 173

Regular singular point of, 346

p-discriminant equation, 326-328

regular singular point of, 358 systems of, 186-187 undetermined coefficients, 139-144 variation of parameters, 156-16] Linear independence of a set of functions, 102 of a set of vector functions, 200

of a set of vectors, 200 Linearity of differential operators, 112-113

of inverse Laplace transform, 275 of Laplace transform, 253

Lipschitz condition, 246 Logarthmic solutions, 381-385

Logistic growth, 69-73 Many-term recurrence relation, 388-392 Matrix algebra

characteristic equation, 198 characteristic polynomial, 198

eigenvalues, 198 eigenvectors, 198 reviewed, 189-197 Mixture problems, 65-69 Networks, electric, 235-241 Newton’s constant of gravitation, 178 Newton’s law of cooling, 64-65 Newton’s law of gravitation, 178 Newton’s second law of motion, 178 Nonelementary integrals, 94-98 Normal linear equation, 99 Null function, 274 Numerical methods continuing methods, 58-61 Euler’s method, 45-48 Euler’s method, modified, 48-49 Runge-Kutta, 54-58 successive approximation, 49-51

Taylor’s theorem, 52-54 Operator

Parameters, variation of, 156-161

Partial differential equations change of variables, 405 defined, 404 of applied mathematics, 404-406

Pendulum, simple, 177-178 Periodic extension, 431 Phase plane complex eigenvalues, 222-226 defined, 220 phase portrait, 220 real distinct eigenvalues, 221] stable node, 221

trajectory, 220 unstable node, 221 unstable saddle, 221 Phase portrait, 220 Planetary motion, 178-179 Polynomials Hermite, 402 Laguerre, 399-400

Legendre, 403 orthogonality of, 419-420

simple sets of, 419 Power series convergence of, 343-345 table of, 468 Recurrence relation defined, 351

many-term, 388-392 Reduction of order

linear equation, 152-156 nonlinear equation, 334-335

Regular singular point, 358 many-term recurrence relation, 388-392 solutions near, 362 Regular singular point of linear equation, 346 Resonance,

169-172

Retarding force, 165 RLC circuit, 233 Runge-Kutta, 54-58

Laplace, 253

Order of a differential equation, 2

Sectionally continuous functions, 257-258

Ordinary differential equation, 3

Separation of variables

Ordinary point defined, 345

solutions near, 347

529

ordinary differential equations, 18-23

partial diferential equations, 406-41| Series, computation with, 443-444

530

Index Shearing force, beams, 307 Simple sets of polynomials, 419 Singular point defined, 346 irregular, 358

regular, 358 solutions near, 362 Singular solutions, 323-325 Solution curve, 11 Solution of a differential equation, 3 Spring, vibration of, 165 constant, 165

critical damping, 173 damped, 172-177 forced, 166 overdamped, 173 resonance in, 169-172 transform methods, 303-307 undamped, 167-169 Square-wave function, 271

real distinct eigenvalues, 197-206 repeated eigenvalues, 212-220 Taylor’s theorem, 52-54 Tractrix, 68

Trajectory, 220 Triangular-wave function, 272

Undamped vibrations, 167-169 Undetermined coefficients, 139-144

Variable dependent, 2 independent, 2

Variation of parameters, 156-161 Velocity of escape, 62

Voltage law, Kirchhoff’s, 233 Wave equation one dimension, 458-461, 485-488

Steady-state temperature, 461

Weight function, 418

String, elastic, 458-461, 485-488 Successive approximation, 49-51

Wronskian Abel’s formula for, 163

Superposition of solutions, 147

of solutions of a system, 200

Systems of equations

of solutions of an equation, 103-106

Laplace transform method, 310-316 matrix methods complex eigenvalues, 206-212

Zeros of orthogonal polynomials, 421

WO i BO01L1-a