Calculus 7eap1501 05

SECTION 15.1 Exact First-Order Equations 1093

Exact Differential Equations Integrating Factors

Exact Differential EquationsIn Section 5.6, you studied applications of differential equations to growth and decayproblems. In Section 5.7, you learned more about the basic ideas of differential equa-tions and studied the solution technique known as separation of variables. In thischapter, you will learn more about solving differential equations and using them inreal-life applications. This section introduces you to a method for solving the first-order differential equation

for the special case in which this equation represents the exact differential of a function

From Section 12.3, you know that if f has continuous second partials, then

This suggests the following test for exactness.

Exactness is a fragile condition in the sense that seemingly minor alterations inan exact equation can destroy its exactness. This is demonstrated in the followingexample.

My

52f

yx5

2fxy

5Nx

.

z 5 f sx, yd.

Msx, yd dx 1 Nsx, yd dy 5 0

15.1S E C T I O N Exact First-Order Equations

Definition of an Exact Differential Equation

The equation is an exact differential equation ifthere exists a function f of two variables x and y having continuous partial deriv-atives such that

and

The general solution of the equation is f sx, yd 5 C.fysx, yd 5 Nsx, yd.fxsx, yd 5 Msx, yd


THEOREM 15.1 Test for Exactness

Let M and N have continuous partial derivatives on an open disc R. The differen-tial equation is exact if and only if

My

5Nx

.


1094 CHAPTER 15 Differential Equations

EXAMPLE 1 Testing for Exactness

a. The differential equation is exact because

and

But the equation is not exact, even though it is obtainedby dividing both sides of the first equation by x.

b. The differential equation is exact because

and

But the equation is not exact, even though it differs from the first equation only by a single sign.

Note that the test for exactness of is the same as thetest for determining whether is the gradient of a poten-tial function (Theorem 14.1). This means that a general solution to anexact differential equation can be found by the method used to find a potential function for a conservative vector field.

EXAMPLE 2 Solving an Exact Differential Equation

Solve the differential equation

Solution The given differential equation is exact because

The general solution, is given by

In Section 14.1, you determined by integrating with respect to y and reconciling the two expressions for An alternative method is to partially differentiate this version of with respect to y and compare the result with

In other words,

Thus, and it follows that Therefore,

and the general solution is Figure 15.1 shows the solution curvesthat correspond to 10, 100, and 1000.C 5 1,

x2y 2 x3 2 y2 5 C.

f sx, yd 5 x2y 2 x3 2 y2 1 C1gsyd 5 2y2 1 C1.g9syd 5 22y,

g9syd 5 22y

fysx, yd 5

y fx2y 2 x3 1 gsydg 5 x2 1 g9syd 5 x2 2 2y.

Nsx, yd

Nsx, yd.f sx, yd

f sx, yd.Nsx, ydgsyd

5 E s2xy 2 3x2d dx 5 x2y 2 x3 1 gsyd. f sx, yd 5 E Msx, yd dx

f sx, yd 5 C,

My

5

y f2xy 2 3x2g 5 2x 5 N

x5

x fx2 2 2yg.

s2xy 2 3x2d dx 1 sx2 2 2yd dy 5 0.

f sx, yd 5 CFsx, yd 5 Msx, yd i 1 Nsx, ydj


cos y dx 1 sy2 1 x sin yd dy 5 0

Nx

5

x f y2 2 x sin yg 5 2sin y.M

y5

y fcos yg 5 2sin y

cos y dx 1 sy2 2 x sin yd dy 5 0

sy2 1 1d dx 1 xy dy 5 0

Nx

5

x f yx2g 5 2xy.M

y5

y fxy2 1 xg 5 2xy

sxy2 1 xd dx 1 yx2 dy 5 0

x

y

44 8

8

8 12

12

16

20

24

12

C = 1C = 10

C = 100

C = 1000

Figure 15.1

NOTE Every differential equation ofthe form

is exact. In other words, a separable vari-ables equation is actually a special typeof an exact equation.

Msxd dx 1 Nsyd dy 5 0


EXAMPLE 3 Solving an Exact Differential Equation

Find the particular solution of

that satisfies the initial condition when

Solution The differential equation is exact because

Because is simpler than it is better to begin by integrating

Thus, and

which implies that , and the general solution is

General solution

Applying the given initial condition produces

which implies that Hence, the particular solution is

Particular solution

The graph of the particular solution is shown in Figure 15.3. Notice that the graphconsists of two parts: the ovals are given by and the y-axis is givenby

In Example 3, note that if the total differential of zis given by

In other words, is called an exact differential equation becauseis exactly the differential of f sx, yd.M dx 1 N dy

M dx 1 N dy 5 0

5 Msx, yd dx 1 Nsx, yd dy. 5 scos x 2 x sin x 1 y2d dx 1 2xy dy

dz 5 fxsx, yd dx 1 fysx, yd dy

z 5 f sx, yd 5 xy2 1 x cos x,

x 5 0.y2 1 cos x 5 0,

xy2 1 x cos x 5 0.

C 5 0.

ps1d2 1 p cos p 5 C

xy2 1 x cos x 5 C.

f sx, yd 5 xy2 1 x cos x 1 C1 5 x cos x 1 C1

gsxd 5 E scos x 2 x sin xd dxg9sxd 5 cos x 2 x sin x

g9sxd 5 cos x 2 x sin x

fxsx, yd 5

x fxy2 1 gsxdg 5 y2 1 g9sxd 5 cos x 2 x sin x 1 y2

Msx, yd

f sx, yd 5 E Nsx, yd dy 5 E 2xy dy 5 xy2 1 gsxdNsx, yd.Msx, yd,Nsx, yd

y fcos x 2 x sin x 1 y2g 5 2y 5

x f2xyg.

Nx

My

x 5 p.y 5 1

scos x 2 x sin x 1 y2d dx 1 2xy dy 5 0

x

y

2

4

2

4

pi 2pi 3pipi3pi 2pi

( , 1)pi

Figure 15.3

TECHNOLOGY You can use agraphing utility to graph a particularsolution that satisfies the initial condi-tion of a differential equation. InExample 3, the differential equationand initial conditions are satisfiedwhen whichimplies that the particular solutioncan be written as or

On a graphing calculator screen, the solution wouldbe represented by Figure 15.2 togetherwith the y-axis.

Figure 15.2

12.57

4

12.57

4

y 5 !2cos x .x 5 0

xy2 1 x cos x 5 0,


Integrating FactorsIf the differential equation is not exact, it may be possi-ble to make it exact by multiplying by an appropriate factor which is called anintegrating factor for the differential equation.

EXAMPLE 4 Multiplying by an Integrating Factor

a. If the differential equation

Not an exact equation

is multiplied by the integrating factor the resulting equation

Exact equation

is exactthe left side is the total differential of b. If the equation

Not an exact equation

is multiplied by the integrating factor the resulting equation

Exact equation

is exactthe left side is the total differential of

Finding an integrating factor can be difficult. However, there are two classes of differential equations whose integrating factors can be found routinelynamely,those that possess integrating factors that are functions of either x alone or y alone.The following theorem, which we present without proof, outlines a procedure forfinding these two special categories of integrating factors.

STUDY TIP If either or is constant, Theorem 15.2 still applies. As an aid toremembering these formulas, note that the subtracted partial derivative identifies both thedenominator and the variable for the integrating factor.

ksydhsxd

xyy.

1y dx 2 x

y2 dy 5 0

usx, yd 5 1yy2,

y dx 2 x dy 5 0

x2y.

2xy dx 1 x2 dy 5 0

usx, yd 5 x,

2y dx 1 x dy 5 0

usx, yd,Msx, yd dx 1 Nsx, yd dy 5 0

THEOREM 15.2 Integrating Factors

Consider the differential equation

1. If

is a function of x alone, then is an integrating factor.2. If

is a function of y alone, then is an integrating factor.eeksyd dy

1Msx, yd fNxsx, yd 2 Mysx, ydg 5 ksyd

eehsxd dx

1Nsx, yd fMysx, yd 2 Nxsx, ydg 5 hsxd

Msx, yd dx 1 Nsx, yd dy 5 0.


EXAMPLE 5 Finding an Integrating Factor


Solution The given equation is not exact because and However, because

it follows that is an integrating factor. Multiplying the given differential equation by produces the exact differential equation

whose solution is obtained as follows.

Therefore, and which implies that

The general solution is or

In the next example, we show how a differential equation can help in sketching aforce field given by

EXAMPLE 6 An Application to Force Fields

Sketch the force field given by

by finding and sketching the family of curves tangent to F.

Solution At the point in the plane, the vector has a slope of

which, in differential form, is

From Example 5, we know that the general solution of this differential equation isor Figure 15.4 shows several representa-

tive curves from this family. Note that the force vector at is tangent to the curvepassing through sx, yd.

sx, ydy2 5 x 2 1 1 Ce2x.y2 2 x 1 1 5 Ce2x,

sy2 2 xd dx 1 2y dy 5 0. 2y dy 5 2sy2 2 xd dx

dydx 5

2sy2 2 xdy!x2 1 y22yy!x2 1 y2

52sy2 2 xd

2y

Fsx, ydsx, yd

Fsx, yd 5 2y!x2 1 y2

i 2 y2 2 x

!x2 1 y2 j

Fsx, yd 5 Msx, yd i 1 Nsx, ydj.

y2 2 x 1 1 5 Ce2x.y2ex 2 xex 1 ex 5 C,

f sx, yd 5 y2ex 2 xex 1 ex 1 C1.gsxd 5 2xex 1 ex 1 C1,g9sxd 5 2xex

g9sxd 5 2xex

fxsx, yd 5 y2ex 1 g9sxd 5 y2ex 2 xexMsx, yd

f sx, yd 5 E Nsx, yd dy 5 E 2yex dy 5 y2ex 1 gsxdsy2ex 2 xexd dx 1 2yex dy 5 0

exeehsxd dx 5 ee dx 5 ex

Mysx, yd 2 Nxsx, ydNsx, yd 5

2y 2 02y 5 1 5 hsxd

Nxsx, yd 5 0.Mysx, yd 5 2y

sy2 2 xd dx 1 2y dy 5 0.

3

2

13

2

3

y

x

jx2

yx

2

22i

2

2x

),,(xFForce field:

yy

F:.toesrve

c

x

uenngtof1ail

2

am

yF

x

yCt

y y

Figure 15.4


In Exercises 110, determine whether the differential equationis exact. If it is, find the general solution.

1.2.3.4.5.6.

7.

8.

9.

10.

In Exercises 11 and 12, (a) sketch an approximate solution ofthe differential equation satisfying the initial condition by handon the direction field, (b) find the particular solution that satis-fies the initial condition, and (c) use a graphing utility to graphthe particular solution. Compare the graph with the hand-drawn graph of part (a).

11.

12.

Figure for 11 Figure for 12

In Exercises 1316, find the particular solution that satisfies theinitial condition.

13.

14.

15.16.

In Exercises 1726, find the integrating factor that is a functionof x or y alone and use it to find the general solution of the differential equation.

17.18.19.20.21.22.23.24.25.26.

In Exercises 2730, use the integrating factor to find the general solution of the differential equation.

27.

28.

29.

30.

31. Show that each of the following is an integrating factor for thedifferential equation

(a) (b) (c) (d)

32. Show that the differential equation

is exact only if If show that is an integrat-ing factor, where

In Exercises 3336, use a graphing utility to graph the family oftangent curves to the given force field.

33.

34.

35.

36. Fsx, yd 5 s1 1 x2d i 2 2xy j

Fsx, yd 5 4x2y i 2 12xy2 1 xy22 jFsx, yd 5 x

!x2 1 y2 i 2 y

!x2 1 y2 j

Fsx, yd 5 y!x2 1 y2

i 2 x!x2 1 y2

j

n 5 22a 1 ba 1 b .m 5 2

2b 1 aa 1 b ,

xmyna b,a 5 b.

saxy2 1 byd dx 1 sbx2y 1 axd dy 5 0

1x2 1 y2

1xy

1y2

1x2

y dx 2 x dy 5 0.

usx, yd 5 x22y222y3 dx 1 sxy2 2 x2d dy 5 0usx, yd 5 x22y23s2y5 1 x2yd dx 1 s2xy4 2 2x3d dy 5 0usx, yd 5 x2ys3y2 1 5x2yd dx 1 s3xy 1 2x3d dy 5 0usx, yd 5 xy2s4x2y 1 2y2d dx 1 s3x3 1 4xyd dy 5 0

s22y3 1 1d dx 1 s3xy2 1 x3d dy 5 02y dx 1 sx 2 sin!yd dy 5 0sx2 1 2x 1 yd dx 1 2 dy 5 0y2 dx 1 sxy 2 1d dy 5 0s2x2y 2 1d dx 1 x3 dy 5 0sx 1 yd dx 1 tan x dy 5 0s5x2 2 y2d dx 1 2y dy 5 0s5x2 2 yd dx 1 x dy 5 0s2x3 1 yd dx 2 x dy 5 0y dx 2 sx 1 6y2d dy 5 0

ys3d 5 1sx2 1 y2d dx 1 2xy dy 5 0ys0d 5 pe3xssin 3y dx 1 cos 3y dyd 5 0

ys0d 5 41x2 1 y2

sx dx 1 y dyd 5 0

ys2d 5 4yx 2 1 dx 1 flnsx 2 1d 1 2yg dy 5 0

Initial ConditionDifferential Equation

x

y

4 2 2 4

4

2

2

4

ys4d 5 31!x2 1 y2

sx dx 1 y dyd 5 0

ys12d 5 py4s2x tan y 1 5d dx 1 sx2sec2yd dy 5 0Initial ConditionDifferential Equation

ey cos xy fydx 1 sx 1 tan xyd dyg 5 0

1sx 2 yd2 sy

2 dx 1 x2 dyd 5 0

e2sx21y2dsx dx 1 y dyd 5 0

1x2 1 y2

sx dy 2 y dxd 5 0

2y2exy2 dx 1 2xyexy2 dy 5 0s4x3 2 6xy2d dx 1 s4y3 2 6xyd dy 5 02 coss2x 2 yd dx 2 coss2x 2 yd dy 5 0s3y2 1 10xy2d dx 1 s6xy 2 2 1 10x2yd dy 5 0yex dx 1 ex dy 5 0s2x 2 3yd dx 1 s2y 2 3xd dy 5 0

E X E R C I S E S F O R S E C T I O N 15 .1

x

y

4 2 2 4

4

2

4

2

LAB SERIES Lab 20

In Exercises 37 and 38, find an equation for the curve with thespecified slope passing through the given point.

37.

38.

39. Cost If represents the cost of producing x units in amanufacturing process, the elasticity of cost is defined as

Find the cost function if the elasticity function is

where and

40. Eulers Method Consider the differential equationwith the initial condition At any point

in the domain of F, yields the slope of the solu-tion at that point. Eulers Method gives a discrete set of esti-mates of the y values of a solution of the differential equationusing the iterative formula

where (a) Write a short paragraph describing the general idea of how

Eulers Method works.(b) How will decreasing the magnitude of affect the accu-

racy of Eulers Method?

41. Eulers Method Use Eulers Method (see Exercise 40) toapproximate for the values of given in the table if

and (Note that the number of iterationsincreases as decreases.) Sketch a graph of the approximatesolution on the direction field in the figure.

The value of accurate to three decimal places, is 4.213.

42. Programming Write a program for a graphing utility orcomputer that will perform the calculations of Eulers Methodfor a specified differential equation, interval, and initialcondition. The output should be a graph of the discrete pointsapproximating the solution.

Eulers Method In Exercises 4346, (a) use the program ofExercise 42 to approximate the solution of the differential equa-tion over the indicated interval with the specified value of and the initial condition, (b) solve the differential equation analytically, and (c) use a graphing utility to graph the particu-lar solution and compare the result with the graph of part (a).

Differential Initial

43. 0.01

44. 0.1

45. 0.05

46. 0.2

47. Eulers Method Repeat Exercise 45 for and discuss how the accuracy of the result changes.

48. Eulers Method Repeat Exercise 46 for and discuss how the accuracy of the result changes.

True or False? In Exercises 4952, determine whether thestatement is true or false. If it is false, explain why or give anexample that shows it is false.

49. The differential equation is exact.50. If is exact, then is also

exact.

51. If is exact, then is also exact.

52. The differential equation is exact.f sxd dx 1 gsyd dy 5 0Ng dy 5 0

f f sxd 1 M g dx 1 fgsyd 1M dx 1 N dy 5 0

xM dx 1 xN dy 5 0M dx 1 N dy 5 02xy dx 1 sy2 2 x2d dy 5 0

Dx 5 0.5

Dx 5 1

ys0d 5 1f0, 5gy9 5 6x 1 y2

ys3y 2 2xd

ys2d 5 1f2, 4gy9 5 2xyx2 1 y2

ys21d 5 21f21, 1gy9 5 p4 sy2 1 1d

ys1d 5 1f1, 2gy9 5 x 3!y

Condition Dx IntervalEquation

Dx

Dx,

x

y

4 3 2 1 1 2

5

4

3

2

1

ys1d,

Dxys0d 5 2.y9 5 x 1 !y

Dxys1d

Dx

Dx 5 xk11 2 xk.

yk11 5 yk 1 Fsxk , ykd Dx

Fsxk , ykdsxk , ykdysx0d 5 y0.y9 5 Fsx, yd

x 100.Cs100d 5 500

Esxd 5 20x 2 y2y 2 10x

Esxd 5 marginal costaverage cost 5

C9sxdCsxdyx 5

x

y

dydx .

y 5 Csxd

s0, 2ddydx 522xy

x2 1 y2

s2, 1ddydx 5y 2 x

3y 2 x

PointSlope

0.50 0.25 0.10

Estimate of ys1d

Dx



15.2S E C T I O N First-Order Linear Differential EquationsFirst-Order Linear Differential Equations Bernoulli Equations Applications

First-Order Linear Differential EquationsIn this section, you will see how integrating factors help to solve a very importantclass of first-order differential equationsfirst-order linear differential equations.

To solve a first-order linear differential equation, you can use an integratingfactor which converts the left side into the derivative of the product Thatis, you need a factor such that

Because you dont need the most general integrating factor, let Multiplying theoriginal equation by produces

The general solution is given by

yeePsxd dx 5 E QsxdeePsxd dx dx 1 C.

ddx 3yeePsxd dx4 5 QsxdeePsxd dx.

y9eePsxd dx 1 yPsxdeePsxd dx 5 QsxdeePsxd dxusxd 5 eePsxddxy9 1 Psxdy 5 Qsxd

C 5 1.

usxd 5 CeePsxd dx.

ln|usxd| 5 E Psxd dx 1 C1 Psxd 5 u9sxd

usxd

usxdPsxdy 5 yu9sxd usxdy9 1 usxdPsxdy 5 usxdy9 1 yu9sxd

usxddydx 1 usxdPsxdy 5dfusxdyg

dx

usxdusxdy.usxd,

Definition of First-Order Linear Differential Equation

A first-order linear differential equation is an equation of the form

where P and Q are continuous functions of x. This first-order linear differentialequation is said to be in standard form.

dydx 1 Psxdy 5 Qsxd

ANNA JOHNSON PELL WHEELER (18831966)

Anna Johnson Pell Wheeler was awarded amasters degree from the University of Iowafor her thesis The Extension of Galois Theoryto Linear Differential Equations in 1904.Influenced by David Hilbert, she worked onintegral equations while studying infinite linearspaces.

SECTION 15.2 First-Order Linear Differential Equations 1101

STUDY TIP Rather than memorizing this formula, just remember that multiplication by theintegrating factor converts the left side of the differential equation into the derivativeof the product

EXAMPLE 1 Solving a First-Order Linear Differential Equation

Find the general solution of

Solution The standard form of the given equation is

Standard form

Thus, and you have

Integrating factor

Therefore, multiplying both sides of the standard form by yields

General solution

Several solution curves are shown in Figure 15.5.sfor C 5 22, 21, 0, 1, 2, 3, and 4d

y 5 x2sln |x| 1 Cd.

yx2

5 ln |x| 1 C

yx2

5 E 1x dx

ddx 3

yx24 5

1x

y9x2

22yx3

51x

1yx2

eePsxd dx 5 e2ln x2

51x2

.

E Psxd dx 5 2E 2x dx 5 2ln x2

Psxd 5 22yx,

y9 2 12x2y 5 x. y9 1 Psxdy 5 Qsxd

xy9 2 2y 5 x2.

yeePsxd dx.eePsxd dx

THEOREM 15.3 Solution of a First-Order Linear Differential Equation

An integrating factor for the first-order linear differential equation

is The solution of the differential equation is

yeePsxd dx5E QsxdeePsxd dx dx 1 C.usxd 5 eePsxd dx.

y9 1 Psxdy 5 Qsxd

Figure 15.5

2 1 1 2

1

2

2

1

y

x

C = 4C = 3C = 2C = 1

C = 0

C = 1

C = 2


EXAMPLE 2 Solving a First-Order Linear Differential Equation


Solution The equation is already in the standard form Thus,and

which implies that the integrating factor is

Integrating factor

A quick check shows that is also an integrating factor. Thus, multiplyingby produces

General solution

Several solution curves are shown in Figure 15.6.

Bernoulli EquationsA well-known nonlinear equation that reduces to a linear one with an appropriatesubstitution is the Bernoulli equation, named after James Bernoulli (16541705).

This equation is linear if and has separable variables if Thus, in thefollowing development, assume that and Begin by multiplying by and to obtain

which is a linear equation in the variable Letting produces the linearequation

Finally, by Theorem 15.3, the general solution of the Bernoulli equation is

dzdx 1 s1 2 ndPsxdz 5 s1 2 ndQsxd.

z 5 y12ny12n.

ddx f y

12ng 1 s1 2 ndPsxdy12n 5 s1 2 ndQsxd

s1 2 ndy2ny9 1 s1 2 ndPsxdy12n 5 s1 2 ndQsxd y2ny9 1 Psxdy12n 5 Qsxd

s1 2 ndy2nn 1.n 0

n 5 1.n 5 0,

y 5 tan t 1 C sec t.

y cos t 5 sin t 1 C

y cos t 5 E cos t dt

ddt f y cos tg 5 cos t

cos ty9 2 y tan t 5 1cos t

5 |cos t|. eePstd dt 5 e ln |cos t|

E Pstd dt 5 2E tan t dt 5 ln |cos t|Pstd 5 2tan t,

y9 1 Pstdy 5 Qstd.

2p

2 < t < p

2.y9 2 y tan t 5 1,

y12nees12ndPsxddx 5 E s1 2 ndQsxdees12ndPsxd dx dx 1 C.

Figure 15.6

y

t

C = 2

C = 1

C = 1

C = 2

C = 0

pi pi2 2

2

1

1

2

y9 1 Psxdy 5 Qsxdyn Bernoulli equation


EXAMPLE 3 Solving a Bernoulli Equation


Solution For this Bernoulli equation, let and use the substitution

Let

Differentiate.

Multiplying the original equation by produces

Original equation

Multiply both sides by

Linear equation:

This equation is linear in z. Using produces

which implies that is an integrating factor. Multiplying the linear equation by thisfactor produces

Linear equation

Exact equation

Write left side as total differential.

Integrate both sides.

Divide both sides by

Finally, substituting the general solution is

General solution

So far you have studied several types of first-order differential equations. Ofthese, the separable variables case is usually the simplest, and solution by an inte-grating factor is usually a last resort.

y4 5 2e2x2 1 Ce22x2.

z 5 y4,

e2x2. z 5 2e2x2 1 Ce22x2.

ze2x2

5 2ex2 1 C

ze2x2

5 E 4xex2 dx

ddx fze

2x2g 5 4xex2 z9e2x

21 4xze2x2 5 4xex2 z9 1 4xz 5 4xe2x2

e2x2

E Psxd dx 5 E 4x dx 5 2x2Psxd 5 4x

z9 1 Psxdz 5 Qsxd z9 1 4xz 5 4xe2x2.4y3. 4y3y9 1 4xy4 5 4xe2x2

y9 1 xy 5 xe2x2y234y3

z9 5 4y3y9.z 5 y12n 5 y12s23d. z 5 y4

n 5 23,

y9 1 xy 5 xe2x2y23.

Summary of First-Order Differential Equations

1. Separable variables:2. Homogeneous: where M and N are nth-degree homogeneous3. Exact: where 4. Integrating factor: is exact5. Linear:6. Bernoulli equation: y9 1 Psxdy 5 Qsxdyn

y9 1 Psxdy 5 Qsxdusx, ydMsx, yddx 1 usx, ydNsx, yddy 5 0

Myy 5 NyxMsx, yddx 1 Nsx, yddy 5 0,Msx, yddx 1 Nsx, yddy 5 0,Msxddx 1 Nsyddy 5 0

Form of Equation Method


ApplicationsOne type of problem that can be described in terms of a differential equation involveschemical mixtures, as illustrated in the next example.

EXAMPLE 4 A Mixture Problem

A tank contains 50 gallons of a solution composed of 90% water and 10% alcohol.A second solution containing 50% water and 50% alcohol is added to the tank atthe rate of 4 gallons per minute. As the second solution is being added, the tankis being drained at the rate of 5 gallons per minute, as shown in Figure 15.7. Assumingthe solution in the tank is stirred constantly, how much alcohol is in the tank after10 minutes?

Solution Let y be the number of gallons of alcohol in the tank at any time t. You knowthat when Because the number of gallons of solution in the tank at anytime is and the tank loses 5 gallons of solution per minute, it must lose

gallons of alcohol per minute. Furthermore, because the tank is gaining 2 gallons ofalcohol per minute, the rate of change of alcohol in the tank is given by

To solve this linear equation, let and obtain

Because you can drop the absolute value signs and conclude that

Thus, the general solution is

Because when you have

which means that the particular solution is

Finally, when the amount of alcohol in the tank is

which represents a solution containing 33.6% alcohol.

y 550 2 10

2 2 20150 2 10

50 25

5 13.45 gal

t 5 10,

y 550 2 t

2 2 20150 2 t

50 25.

220505 5 C5 5

502 1 Cs50d

5

t 5 0,y 5 5

y 550 2 t

2 1 Cs50 2 td5.

ys50 2 td5 5 E 2s50 2 td5 dt 5 12s50 2 td4 1 C

eePstddt 5 e25 lns502 td 51

s50 2 td5.

t < 50,

EPstd dt 1 E 550 2 t dt 5 25 ln |50 2 t|.Pstd 5 5ys50 2 td

dydt 1 1

550 2 t2y 5 2.

dydt 5 2 2 1

550 2 t2y

1 550 2 t2y50 2 t,

t 5 0.y 5 5Figure 15.7

4 gal/min

5 gal/min


In most falling-body problems discussed so far in the text, we have neglected airresistance. The next example includes this factor. In the example, the air resistance onthe falling object is assumed to be proportional to its velocity v. If g is the gravita-tional constant, the downward force F on a falling object of mass m is given by thedifference But by Newtons Second Law of Motion, you know that

which yields the following differential equation.

EXAMPLE 5 A Falling Object with Air Resistance

An object of mass m is dropped from a hovering helicopter. Find its velocity as afunction of time t, assuming that the air resistance is proportional to the velocity ofthe object.

Solution The velocity v satisfies the equation

where g is the gravitational constant and k is the constant of proportionality. Lettingyou can separate variables to obtain

Because the object was dropped, when thus and it follows that

NOTE Notice in Example 5 that the velocity approaches a limit of as a result of the air resistance. For falling-body problems in which air resistance is neglected, the velocity increases without bound.

A simple electrical circuit consists of electric current I (in amperes), a resistanceR (in ohms), an inductance L (in henrys), and a constant electromotive force E (involts), as shown in Figure 15.8. According to Kirchhoffs Second Law, if the switchS is closed when the applied electromotive force (voltage) is equal to the sumof the voltage drops in the rest of the circuit. This in turn means that the current Isatisfies the differential equation

L dIdt 1 RI 5 E.

t 5 0,

mgyk

v 5g 2 ge2bt

b 5mgk s1 2 e

2ktymd.2bv 5 2g 1 ge2bt

g 5 C,t 5 0;v 5 0

g 2 bv 5 Ce2bt. ln |g 2 bv| 5 2bt 2 bC1

21b ln |g 2 bv| 5 t 1 C1

E dvg 2 bv 5 E dt dv 5 sg 2 bvd dt

b 5 kym,

dvdt 1

kvm

5 g

dvdt 1

km

v 5 gm dvdt 5 mg 2 kv

F 5 ma 5 msdvydtd,mg 2 kv.

Figure 15.8

E

S

R I

L


True or False? In Exercises 1 and 2, determine whether thestatement is true or false. If it is false, explain why or give anexample that shows it is false.

1. is a first-order linear differential equation.2. is a first-order linear differential equation.

In Exercises 3 and 4, (a) sketch an approximate solution of thedifferential equation satisfying the initial condition by hand onthe direction field, (b) find the particular solution that satisfiesthe initial condition, and (c) use a graphing utility to graph theparticular solution. Compare the graph with the hand-drawngraph of part (a).

3.

4.

In Exercises 512, solve the first-order linear differentialequation.

5.

6.

7.8.9.

10.11.12.

In Exercises 1318, find the particular solution of the differen-tial equation that satisfies the boundary condition.

13.14.15.16.

17.

18. ys1d 5 2y9 1 s2x 2 1dy 5 0

ys2d 5 2y9 1 11x2y 5 0ys0d 5 4y9 1 y sec x 5 sec xys0d 5 1y9 1 y tan x 5 sec x 1 cos xys1d 5 ex3y9 1 2y 5 e1yx2ys0d 5 5y9 cos2 x 1 y 2 1 5 0

Boundary ConditionDifferential Equation

y9 1 5y 5 e5xsx 2 1dy9 1 y 5 x2 2 1sy 2 1dsin x dx 2 dy 5 0s3y 1 sin 2xd dx 2 dy 5 0y9 1 2xy 5 2xy9 2 y 5 cos x

dydx 1 1

2x2y 5 3x 1 1

dydx 1 1

1x2y 5 3x 1 4

s0, 4dy9 1 2y 5 sin x

s0, 1ddydx 5 ex 2 y

Initial ConditionDifferential Equation

y9 1 xy 5 exyy9 1 x!y 5 x2

E X E R C I S E S F O R S E C T I O N 15 . 2

EXAMPLE 6 An Electric Circuit Problem

Find the current I as a function of time t (in seconds), given that I satisfies thedifferential equation where R and L are nonzero constants.

Solution In standard form, the given linear equation is

Let so that and, by Theorem 15.3,

Thus, the general solution is

I 5 14L2 1 R2 sR sin 2t 2 2L cos 2td 1 Ce2sRyLdt

.

I 5 e2sRyLdt 3 14L2 1 R2 esRyLdtsR sin 2t 2 2L cos 2td 1 C4

51

4L2 1 R2 esRyLdtsR sin 2t 2 2L cos 2td 1 C.

IesRyLdt 5 1L E esRyLdt sin 2t dteePstddt 5 esRyLdt,Pstd 5 RyL,

dIdt 1

RL I 5

1L sin 2t.

LsdIydtd 1 RI 5 sin 2t,

x

y

4 2 2 42

2

4

x

y

4 2 2 42

4

2

4

Figure for 3 Figure for 4


In Exercises 1924, solve the Bernoulli differential equation.

19. 20.

21. 22.

23. 24.

In Exercises 2528, (a) use a graphing utility to graph thedirection field for the differential equation, (b) find theparticular solutions of the differential equation passing throughthe specified points, and (c) use a graphing utility to graph theparticular solutions on the direction field.

25.

26.

27.

28.

Electrical Circuits In Exercises 2932, use the differentialequation for electrical circuits given by

In this equation, I is the current, R is the resistance, L is theinductance, and E is the electromotive force (voltage).29. Solve the differential equation given a constant voltage 30. Use the result of Exercise 29 to find the equation for the current

if volts, ohms, and henrys.When does the current reach 90% of its limiting value?

31. Solve the differential equation given a periodic electromotiveforce

32. Verify that the solution of Exercise 31 can be written in theform

where the phase angle, is given by (Notethat the exponential term approaches 0 as This impliesthat the current approaches a periodic function.)

33. Population Growth When predicting population growth,demographers must consider birth and death rates as well as thenet change caused by the difference between the rates of immi-gration and emigration. Let P be the population at time t and letN be the net increase per unit time resulting from the differencebetween immigration and emigration. Thus, the rate of growthof the population is given by

N is constant.

Solve this differential equation to find P as a function of time ifat time the size of the population is

34. Investment Growth A large corporation starts at time to continuously invest part of its receipts at a rate of P dollarsper year in a fund for future corporate expansion. Assume thatthe fund earns r percent interest per year compounded continu-ously. Thus, the rate of growth of the amount A in the fund isgiven by

where when Solve this differential equation for Aas a function of t.

Investment Growth In Exercises 35 and 36, use the result ofExercise 34.

35. Find A for the following.(a) and years(b) and years

36. Find t if the corporation needs $800,000 and it can invest$75,000 per year in a fund earning 8% interest compoundedcontinuously.

37. Intravenous Feeding Glucose is added intravenously tothe bloodstream at the rate of q units per minute, and the bodyremoves glucose from the bloodstream at a rate proportional tothe amount present. Assume is the amount of glucose inthe bloodstream at time t.(a) Determine the differential equation describing the rate of

change with respect to time of glucose in the bloodstream.(b) Solve the differential equation from part (a), letting

when (c) Find the limit of as

38. Learning Curve The management at a certain factory hasfound that the maximum number of units a worker can producein a day is 30. The rate of increase in the number of units Nproduced with respect to time t in days by a new employee isproportional to (a) Determine the differential equation describing the rate of

change of performance with respect to time.(b) Solve the differential equation from part (a).(c) Find the particular solution for a new employee who

produced ten units on the first day at the factory and 19units on the twentieth day.

30 2 N.

t `.Qstdt 5 0.Q 5 Q0

Qstd

t 5 10P 5 $250,000, r 5 5%,t 5 5P 5 $100,000, r 5 6%,

t 5 0.A 5 0

dAdt 5 rA 1 P

t 5 0

P0.t 5 0

dPdt 5 kP 1 N,

t `.arctans2vLyRd.f,

I 5 ce2sRyLdt 1 E0!R2 1 v2L2

sinsvt 1 fd

E0 sin vt.

L 5 4R 5 550Is0d 5 0, E0 5 110

E0.

L dIdt 1 RI 5 E.

s0, 3d, s0, 1ddydx 1 2xy 5 xy2

s1, 1d, s3, 21ddydx 1 scot xdy 5 x

s0, 72d, s0, 212ddydx 1 2xy 5 x

3

s22, 4d, s2, 8ddydx 21x y 5 x2

Points Differential Equation

yy9 2 2y2 5 exy9 2 y 5 x3 3!y

y9 1 11x2y 5 x!yy9 1 11x2y 5 xy2

y9 1 2xy 5 xy2y9 1 3x2y 5 x2y3


Mixture In Exercises 3944, consider a tank that at timecontains gallons of a solution of which, by weight,

pounds is soluble concentrate. Another solution containing pounds of the concentrate per gallon is running into the tank atthe rate of gallons per minute. The solution in the tank is keptwell stirred and is withdrawn at the rate of gallons perminute.

39. If Q is the amount of concentrate in the solution at any time t,show that

40. If Q is the amount of concentrate in the solution at any time t,write the differential equation for the rate of change of Q withrespect to t if

41. A 200-gallon tank is full of a solution containing 25 pounds ofconcentrate. Starting at time distilled water is admittedto the tank at a rate of 10 gallons per minute, and thewell-stirred solution is withdrawn at the same rate.(a) Find the amount of concentrate Q in the solution as a

function of t.(b) Find the time at which the amount of concentrate in the

tank reaches 15 pounds.(c) Find the quantity of the concentrate in the solution as

42. Repeat Exercise 41, assuming that the solution entering thetank contains 0.05 pound of concentrate per gallon.

43. A 200-gallon tank is half full of distilled water. At time a solution containing 0.5 pound of concentrate per gallon entersthe tank at the rate of 5 gallons per minute, and the well-stirredmixture is withdrawn at the rate of 3 gallons per minute.(a) At what time will the tank be full?(b) At the time the tank is full, how many pounds of concen-

trate will it contain?44. Repeat Exercise 43, assuming that the solution entering the

tank contains 1 pound of concentrate per gallon.

In Exercises 4548, match the differential equation with itssolution.

45. (a)46. (b)47. (c)48. (d)

In Exercises 4964, solve the first-order differential equation byany appropriate method.

49. 50.

51.52.

53.

54.

55.

56.57.58.59.60.61.62.63.64. x dx 1 sy 1 eydsx2 1 1ddy 5 0

3ydx 2 sx2 1 3x 1 y2ddy 5 0ydx 1 s3x 1 4yddy 5 0sx2y4 2 1ddx 1 x3y3 dy 5 0sy2 1 xyddx 2 x2 dy 5 0s2y 2 exddx 1 x dy 5 0sx 1 yddx 2 x dy 5 0s3y2 1 4xyddx 1 s2xy 1 x2ddy 5 0y9 5 2x!1 2 y2

sx2 1 cos yd dydx 5 22xy

sx 1 1d dydx 5 ex 2 y

y cos x 2 cos x 1dydx 5 0

s1 1 2e2x1yddx 1 e2x1y dy 5 0s1 1 y2ddx 1 s2xy 1 y 1 2d dy 5 0

dydx 5

x 1 1ysy 1 2d

dydx 5

e2x1y

ex2y

y 5 Ce2xy9 2 2xy 5 xy 5 x2 1 Cy9 2 2xy 5 0y 5 212 1 Cex

2y9 2 2y 5 0y 5 Cex2y9 2 2x 5 0

Solution Differential Equation

t 5 0,

t `.

t 5 0,

r1 5 r2 5 r.

dQdt 1

r2Qv0 1 sr1 2 r2dt

5 q1r1.

r2

r1

q1q0v0t 5 0

S E C T I O N P R O J E C T

Weight Loss A persons weight depends on both the amountof calories consumed and the energy used. Moreover, the amountof energy used depends on a persons weightthe averageamount of energy used by a person is 17.5 calories per poundper day. Thus, the more weight a person loses, the less energythe person uses (assuming that the person maintains a constantlevel of activity). An equation that can be used to model weightloss is

where w is the persons weight (in pounds), t is the time in days,and C is the constant daily calorie consumption.(a) Find the general solution of the differential equation.(b) Consider a person who weighs 180 pounds and begins a diet

of 2500 calories per day. How long will it take the person tolose 10 pounds? How long will it take the person to lose 35pounds?

(c) Use a graphing utility to graph the solution. What is thelimiting weight of the person?

(d) Repeat parts (b) and (c) for a person who weighs 200pounds when the diet is started.

1dwdt 2 5C

3500 217.53500 w

SECTION 15.3 Second-Order Homogeneous Linear Equations 1109

Second-Order Linear Differential Equations Higher-Order Linear Differential Equations Applications

Second-Order Linear Differential EquationsIn this section and the following section, we discuss methods for solving higher-orderlinear differential equations.

NOTE Notice that this use of the term homogeneous differs from that in Section 5.7.

We discuss homogeneous equations in this section, and leave the nonhomoge-neous case for the next section.

The functions are linearly independent if the only solution of theequation

is the trivial one, Otherwise, this set of functions islinearly dependent.

EXAMPLE 1 Linearly Independent and Dependent Functions

a. The functions and are linearly independent because the onlyvalues of and for which

for all x are and b. It can be shown that two functions form a linearly dependent set if and only if one

is a constant multiple of the other. For example, and arelinearly dependent because

has the nonzero solutions and C2 5 1.C1 5 23

C1x 1 C2s3xd 5 0

y2sxd 5 3xy1sxd 5 x

C2 5 0.C1 5 0

C1 sin x 1 C2x 5 0

C2C1y2 5 xy1sxd 5 sin x

C1 5 C2 5 . . . 5 Cn 5 0.

y1, y2, . . . , yn

15.3S E C T I O N Second-Order Homogeneous Linear Equations

Definition of Linear Differential Equation of Order

Let and f be functions of x with a common (interval) domain. Anequation of the form

is called a linear differential equation of order n. If the equation ishomogeneous; otherwise, it is nonhomogeneous.

f sxd 5 0,ysnd 1 g1sxdysn21d 1 g2sxdysn22d 1 . . . 1 gn21sxdy9 1 gnsxdy 5 f sxd

g1, g2, . . . , gn

n

C1y1 1 C2y2 1 . . . 1 Cnyn 5 0


The following theorem points out the importance of linear independence inconstructing the general solution of a second-order linear homogeneous differentialequation with constant coefficients.

Proof We prove this theorem in only one direction. If and are solutions, you canobtain the following system of equations.

Multiplying the first equation by , multiplying the second by and adding theresulting equations together produces

which means that

is a solution, as desired. The proof that all solutions are of this form is best left to afull course on differential equations.

Theorem 15.4 states that if you can find two linearly independent solutions, youcan obtain the general solution by forming a linear combination of the two solutions.

To find two linearly independent solutions, note that the nature of the equationsuggests that it may have solutions of the form If so, then

and Thus, by substitution, is a solution if and only if

Because is never 0, is a solution if and only if

Characteristic equation

This is the characteristic equation of the differential equation

Note that the characteristic equation can be determined from its differential equationsimply by replacing with with and with 1.ym,y9m2,y0

y0 1 ay9 1 by 5 0.

y 5 emxemx emxsm2 1 am 1 bd 5 0.

m2emx 1 amemx 1 bemx 5 0 y0 1 ay9 1 by 5 0

y 5 emxy0 5 m2emx.y9 5 memxy 5 emx.y0 1 ay9 1 by 5 0

y 5 C1y1 1 C2y2

fC1y10 sxd 1 C2y20 sxdg 1 afC1y19sxd 1 C2y29sxdg 1 bfC1y1sxd 1 C2y2sxdg 5 0

C2,C1

y20 sxd 1 ay29 sxd 1 by2sxd 5 0y10 sxd 1 ay19 sxd 1 by1sxd 5 0

y2y1

THEOREM 15.4 Linear Combinations of Solutions

If and are linearly independent solutions of the differential equationthen the general solution is

where and are constants.C2C1

y 5 C1y1 1 C2y2

y0 1 ay9 1 by 5 0,y2y1

m2 1 am 1 b 5 0.


EXAMPLE 2 Characteristic Equation with Distinct Real Roots


Solution In this case, the characteristic equation is

Characteristic equation

so Thus, and are particular solutions ofthe given differential equation. Furthermore, because these two solutions are linearlyindependent, you can apply Theorem 15.4 to conclude that the general solution is

General solution

The characteristic equation in Example 2 has two distinct real roots. Fromalgebra, you know that this is only one of three possibilities for quadratic equations.In general, the quadratic equation has roots

and

which fall into one of three cases.

1. Two distinct real roots, 2. Two equal real roots, 3. Two complex conjugate roots, and In terms of the differential equation these three cases correspondto three different types of general solutions.

y0 1 ay9 1 by 5 0,

m2 5 a 2 bim1 5 a 1 bim1 5 m2

m1 m2

m2 52a 2 !a2 2 4b

2m1 52a 1 !a2 2 4b

2

m2 1 am 1 b 5 0

y 5 C1e2x 1 C2e22x.

y2 5 em2x 5 e22xy1 5 em1x 5 e2xm 5 2.

m2 2 4 5 0

y0 2 4y 5 0.

E X P LO R AT I O NFor each differential equation belowfind the characteristic equation. Solvethe characteristic equation for m, anduse the values of m to find a generalsolution to the differential equation.Using your results, develop a generalsolution to differential equations withcharacteristic equations that havedistinct real roots.

(a)(b) y0 2 6y9 1 8y 5 0

y0 2 9y 5 0

THEOREM 15.5 Solutions of

The solutions of

fall into one of the following three cases, depending on the solutions of thecharacteristic equation,

1. Distinct Real Roots If are distinct real roots of the characteristicequation, then the general solution is

2. Equal Real Roots If are equal real roots of the characteristicequation, then the general solution is

3. Complex Roots If and are complex roots of thecharacteristic equation, then the general solution is

y 5 C1eax cos bx 1 C2eax sin bx.

m2 5 a 2 bim1 5 a 1 bi

y 5 C1em1x 1 C2xem1x 5 sC1 1 C2xdem1x.

m1 5 m2

y 5 C1em1x 1 C2em2x.

m1 m2

m2 1 am 1 b 5 0.

y0 1 ay9 1 by 5 0

y0 1 ay9 1 by 5 0


EXAMPLE 3 Characteristic Equation with Complex Roots

Find the general solution of the differential equation

Solution The characteristic equation

has two complex roots, as follows.

Thus, and and the general solution is

NOTE In Example 3, note that although the characteristic equation has two complex roots, thesolution of the differential equation is real.

EXAMPLE 4 Characteristic Equation with Repeated Roots


subject to the initial conditions and

Solution The characteristic equation

has two equal roots given by Thus, the general solution is

General solution

Now, because when we have

Furthermore, because when we have

Therefore, the solution is

Particular solution

Try checking this solution in the original differential equation.

y 5 2e22x 1 5xe22x.

5 5 C2. 1 5 22s2ds1d 1 C2f22s0ds1d 1 1g y9 5 22C1e22x 1 C2s22xe22x 1 e22xd

x 5 0,y9 5 1

2 5 C1s1d 1 C2s0ds1d 5 C1.

x 5 0,y 5 2

y 5 C1e22x 1 C2xe22x.

m 5 22.

m2 1 4m 1 4 5 sm 1 2d2 5 0

y9s0d 5 1.ys0d 5 2

y0 1 4y9 1 4y 5 0

y 5 C1e23x cos!3x 1 C2e23x sin!3x.

b 5 !3,a 5 23

5 23 !3i 5 23 !23

526 !212

2

m 526 !36 2 48

2

m2 1 6m 1 12 5 0

y0 1 6y9 1 12y 5 0.


Higher-Order Linear Differential EquationsFor higher-order homogeneous linear differential equations, you can find the generalsolution in much the same way as you do for second-order equations. That is, you beginby determining the n roots of the characteristic equation. Then, based on these n roots,you form a linearly independent collection of n solutions. The major difference is thatwith equations of third or higher order, roots of the characteristic equation may occurmore than twice. When this happens, the linearly independent solutions are formed bymultiplying by increasing powers of x, as demonstrated in Examples 6 and 7.

EXAMPLE 5 Solving a Third-Order Equation


Solution The characteristic equation is

Because the characteristic equation has three distinct roots, the general solution is

General solution



Solution The characteristic equation is

Because the root occurs three times, the general solution is

General solution

EXAMPLE 7 Solving a Fourth-Order Equation


Solution The characteristic equation is as follows.

Because each of the roots and occurstwice, the general solution is

General solutiony 5 C1 cos x 1 C2 sin x 1 C3x cos x 1 C4x sin x.

m2 5 a 2 bi 5 0 2 im1 5 a 1 bi 5 0 1 i

m 5 i sm2 1 1d2 5 0

m4 1 2m2 1 1 5 0

ys4d 1 2y0 1 y 5 0.

y 5 C1e2x 1 C2xe2x 1 C3x2e2x.

m 5 21

m 5 21. sm 1 1d3 5 0

m3 1 3m2 1 3m 1 1 5 0

y999 1 3y0 1 3y9 1 y 5 0.

y 5 C1 1 C2e2x 1 C3ex.

m 5 0, 1, 21. msm 2 1dsm 1 1d 5 0

m3 2 m 5 0

y999 2 y9 5 0.


ApplicationsOne of the many applications of linear differential equations is describing the motionof an oscillating spring. According to Hookes Law, a spring that is stretched (orcompressed) y units from its natural length l tends to restore itself to its natural lengthby a force F that is proportional to y. That is, where k is the springconstant and indicates the stiffness of the given spring.

Suppose a rigid object of mass m is attached to the end of a spring and causes adisplacement, as shown in Figure 15.9. Assume that the mass of the spring isnegligible compared with m. If the object is pulled down and released, the resultingoscillations are a product of two opposing forcesthe spring force andthe weight mg of the object. Under such conditions, you can use a differentialequation to find the position y of the object as a function of time t. According toNewtons Second Law of Motion, the force acting on the weight is where

is the acceleration. Assuming that the motion is undampedthat is,there are no other external forces acting on the objectit follows that

and you have

Undamped motion of a spring

EXAMPLE 8 Undamped Motion of a Spring

Suppose a 4-pound weight stretches a spring 8 inches from its natural length. Theweight is pulled down an additional 6 inches and released with an initial upwardvelocity of 8 feet per second. Find a formula for the position of the weight as afunction of time t.

Solution By Hookes Law, so Moreover, because the weight w isgiven by mg, it follows that Hence, the resulting differentialequation for this undamped motion is

Because the characteristic equation has complex roots the general solution is

Using the initial conditions, you have

Consequently, the position at time t is given by

y 512 cos 4

!3 t 1 2!33 sin 4

!3 t.

y9s0d 5 8C2 52!3

3 .8 5 24!3 1122s0d 1 4!3 C2s1d

y9std 5 24!3 C1 sin 4!3 t 1 4!3 C2 cos 4!3 t

ys0d 5 12C1 512

12 5 C1s1d 1 C2s0d

y 5 C1e0 cos 4!3 t 1 C2e0 sin 4!3 t 5 C1 cos 4!3 t 1 C2 sin 4!3 t.

m 5 0 4!3i,m2 1 48 5 0

d2ydt2 1 48y 5 0.

m 5 wyg 5 432 518 .

k 5 6.4 5 ks23d,

d2ydt2 1 1 km 2y 5 0.

2ky,msd2yydt2d 5

a 5 d2yydt2F 5 ma,

Fsyd 5 2ky

Fs yd 5 2ky,

A rigid object of mass m attached to the endof the spring causes a displacement of y.Figure 15.9

m

l = naturallength

y = displacement


In Exercises 14, verify the solution of the differential equation.

1.2.3.4.

In Exercises 530, find the general solution of the lineardifferential equation.

5. 6.7. 8.9. 10.

11. 12.13. 14.15. 16.17. 18.19. 20.21. 22.23. 24.25. 26.27.28.29.30.

31. Consider the differential equation and thesolution Find the particularsolution satisfying each of the following initial conditions.(a)(b)(c)

32. Determine C and such that is a particularsolution of the differential equation where

In Exercises 3336, find the particular solution of the lineardifferential equation.

33. 34.

35. 36.

Think About It In Exercises 37 and 38, give a geometricargument to explain why the graph cannot be a solution of thedifferential equation. It is not necessary to solve the differentialequation.

37. 38. y0 5 212 y9y0 5 y9

ys0d 5 2, y9s0d 5 1ys0d 5 0, y9s0d 5 2y0 1 2y9 1 3y 5 0y0 1 16y 5 0ys0d 5 2, y9s0d 5 1ys0d 5 1, y9s0d 5 24y0 1 2y9 1 3y 5 0y0 2 y9 2 30y 5 0

y9s0d 5 25.y0 1 vy 5 0,

y 5 C sin!3 tvys0d 5 21, y9s0d 5 3ys0d 5 0, y9s0d 5 2ys0d 5 2, y9s0d 5 0

y 5 C1 cos 10x 1 C2 sin 10x.y0 1 100y 5 0

y999 2 3y0 1 3y9 2 y 5 0y999 2 3y0 1 7y9 2 5y 5 0y999 2 y0 2 y9 1 y 5 0y999 2 6y0 1 11y9 2 6y 5 0

ys4d 2 y0 5 0ys4d 2 y 5 02y0 2 6y9 1 7y 5 09y0 2 12y9 1 11y 5 03y0 1 4y9 2 y 5 0y0 2 3y9 1 y 5 0y0 2 4y9 1 21y 5 0y0 2 2y9 1 4y 5 0y0 2 2y 5 0y0 2 9y 5 0y0 1 4y 5 0y0 1 y 5 09y0 2 12y9 1 4y 5 016y0 2 8y9 1 y 5 0y0 2 10y9 1 25y 5 0y0 1 6y9 1 9y 5 016y0 2 16y9 1 3y 5 02y0 1 3y9 2 2y 5 0y0 1 6y9 1 5y 5 0y0 2 y9 2 6y 5 0y0 1 2y9 5 0y0 2 y9 5 0

y0 1 2y9 1 10y 5 0y 5 e2x sin 3xy0 1 4y 5 0y 5 C1 cos 2x 1 C2 sin 2xy0 2 4y 5 0y 5 C1e2x 1 C2e22xy0 1 6y9 1 9y 5 0y 5 sC1 1 C2xde23xDifferential EquationSolution


Suppose the object in Figure 15.10 undergoes an additional damping or frictionalforce that is proportional to its velocity. A case in point would be the damping forceresulting from friction and movement through a fluid. Considering this dampingforce, the differential equation for the oscillation is

or, in standard linear form,

Damped motion of a springd2ydt2 1

pm

1dydt2 1km

y 5 0.

m d2ydt2 5 2ky 2 p

dydt

2psdyydtd,

A damped vibration could be caused byfriction and movement through a liquid.Figure 15.10

3 2 1 1 2 3

1

32

54

x

y

3 2 1 2 3

1

32

x

y

3


Vibrating Spring In Exercises 3944, describe the motion ofa 32-pound weight suspended on a spring. Assume that theweight stretches the spring foot from its natural position.

39. The weight is pulled foot below the equilibrium position andreleased.

40. The weight is raised foot above the equilibrium position andreleased.

41. The weight is raised foot above the equilibrium positionand started off with a downward velocity of foot per second.

42. The weight is pulled foot below the equi-librium position and started off with an upward velocity of foot per second.

43. The weight is pulled foot below the equilibrium position andreleased. The motion takes place in a medium that furnishes adamping force of magnitude speed at all times.

44. The weight is pulled footbelow the equilibrium position and released. The motion takesplace in a medium that furnishes a damping force of magnitude

at all times.

Vibrating Spring In Exercises 4548, match the differentialequation with the graph of a particular solution. [The graphsare labeled (a), (b), (c), and (d).] The correct match can be madeby comparing the frequency of the oscillations or the rate atwhich the oscillations are being damped with the appropriatecoefficient in the differential equation.

(a) (b)

(c) (d)

45. 46.47. 48.

49. If the characteristic equation of the differential equation

has two equal real roots given by show that

is a solution.

50. If the characteristic equation of the differential equation

has complex roots given by and show that

is a solution.

True or False? In Exercises 5154, determine whether thestatement is true or false. If it is false, explain why or give anexample that shows it is false.

51. is the general solution of

52. is the general solu-tion of

53. is a solution ofif and only if

54. It is possible to choose a and b such that is a solutionof

The Wronskian of two differentiable functions f and g, denotedby W( f, g), is defined as the function given by the determinant

The functions f and g are linearly independent if there exists atleast one value of x for which In Exercises 5558,use the Wronskian to verify the linear independence of the twofunctions.

55. 56.

57. 58.

59. Eulers differential equation is of the form

where a and b are constants.

(a) Show that this equation can be transformed into a second-order linear equation with constant coefficients by using thesubstitution

(b) Solve 60. Solve

where A is constant, subject to the conditions andyspd 5 0.

ys0d 5 0

y0 1 Ay 5 0

x2y0 1 6xy9 1 6y 5 0.x 5 et.

x2y0 1 axy9 1 by 5 0, x > 0

y2 5 x2y2 5 eax cos bx, b 0y1 5 xy1 5 eax sin bxy2 5 xeaxy2 5 ebx, a by1 5 eaxy1 5 eax

Wx f, gc 0.

Wx f, gc 5 | ff9 gg9|.

y0 1 ay9 1 by 5 0.y 5 x2ex

a1 5 a0 5 0.a0y 5 0anysnd 1 an21ysn21d 1 . . . 1 a1y9 1y 5 x

ys4d 1 2y0 1 y 5 0.y 5 sC1 1 C2xdsin x 1 sC3 1 C4xdcos x

9 5 0.y0 2 6y9 1y 5 C1e3x 1 C2e23x

y 5 C1eax cos bx 1 C2eax sin bx

m2 5 a 2 bi,m1 5 a 1 bi

y0 1 ay9 1 by 5 0

y 5 C1erx 1 C2xerxm 5 r,

y0 1 ay9 1 by 5 0

y0 1 y9 1 374 y 5 0y0 1 2y9 1 10y 5 0y0 1 25y 5 0y0 1 9y 5 0

14 |v|

12

18

12

12

12

12

23

23

12

23

5 6

3

421x

y

6

3

42 31x

y

61

3

3

x

y

5 642 3

3

x

y

SECTION 15.4 Second-Order Nonhomogeneous Linear Equations 1117

15.4S E C T I O N Second-Order Nonhomogeneous Linear EquationsNonhomogeneous Equations Method of Undetermined Coefficients Variation of Parameters

Nonhomogeneous EquationsIn the preceding section, we represented damped oscillations of a spring by the homo-geneous second-order linear equation

Free motion

This type of oscillation is called free because it is determined solely by the spring andgravity and is free of the action of other external forces. If such a system is also subject to an external periodic force such as caused by vibrations at the oppo-site end of the spring, the motion is called forced, and it is characterized by the nonhomogeneous equation

Forced motion

In this section, you will study two methods for finding the general solution of anonhomogeneous linear differential equation. In both methods, the first step is to findthe general solution of the corresponding homogeneous equation.

General solution of homogeneous equation

Having done this, you try to find a particular solution of the nonhomogeneous equation.

Particular solution of nonhomogeneous equation

By combining these two results, you can conclude that the general solution of the nonhomogeneous equation is as stated in the following theorem.y 5 yh 1 yp,

y 5 yp

y 5 yh

d2ydt2 1

pm

1dydt2 1km

y 5 a sin bt.

a sin bt,

d2ydt2 1

pm

1dydt2 1km

y 5 0.

THEOREM 15.6 Solution of Nonhomogeneous Linear Equation

Let

be a second-order nonhomogeneous linear differential equation. If is a partic-ular solution of this equation and is the general solution of the correspondinghomogeneous equation, then

is the general solution of the nonhomogeneous equation.

y 5 yh 1 yp

yhyp

y0 1 ay9 1 by 5 Fsxd

SOPHIE GERMAIN (17761831)

Many of the early contributors to calculuswere interested in forming mathematicalmodels for vibrating strings and membranes,oscillating springs, and elasticity. One of thesewas the French mathematician SophieGermain, who in 1816 was awarded a prize bythe French Academy for a paper entitledMemoir on the Vibrations of Elastic Plates.

The

Gra

nger

Col

lect

ion


Method of Undetermined CoefficientsYou already know how to find the solution of a linear homogeneous differentialequation. The remainder of this section looks at ways to find the particular solution

If in

consists of sums or products of or you can find a particularsolution by the method of undetermined coefficients. The gist of this method is toguess that the solution is a generalized form of Here are some examples.

1. If choose 2. If choose 3. If choose

Then, by substitution, determine the coefficients for the generalized solution.

EXAMPLE 1 Method of Undetermined Coefficients

Find the general solution of the equation

Solution To find solve the characteristic equation.

or

Thus, Next, let be a generalized form of

Substitution into the original differential equation yields

By equating coefficients of like terms, you obtain

and

with solutions and B Therefore,

and the general solution is

5 C1e2x 1 C2e3x 115 cos x 2

25 sin x.

y 5 yh 1 yp

yp 515 cos x 2

25 sin x

5 225 .A 5

15

2A 2 4B 5 224A 2 2B 5 0

s24A 2 2Bdcos x 1 s2A 2 4Bdsin x 5 2 sin x. 2A cos x 2 B sin x 1 2A sin x 2 2B cos x 2 3A cos x 2 3B sin x 5 2 sin x

y0 2 2y9 2 3y 5 2 sin x

yp0 5 2A cos x 2 B sin x yp9 5 2A sin x 1 B cos x yp 5 A cos x 1 B sin x

2 sin x.ypyh 5 C1e2x 1 C2e3x.

m 5 3 m 5 21 sm 1 1dsm 2 3d 5 0

m2 2 2m 2 3 5 0

yh,

y0 2 2y9 2 3y 5 2 sin x.

yp 5 sAx 1 Bd 1 C sin 2x 1 D cos 2x.Fsxd 5 x 1 sin 2x,yp 5 Axex 1 Bex.Fsxd 5 4xex,

yp 5 Ax2 1 Bx 1 C.Fsxd 5 3x2,

Fsxd.ypyp

sin bx,xn, emx, cos bx,


Fsxdyp.

yh


In Example 1, the form of the homogeneous solution

has no overlap with the function in the equation

However, suppose the given differential equation in Example 1 were of the form

Now, it would make no sense to guess that the particular solution were because you know that this solution would yield 0. In such cases, you should alteryour guess by multiplying by the lowest power of x that removes the duplication. Forthis particular problem, you would guess

EXAMPLE 2 Method of Undetermined Coefficients


Solution The characteristic equation has solutions and Thus,

Because your first choice for would be However,because already contains a constant term you should multiply the polynomialpart by x and use

Substitution into the differential equation produces

Equating coefficients of like terms yields the system

with solutions and Therefore,


5 C1 1 C2e2x 214 x 2

14 x

2 2 2ex.

y 5 yh 1 yp

yp 5 214 x 2

14 x

2 2 2ex

C 5 22.A 5 B 5 214

2C 5 224B 5 1,2B 2 2A 5 0,

s2B 2 2Ad 2 4Bx 2 Cex 5 x 1 2ex. s2B 1 Cexd 2 2sA 1 2Bx 1 Cexd 5 x 1 2ex

y0 2 2y9 5 x 1 2ex

yp0 5 2B 1 Cex. yp9 5 A 1 2Bx 1 Cex yp 5 Ax 1 Bx2 1 Cex

C1,yhsA 1 Bxd 1 Cex.ypFsxd 5 x 1 2ex,

yh 5 C1 1 C2e2x.

m 5 2.m 5 0m2 2 2m 5 0

y0 2 2y9 5 x 1 2ex.

yp 5 Axe2x.

y 5 Ae2x,

y0 2 2y9 2 3y 5 e2x.


Fsxd

yh 5 C1e2x 1 C2e3x


In Example 2, the polynomial part of the initial guess

for overlapped by a constant term with and it was necessary tomultiply the polynomial part by a power of x that removed the overlap. The nextexample further illustrates some choices for that eliminate overlap with Remember that in all cases the first guess for should match the types of functionsoccurring in

EXAMPLE 3 Choosing the Form of the Particular Solution

Determine a suitable choice for for each of the following.

a.b.c.

Solution

a. Because the normal choice for would be However,because already contains a linear term, you should multiply by to obtain

b. Because and each term in contains a factor of you can simply let

c. Because the normal choice for would be However, becausealready contains an term, you should multiply by to

get



Solution From Example 6 in the preceding section, you know that the homogeneoussolution is

Because let and obtain and Thus, by substi-tution, you have

Thus, and which implies that Therefore, the generalsolution is

5 C1e2x 1 C2xe2x 1 C3x2e2x 2 3 1 x. y 5 yh 1 yp

yp 5 23 1 x.A 5 23,B 5 1

s0d 1 3s0d 1 3sBd 1 sA 1 Bxd 5 s3B 1 Ad 1 Bx 5 x.

yp0 5 0.yp9 5 Byp 5 A 1 BxFsxd 5 x,

yh 5 C1e2x 1 C2xe2x 1 C3x2e2x.

y999 1 3y0 1 3y9 1 y 5 x.

yp 5 Ax2e2x.

x2xe2xyh 5 C1e2x 1 C2xe2xAe2x.ypFsxd 5 e2x,

yp 5 A cos 3x 1 B sin 3x.

e2x,yhFsxd 5 4 sin 3x

yp 5 Ax2 1 Bx3 1 Cx4.

x2yh 5 C1 1 C2xA 1 Bx 1 Cx2.ypFsxd 5 x2,

C1e2x 1 C2xe2xy0 2 4y9 1 4 5 e2xC1e2x cos 3x 1 C2e2x sin 3xy0 1 2y9 1 10y 5 4 sin 3xC1 1 C2xy0 5 x2yh y0 1 ay9 1 by 5 Fsxd

yp

Fsxd.yp

yh.yp

yh 5 C1 1 C2e2x,yp

sA 1 Bxd 1 Cex


Variation of ParametersThe method of undetermined coefficients works well if is made up of polynomi-als or functions whose successive derivatives have a cyclic pattern. For functions suchas and which do not have such characteristics, it is better to use a more general method called variation of parameters. In this method, you assume that has the same form as except that the constants in are replaced by variables.

EXAMPLE 5 Variation of Parameters


Solution The characteristic equation has one solution, Thus, the homogeneous solution is

Replacing and by and produces

The resulting system of equations is

Subtracting the second equation from the first produces Then, by substitution in the first equation, you have Finally, integration yields

and

From this result it follows that a particular solution is


y 5 C1ex 1 C2xex 212 xe

x 1 xex ln

!x.

yp 5 212 xe

x 1 sln !x dxex

u2 512 E 1x dx 5 12 ln x 5 ln !x.u1 5 2E 12 dx 5 2x2

u19 5 212 .

u29 5 1ys2xd.

u19 ex 1 u29 sxex 1 exd 5

ex

2x.

u19 ex 1 u29 xe

x 5 0

yp 5 u1y1 1 u2y2 5 u1ex 1 u2xex.

u2u1C2C1

yh 5 C1y1 1 C2y2 5 C1ex 1 C2xex.

m 5 1.m2 2 2m 1 1 5 sm 2 1d2 5 0

x > 0.y0 2 2y9 1 y 5 ex

2x,

yhyh,yp

tan x,1yx

Fsxd

Variation of Parameters

To find the general solution to the equation use the following steps.

1. Find 2. Replace the constants by variables to form 3. Solve the following system for and

4. Integrate to find and The general solution is y 5 yh 1 yp.u2.u1

u19y19 1 u29y29 5 Fsxdu19y1 1 u29y2 5 0

u29.u19

yp 5 u1y1 1 u2y2.yh 5 C1y1 1 C2y2.

y0 1 ay9 1 by 5 Fsxd,


EXAMPLE 6 Variation of Parameters


Solution Because the characteristic equation has solutions thehomogeneous solution is

Replacing and by and produces

The resulting system of equations is

Multiplying the first equation by and the second by produces

Adding these two equations produces which implies that

Integration yields

and

so that


5 C1 cos x 1 C2 sin x 2 cos x ln |sec x 1 tan x|. y 5 yh 1 yp

5 2cos x ln |sec x 1 tan x| yp 5 sin x cos x 2 cos x ln |sec x 1 tan x| 2 sin x cos x

5 2cos x

u2 5 E sin x dx 5 sin x 2 ln |sec x 1 tan x|

u1 5 E scos x 2 sec xd dx 5 cos x 2 sec x.

5cos2 x 2 1

cos x

u19 5 2sin2 xcos x

u29 5 sin x,

2u19 sin x cos x 1 u29 cos2x 5 sin x. u19 sin x cos x 1 u29 sin2x 5 0

cos xsin x

2u19 sin x 1 u29 cos x 5 tan x. u19 cos x 1 u29 sin x 5 0

yp 5 u1 cos x 1 u2 sin x.

u2u1C2C1

yh 5 C1 cos x 1 C2 sin x.

m 5 i,m2 1 1 5 0

y0 1 y 5 tan x.


In Exercises 14, verify the solution of the differential equation.

1.2.3.4.

In Exercises 520, solve the differential equation by the methodof undetermined coefficients.

5. 6.7. 8.

9. 10.11.12.13. 14.

15.16.17.18.

19.

20.

21. Think About It(a) Explain how, by observation, you know that a particular

solution of the differential equation is

(b) Use the explanation of part (a) to give a particular solutionof the differential equation

(c) Use the explanation of part (a) to give a particular solutionof the differential equation

22. Think About It(a) Explain how, by observation, you know that a form

of a particular solution of the differential equationis

(b) Use the explanation of part (a) to find a particular solutionof the differential equation

(c) Compare the algebra required to find particular solutions inparts (a) and (b) with that required if the form of the partic-ular solution were

In Exercises 2328, solve the differential equation by themethod of variation of parameters.

23. 24.25. 26.

27. 28.

Electrical Circuits In Exercises 29 and 30, use the electricalcircuit differential equation

where R is the resistance (in ohms), C is the capacitance (infarads), L is the inductance (in henrys), is the electromotiveforce (in volts), and q is the charge on the capacitor (incoulombs). Find the charge q as a function of time for the electrical circuit described. Assume that and

29.

30.

Vibrating Spring In Exercises 3134, find the particularsolution of the differential equation

for the oscillating motion of an object on the end of a spring. Usea graphing utility to graph the solution. In the equation, y is thedisplacement from equilibrium (positive direction is down-ward) measured in feet, and t is time in seconds (see figure). Theconstant w is the weight of the object, g is the acceleration dueto gravity, b is the magnitude of the resistance to the motion, kis the spring constant from Hookes Law, and is the accel-eration imposed on the system.

31.

32.

33.

34.

m

l = naturallength

y = displacement

Spring displacement

ys0d 5 12 , y9s0d 5 24

432 y0 1

12 y9 1

252 y 5 0

ys0d 5 14 , y9s0d 5 23

232 y0 1 y9 1 4y 5

232 s4 sin 8td

ys0d 5 14 , y9s0d 5 0

232 y0 1 4y 5

232 s4 sin 8td

ys0d 5 14 , y9s0d 5 0

2432 y0 1 48y 5

2432 s48 sin 4td

Fxtc

w

g y0 xtc 1 by9xtc 1 kyxtc 5w

g Fxtc

Estd 5 10 sin 5tR 5 20, C 5 0.02, L 5 1Estd 5 12 sin 5tR 5 20, C 5 0.02, L 5 2

q9x0c 5 0.qx0c 5 0

Estd

d 2qdt2 1 _

RL+

dqdt 1 _

1LC+q 5 _

1L+Extc

y0 2 4y9 1 4y 5 e2x

xy0 2 2y9 1 y 5 ex ln x

y0 2 4y9 1 4y 5 x2e2xy0 1 4y 5 csc 2xy0 1 y 5 sec x tan xy0 1 y 5 sec x

yp 5 A cos x 1 B sin x.

y0 1 5y 5 10 cos x.

yp 5 A sin x.y0 1 3y 5 12 sin x

y0 1 2y9 1 2y 5 8.

y0 1 5y 5 10.

yp 5 4.y0 1 3y 5 12

y1p22 525

y9 1 2y 5 sin x

ys0d 5 13

y9 2 4y 5 xex 2 xe4xys0d 5 1, y9s0d 5 1, y0s0d 5 1y999 2 y0 5 4x2y999 2 3y9 1 2y 5 2e22xy0 1 4y9 1 5y 5 sin x 1 cos xy0 1 9y 5 sin 3x

ys0d 5 21, y9s0d 5 2ys0d 5 0, y9s0d 5 23y0 1 y9 2 2y 5 3 cos 2xy0 1 y9 5 2sin x

16y0 2 8y9 1 y 5 4sx 1 exdy0 2 10y9 1 25y 5 5 1 6ex

y0 2 9y 5 5e3xy0 1 2y9 5 2exys0d 5 1, y9s0d 5 6ys0d 5 1, y9s0d 5 0y0 1 4y 5 4y0 1 y 5 x3y0 2 2y9 2 3y 5 x2 2 1y0 2 3y9 1 2y 5 2x

y0 1 y 5 csc x cot xy 5 s5 2 ln |sin x|dcos x 2 x sin xy0 1 y 5 tan xy 5 3 sin x 2 cos x ln |sec x 1 tan x|y0 1 y 5 cos xy 5 s2 1 12 xdsin xy0 1 y 5 10e2xy 5 2se2x 2 cos xd

Differential EquationSolution



35. Vibrating Spring Rewrite in the solution for Exercise 31by using the identity

where

36. Vibrating Spring The figure shows the particular solutionof the differential equation

for values of the resistance component b in the interval (Note that when the problem is identical to that ofExercise 34.)(a) If there is no resistance to the motion describe the

motion.(b) If what is the ultimate effect of the retarding force?(c) Is there a real number M such that there will be no oscilla-

tions of the spring if Explain your answer.

37. Parachute Jump The fall of a parachutist is described bythe second-order linear differential equation

where w is the weight of the parachutist, y is the height at timet, g is the acceleration due to gravity, and k is the drag factor ofthe parachute. If the parachute is opened at 2000 feet,

and at that time the velocity is feetper second, then for a 160-pound parachutist, using thedifferential equation is

Using the given initial conditions, verify that the solution of thedifferential equation is

38. Parachute Jump Repeat Exercise 37 for a parachutist whoweighs 192 pounds and has a parachute with a drag factor of

39. Solve the differential equation

given that and are solutions of the corre-sponding homogeneous equation.

40. True or False? is a particular solutionof the differential equation

y0 2 3y9 1 2y 5 cos e2x.

yp 5 2e2x cos e2x

y2 5 x ln xy1 5 x

x2y0 2 xy9 1 y 5 4x ln x

k 5 9.

y 5 1950 1 50e21.6t 2 20t.

25y0 2 8y9 5 160.

k 5 8,y9s0d 5 2100ys0d 5 2000,

w

g

d 2ydt2 2 k

dydt 5 w

y

t

bGenerated by Maple

b = 0

b = 1

b = 12

b > M ?

b > 0,

sb 5 0d,

b 5 12 ,f0, 1g.

ys0d 5 12, y9s0d 5 24

432 y0 1 by9 1

252 y 5 0

f 5 arctan ayb.

a cos vt 1 b sin vt 5 !a2 1 b2 sinsvt 1 fd

yh

SECTION 15.5 Series Solutions of Differential Equations 1125

15.5S E C T I O N Series Solutions of Differential EquationsPower Series Solution of a Differential Equation Approximation by Taylor Series

Power Series Solution of a Differential EquationWe conclude this chapter by showing how power series can be used to solve certaintypes of differential equations. We begin with the general power series solutionmethod.

Recall from Chapter 8 that a power series represents a function f on an interval ofconvergence, and that you can successively differentiate the power series to obtain aseries for and so on. These properties are used in the power series solutionmethod demonstrated in the first two examples.

EXAMPLE 1 Power Series Solution

Use a power series to solve the differential equation

Solution Assume that is a solution. Then, Substitutingfor and you obtain the following series form of the differential equation.(Note that, from the third step to the fourth, the index of summation is changed toensure that occurs in both sums.)

Now, by equating coefficients of like terms, you obtain the recursion formulawhich implies that

This formula generates the following results.

. . .

. . .

Using these values as the coefficients for the solution series, you have

y 5 o`

n50

2na0n! x

n 5 a0 o`

n50

s2xdnn! 5 a0e

2x.

25a05!

24a04!

23a03!

22a022a0a0

a5a4a3a2a1a0

n 0.an11 52an

n 1 1,

sn 1 1dan11 5 2an,

o`

n50 sn 1 1dan11xn 5 o

`

n50 2anxn

o`

n51 nanx

n21 5 o`

n50 2anxn

o`

n51 nanx

n21 2 2 o`

n50 anx

n 5 0

y9 2 2y 5 0

xn

22y,y9y9 5 onanxn21.y 5 oanxn

y9 2 2y 5 0.

f9, f 0,


In Example 1, the differential equation could be solved easily without using aseries. The differential equation in Example 2 cannot be solved by any of the methodsdiscussed in previous sections.

EXAMPLE 2 Power Series Solution

Use a power series to solve the differential equation

Solution Assume that is a solution. Then you have

Substituting for and y in the given differential equation, you obtain the fol-lowing series.

To obtain equal powers of x, adjust the summation indices by replacing n by inthe left-hand sum, to obtain

By equating coefficients, you have from whichyou obtain the recursion formula

and the coefficients of the solution series are as follows.

Thus, you can represent the general solution as the sum of two seriesone for theeven-powered terms with coefficients in terms of and one for the odd-poweredterms with coefficients in terms of

The solution has two arbitrary constants, and as you would expect in the general solution of a second-order differential equation.

a1,a0

5 a0 o`

k50

s21dkx2k2ksk!d 1 a1 o

`

k50

s21dkx2k113 ? 5 ? 7 . . . s2k 1 1d

y 5 a011 2 x2

2 1x4

2 ? 42 . . .2 1 a11x 2 x

3

3 1x5

3 ? 52 . . .2

a1.

a0

a2k11 5s21dk a1

3 ? 5 ? 7 . . . s2k 1 1d a2k 5

s21dk a02 ? 4 ? 6 . . . s2kd

5s21dka02ksk!d

::

a7 5 2a57 5 2

a13 ? 5 ? 7

a6 5 2a46 5 2

a02 ? 4 ? 6

a5 5 2a35 5

a13 ? 5

a4 5 2a24 5

a02 ? 4

a3 5 2a1

3 a2 5 2a02

n 0,an12 5 2sn 1 1d

sn 1 2dsn 1 1d an 5 2an

n 1 2,

sn 1 2dsn 1 1dan12 5 2sn 1 1dan,

o`

n50 sn 1 2dsn 1 1dan12xn 5 2 o

`

n50 sn 1 1danxn.

n 1 2

o`

n52 nsn 2 1danxn22 5 2 o

`

n50 sn 1 1danxn

o`

n52nsn 2 1danxn22 1 o

`

n50 nanx

n 1 o`

n50 anx

n 5 0

y0, xy9,

y0 5 o`

n52 nsn 2 1danxn22.xy9 5 o

`

n51 nanx

n,y9 5 o

`

n51 nanx

n21,

o`

n50anx

n

y0 1 xy9 1 y 5 0.

x y

0.0 1.0000

0.1 1.1057

0.2 1.2264

0.3 1.3691

0.4 1.5432

0.5 1.7620

0.6 2.0424

0.7 2.4062

0.8 2.8805

0.9 3.4985

1.0 4.3000

SECTION 15.5 Series Solutions of Differential Equations 1127

In Exercises 16, verify that the power series solution of the differential equation is equivalent to the solution found usingthe techniques in Sections 5.7 and 15.115.4.

1. 2.3. 4.5. 6.

In Exercises 710, use power series to solve the differentialequation and find the interval of convergence of the series.

7. 8.9. 10.

In Exercises 11 and 12, find the first three terms of each of thepower series representing independent solutions of the differen-tial equation.

11. 12.

In Exercises 13 and 14, use Taylors Theorem to find the seriessolution of the differential equation under the specified initialconditions. Use n terms of the series to approximate y for thegiven value of x and compare the result with the approximationgiven by Eulers Method for

13.14. y9 2 2xy 5 0, ys0d 5 1, n 5 4, x 5 1

y9 1 s2x 2 1dy 5 0, ys0d 5 2, n 5 5, x 5 12 ,

Dx 5 0.1.

y0 1 x2y 5 0sx2 1 4dy0 1 y 5 0

y0 2 xy9 2 y 5 0y0 2 xy9 5 0y9 2 2xy 5 0y9 1 3xy 5 0

y0 1 k2y 5 0y0 1 4y 5 0y0 2 k2y 5 0y0 2 9y 5 0y9 2 ky 5 0y9 2 y 5 0


Approximation by Taylor SeriesA second type of series solution method involves a differential equation with initialconditions and makes use of Taylor series, as given in Section 8.10.

EXAMPLE 3 Approximation by Taylor Series

Use a Taylor series to find the series solution of

given the initial condition when Then, use the first six terms of thisseries solution to approximate values of y for

Solution Recall from Section 8.10 that, for

Because and you obtain the following.

Therefore, you can approximate the values of the solution from the series

Using the first six terms of this series, you can compute values for y in the intervalas shown in the table at the left.0 x 1,

5 1 1 x 1 12 x2 1

43! x

3 1144! x

4 1665! x

5 1 . . . .

y 5 ys0d 1 y9s0dx 1 y0 s0d2! x2 1

y999s0d3! x

3 1ys4ds0d

4! x4 1

ys5ds0d5! x

5 1 . . .

ys5ds0d 5 28 1 32 1 6 5 66 ys5d 5 2yys4d 1 8y9y999 1 6sy0 d2 ys4ds0d 5 8 1 6 5 14 ys4d 5 2yy999 1 6y9y0 y999s0d 5 2 1 2 5 4 y999 5 2yy0 1 2sy9d2 y0 s0d 5 2 2 1 5 1 y9 5 2yy9 2 1 y9s0d 5 1 y9 5 y2 2 x ys0d 5 1

y9 5 y2 2 x,ys0d 5 1

y 5 ys0d 1 y9s0dx 1 y0 s0d2! x2 1

y999s0d3! x

3 1 . . . .

c 5 0,

0 x 1.x 5 0.y 5 1

y9 5 y2 2 x


15. Investigation Consider the differential equation with initial conditions and

(a) Find the solution of the differential equation using the techniques of Section 15.3.

(b) Find the series solution of the differential equation.(c) The figure shows the graph of the solution of the

differential equation and the third-degree and fifth-degreepolynomial approximations of the solution. Identify each.

16. Consider the differential equation with the initialconditions and

(See Exercise 9.)(a) Find the series solution satisfying the initial conditions.(b) Use a graphing utility to graph the third-degree and fifth-

degree series approximations of the solution. Identify theapproximations.

(c) Identify the symmetry of the solution.

In Exercises 17 and 18, use Taylors Theorem to find the seriessolution of the differential equation under the specified initialconditions. Use n terms of the series to approximate y for thegiven value of x.

17.18.

In Exercises 1922, verify that the series converges to the givenfunction on the indicated interval. (Hint: Use the given differ-ential equation.)

19.

Differential equation:

20.


21.


22.


23. Find the first six terms in the series solution of Airys equationy0 2 xy 5 0.

s1 2 x2dy0 2 xy9 5 0

o`

n50

s2nd!x2n11s2nn!d2s2n 1 1d 5 arcsin x, s21, 1d

sx2 1 1dy0 1 2xy9 5 0

o`

n50

s21dnx2n112n 1 1 5 arctan x, s21, 1d

y0 1 y 5 0

o`

n50

s21dnx2ns2nd! 5 cos x, s2`, `d

y9 2 y 5 0

o`

n50

xn

n! 5 ex, s2`, `d

y0 2 2xy9 1 y 5 0, ys0d 5 1, y9s0d 5 2, n 5 8, x 5 12

y0 2 2xy 5 0, ys0d 5 1, y9s0d 5 23, n 5 6, x 5 14

y9s0d 5 2.ys0d 5 0

y0 2 xy9 5 0

x

y

1 2

321

y9s0d 5 6.ys0d 5 29y 5 0y0 1

In Exercises 14, classify the differential equation according totype and order.

1. 2.

3. 4.

In Exercises 5 and 6, use the given differential equation and itsdirection field.

5.

(a) Sketch several solution curves for the differential equationon the direction field.

(b) Find the general solution of the differential equation.Compare the result with the sketches from part (a).

6.

(a) Sketch several solution curves for the differential equationon the direction field.

(b) When is the rate of change of the solution greatest? When isit least?

(c) Find the general solution of the differential equation.Compare the result with the sketches from part (a).

x

y

2

2

2

2

dydx 5

!1 2 y 2

x

y

1

432

321 44 3 2

432

1

dydx 5

yx

s y0 d2 1 4y9 5 0y0 1 3y9 2 10 5 0

yy0 5 x 1 12u

t25 c2

2u

x2

R E V I E W E X E R C I S E S F O R C H A P T E R 15

REVIEW EXERCISES 1129

In Exercises 710, match the differential equation with its solution.

7. (a)8. (b)9. (c)

10. (d)

In Exercises 1132, find the general solution of the first-orderdifferential equation.

11. 12.

13. 14.

15. 16.

17.18.19.

20.21.22.23.24.25.26.27.28.29.30.31.32.

In Exercises 3340, find the particular solution of the differen-tial equation that satisfies the boundary condition.

33.

34.

35.

36.

37.

38.

39.

40.

In Exercises 41 and 42, find the orthogonal trajectories of thegiven family and sketch several members of each family.

41.42.

43. Snow Removal Assume that the rate of change in the number of miles s of road cleared per hour by a snowplow isinversely proportional to the height h of snow.(a) Write and solve the differential equation to find s as a func-

tion of h.(b) Find the particular solution if miles when

inches and miles when inches

44. Growth Rate Let x and y be the sizes of two internal organsof a particular mammal at time t. Empirical data indicate thatthe relative growth rates of these two organs are equal, andhence we have

Solve this differential equation, writing y as a function of x.

45. Population Growth The rate of growth in the number N ofdeer in a state park varies jointly over time t as N and where is the estimated limiting size of the herd. WriteN as a function of t if when and when

46. Population Growth The rate of growth in the number N ofelk in a game preserve varies jointly over time t (in years) as Nand where 300 is the estimated limiting size of theherd.(a) Write and solve the differential equation for the population

model if when and when

(b) Use a graphing utility to graph the direction field of the differential equation and the particular solution of part (a).

(c) At what time is the population increasing most rapidly?(d) If 400 elk had been placed in the preserve initially, use the

direction field to describe the change in the population overtime.

t 5 1.N 5 75t 5 0N 5 50

300 2 N

t 5 4.N 5 200t 5 0N 5 100

L 5 500L 2 N,

1x

dxdt 5

1y

dydt .

s2 h 15d.h 5 10s 5 12h 5 2

s 5 25

y 2 2x 5 Csx 2 Cd2 1 y 2 5 C2

ys4d 5 22xy9 2 y 5 x3 2 x

ys0d 5 3s1 1 ed1 2 e

y9 5 x2y 2 2 9x2ys2d 5 0s2x 1 y 2 3d dx 1 sx 2 3y 1 1d dy 5 0ys0d 5 2s1 1 yd ln s1 1 yd dx 1 dy 5 0

ys22d 5 25yexydx 1 xexy dy 5 0ys1d 5 10xdy 5 sx 1 y 1 2d dxys1d 5 2

y9 12yx

5 2x9y5

ys0d 5 4y9 2 2y 5 ex

y9 5 y 1 2xs y 2 ex dxy9 2 ay 5 bx4x3yy9 5 x4 1 3x2y 2 1 y4s1 1 x2d dy 5 s1 1 y 2d dx2xdx 1 2ydy 5 sx2 1 y 2d dxyy9 1 y 2 5 1 1 x2xy9 1 y 5 sin xx 1 yy9 5 !x2 1 y 2y9 5 2x!1 2 y 2sx 2 y 2 5d dx 2 sx 1 3y 2 2d dy 5 0ydx 2 sx 1 !xy d dy 5 0dy 5 sy tan x 1 2ex d dx3x2y2 dx 1 s2x3y 1 x3y4d dy 5 0s2x 2 2y3 1 yd dx 1 sx 2 6xy 2d dy 5 0s y 1 x3 1 xy 2d dx 2 xdy 5 0s10x 1 8y 1 2d dx 1 s8x 1 5y 1 2d dy 5 0

dydx 2

3yx2

51x2

dydx 2

yx

5x

y

dydx 2 3x

2y 5 ex3y9 22yx

5y9x

dydx 1 xy 5 2y

dydx 2

yx

5 2 1 !x

y 5 Ce4xy0 1 4y 5 0y 5 C1 cos 2x 1 C2 sin 2xy0 2 4y 5 0y 5 4x 1 Cy9 2 4y 5 0y 5 C1e2x 1 C2e22xy9 2 4 5 0

Solution Differential Equation


47. Slope The slope of a graph is given by Find the equation of the graph if the graph passes through thepoint Use a graphing utility to graph the solution.

48. Investment Let be the amount in a fund earning interestat an annual rate r compounded continuously. If a continuouscash flow of P dollars per year is withdrawn from the fund, therate of change of A is given by the differential equation

where when Solve this differential equation forA as a function of t.

49. Investment A retired couple plans to withdraw P dollars peryear from a retirement account of $500,000 earning 10% compounded continuously. Use the result of Exercise 48 and agraphing utility to graph the function A for each of the follow-ing continuous annual cash flows. Use the graphs to describewhat happens to the balance in the fund for each of the cases.(a)(b)(c)

50. Investment Use the result of Exercise 48 to find the timenecessary to deplete a fund earning 14% interest compoundedcontinuously if and

In Exercises 5154, find the particular solution of the differen-tial equation that satisfies the initial conditions. Use a graphingutility to graph the solution.

51.52.53.54.

In Exercises 5560, find the general solution of the second-orderdifferential equation.

55.56.57.58.

59.

60.

In Exercises 6164, find the particular solution of the differen-tial equation that satisfies the initial conditions.

61.62.63.64.

Vibrating Spring In Exercises 65 and 66, describe the motionof a 64-pound weight suspended on a spring. Assume that theweight stretches the spring feet from its natural position.

65. The weight is pulled foot below the equilibrium position andreleased.

66. The weight is pulled foot below the equilibrium position andreleased. The motion takes place in a medium that furnishes adamping force of magnitude speed at all times.

67. Investigation The differential equation

models the motion of a weight suspended on a spring.(a) Solve the differential equation and use a graphing utility to

graph the solution for each of the assigned quantities for b,k, and (i)(ii)(iii)(iv)

(b) Describe the effect of increasing the resistance to motion b.(c) Explain how the motion of the object would change if a

stiffer spring (increased k) were used.(d) Matching the input and natural frequencies of a system is

known as resonance. In which case of part (a) does thisoccur, and what is the result?

68. Think About It Explain how you can find a particular solu-tion of the differential equation

by observation.

In Exercises 69 and 70, find th

Calculus 7eap1501 05

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