16 Additional Topics in Differential Equations - Cengage · Use an integrating factor to make a differential equation exact. Exact Differential Equations ... Solution The differential

16.1 Exact First-Order Equations16.2 Second-Order Homogeneous Linear Equations16.3 Second-Order Nonhomogeneous Linear Equations16.4 Series Solutions of Differential Equations

Electrical Circuits (Exercises 33 and 34, p. 1147)

Parachute Jump(Section Project, p. 1148)

Motion of a Spring(Example 8, p. 1138)

Cost (Exercise 45, p. 1132)

Undamped or Damped Motion? (Exercise 47, p. 1140)

1125

16

Clockwise from top left, nsj-images/iStockphoto; Danshutter/Shutterstock.com;ICHIRO/Photodisc/Getty Images; Mircea BEZERGHEANU/Shutterstock.com; Monty Rakusen/Cultura/Getty Images

Additional Topics inDifferential Equations

9781285060293_1600.qxp 10/15/12 5:09 PM Page 1125

1126 Chapter 16 Additional Topics in Differential Equations

16.1 Exact First-Order Equations

Solve an exact differential equation.Use an integrating factor to make a differential equation exact.

Exact Differential EquationsIn Chapter 6, you studied applications of differential equations to growth and decayproblems. You also learned more about the basic ideas of differential equations andstudied the solution technique known as separation of variables. In this chapter, you willlearn more about solving differential equations and using them in real-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 13.3, you know that if has continuous second partials, then

This suggests the following test for exactness.

Every differential equation of the form

is exact. In other words, a separable differential equation is actually a special type of anexact equation.

Exactness is a fragile condition in the sense that seemingly minor alterations in anexact equation can destroy its exactness. This is demonstrated in the next example.

M�x� dx � N�y� dy � 0

�M�y

��2f

�y�x�

�2f�x�y

��N�x

.

f

z � f �x, y�.

M�x, y� dx � N�x, y� dy � 0

Definition of an Exact Differential Equation

The equation

is an exact differential equation when there exists a function of two variables and having continuous partial derivatives such that

and

The general solution of the equation is f �x, y� � C.

fy�x, y� � N�x, y�.fx�x, y� � M�x, y�

yxf

M�x, y� dx � N�x, y� dy � 0

THEOREM 16.1 Test for Exactness

Let and have continuous partial derivatives on an open disk The differential equation

is exact if and only if

�M�y

��N�x

.

M�x, y� dx � N�x, y� dy � 0

R.NM

9781285060293_1601.qxp 10/15/12 5:09 PM Page 1126

Testing for Exactness

Determine whether each differential equation is exact.

a. b.

Solution

a. This differential equation is exact because

and

Notice that the equation is not exact, even though it isobtained by dividing each side of the first equation by

b. This differential equation is exact because

and

Notice that the equation is not exact, even thoughit 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 potentialfunction (Theorem 15.1). This means that a general solution to an exact differential equation can be found by the method used to find a potential function for aconservative vector field.

Solving an Exact Differential Equation

See LarsonCalculus.com for an interactive version of this type of example.

Solve the differential equation

Solution This differential equation is exact because

and

The general solution, is

In Section 15.1, you determined by integrating with respect to and reconciling the two expressions for An alternative method is to partially differentiate this version of with respect to and compare the result with In other words,

So, and it follows that Therefore,

and the general solution is Figure 16.1 shows the solution curvesthat correspond to 10, 100, and 1000.C � 1,

x2y � x3 � y2 � C.

f �x, y� � x2y � x3 � y2 � C1

g�y� � �y2 � C1.g��y� � �2y,

fy�x, y� ��

�y �x2y � x3 � g�y�� x2 � g��y� � x2 � 2y.

N�x, y�

N�x, y�.yf �x, y�f �x, y�.

yN�x, y�g�y�

� � �2xy � 3x2� dx � x2y � x3 � g�y�. f �x, y� � � M�x, y� dx

f �x, y� � C,

�N�x

��

�x�x2 � 2y� � 2x.

�M�y

��

�y �2xy � 3x2� � 2x

�2xy � 3x2� dx � �x2 � 2y� dy � 0.

f �x, y� � CF�x, y� � M�x, y� i � N�x, y�j

M�x, y� dx � N�x, y� dy � 0

cos y dx � �y2 � x sin y� dy � 0

�N�x

��

�x � y2 � x sin y� � �sin y.

�M�y

��

�y �cos y� � �sin y

x.�y2 � 1� dx � xy dy � 0

�N�x

��

�x � yx2� � 2xy.

�M�y

��

�y �xy2 � x� � 2xy

cos y dx � �y2 � x sin y� dy � 0�xy2 � x� dx � yx2 dy � 0

16.1 Exact First-Order Equations 1127

g��y� � �2y

x

y

4−4 8

8

−8 12

12

16

20

24

−12

C = 1C = 10

C = 100

C = 1000

Figure 16.1

9781285060293_1601.qxp 10/15/12 5:09 PM Page 1127

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

So, and

which implies that , and the general solution is

General solution

Applying the given initial condition produces

which implies that So, the particular solution is

The graph of the particular solution is shown in Figure 16.3. Notice that the graph consists of two parts: the ovals are given by and the

axis is given by

In Example 3, note that for the total differential of is given by

In other words, is called an exact differential equation becauseis exactly the differential of f �x, y�.M dx � N dy

M dx � N dy � 0

� M�x, y� dx � N�x, y� dy.

� �cos x � x sin x � y2� dx � 2xy dy

dz � fx�x, y� dx � fy�x, y� dy

zz � f �x, y� � xy2 � x cos x,

x � 0.y-y2 � cos x � 0,

xy2 � x cos x � 0.

C � 0.

��1�2 � � cos � � C

xy2 � x cos x � C.

f �x, y� � xy2 � x cos x � C1

� x cos x � C1

g�x� � � �cos x � x sin x� dx

g��x� � cos x � x sin x

fx�x, y� ��

�x �xy2 � g�x�� y2 � g��x� � cos x � x sin x � y2

M�x, y�

f �x, y� � � N�x, y� dy � � 2xy dy � xy2 � g�x�

N�x, y�.M�x, y�,N�x, y�

�

�y �cos x � x sin x � y2� � 2y �

�

�x �2xy�.

�N�x

�M�y

x � �.y � 1

�cos x � x sin x � y2� dx � 2xy dy � 0


g��x� � cos x � x sin x

x

y

2

4

−2

−4

π 2π 3ππ−3π− 2π−

( , 1)π

Figure 16.3

TECHNOLOGY A graphingutility can be used to graph a particular solution that satisfies the initial condition of a differential equation. In Example 3, the differentialequation and initial conditionsare satisfied when

which implies that the particular solution can be written as

or On a graphing utility screen,the solution would be represented by Figure 16.2together with the axis.y-

y � ±��cos x.x � 0

xy2 � x cos x � 0,

−4

−4 4

4

� �

Figure 16.2

9781285060293_1601.qxp 10/15/12 5:09 PM Page 1128

Integrating FactorsWhen the differential equation is not exact, it may be possible to make it exact by multiplying by an appropriate factor which is calledan integrating factor for the differential equation.

Multiplying by an Integrating Factor

a. When the differential equation

Not an exact equation

is multiplied by the integrating factor the resulting equation

Exact equation

is exact—the left side is the total differential of

b. When the equation

Not an exact equation

is multiplied by the integrating factor the resulting equation

Exact equation

is exact—the left side is the total differential of

Finding an integrating factor can be difficult. There are two classes of differential equations, however, whose integrating factors can be found routinely—namely, those that possess integrating factors that are functions of either alone or alone. The next theorem, which is presented without proof, outlines a procedure forfinding these two special categories of integrating factors.

yx

x�y.

1y dx �

xy2 dy � 0

u�x, y� � 1�y2,

y dx � x dy � 0

x2y.

2xy dx � x2 dy � 0

u�x, y� � x,

2y dx � x dy � 0

u�x, y�,M�x, y� dx � N�x, y� dy � 0


THEOREM 16.2 Integrating Factors

Consider the differential equation

1. If

is a function of alone, then is an integrating factor.

2. If

is a function of alone, then is an integrating factor.ek�y� dyy

1M�x, y� �Nx�x, y� � My�x, y�� k�y�

eh�x� dxx

1N�x, y� �My�x, y� � Nx�x, y�� h�x�

M�x, y� dx � N�x, y� dy � 0.

Exploration

In Chapter 6, you solved the first-order linear differential equation

by using the integrating factor Show that you can obtain this integrating factor by using the methods of this section.

u�x� � eP�x� dx.

dydx

� P�x�y � Q�x�

REMARK When either or is constant,

Theorem 16.2 still applies. As an aid to remembering these formulas, note that thesubtracted partial derivativeidentifies both the denominatorand the variable for the integrating factor.

k�x�h�x�

9781285060293_1601.qxp 10/15/12 5:09 PM Page 1129

Finding an Integrating Factor


Solution This equation is not exact because

and

However, because

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

whose solution is obtained as follows.

Therefore, and which implies that

The general solution is or

General solution

The next example shows how a differential equation can help in sketching a forcefield given by

An Application to Force Fields

Sketch the force field

by finding and sketching the family of curves tangent to

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

which, in differential form, is

From Example 5, you know that the general solution of this differential equation is Figure 16.4 shows several representative curves from this family. Note that the force vector at is tangent to the curve passing through �x, y�.�x, y�

y2 � x � 1 � Ce�x.

�y2 � x� dx � 2y dy � 0.

2y dy � ��y2 � x� dx

dydx

��y2 � x��x2 � y2

2y��x2 � y2�

��y2 � x�2y

F�x, y��x, y�

F.

F�x, y� �2y

�x2 � y2 i �

y2 � x�x2 � y2

j

F�x, y� � M�x, y� i � N�x, y�j.

y2 � x � 1 � Ce�x.

y2ex � xex � ex � C,

f �x, y� � y2ex � xex � ex � C1.

g�x� � �xex � ex � C1,g��x� � �xex

fx�x, y� � y2ex � g��x� � y2ex � xex

M�x, y�

f �x, y� � � N�x, y� dy � � 2yex dy � y2ex � g�x�

�y2ex � xex� dx � 2yex dy � 0

exeh�x� dx � e dx � ex

My�x, y� � Nx�x, y�N�x, y� �

2y � 02y

� 1 � h�x�

Nx�x, y� � 0.My�x, y� � 2y

�y2 � x� dx � 2y dy � 0.


g��x� � �xex

3

2

−1−3

−2

−3

y

x

F(x, y) = i −

Force field:

y2 = x − 1 + Ce−x

Family of curves tangent to F:

2y

x2 + y2j

y2 − x

x2 + y2

Figure 16.4

9781285060293_1601.qxp 10/15/12 5:09 PM Page 1130


16.1 Exercises See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.

Testing for Exactness In Exercises 1–4, determinewhether the differential equation is exact. Explain your reasoning.

1.

2.

3.

4.

Solving an Exact Differential Equation In Exercises5–14, determine whether the differential equation is exact. If itis, find the general solution.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

Graphical and Analytic Analysis In Exercises 15 and 16,(a) sketch an approximate solution of the differential equationsatisfying the initial condition on the slope field, (b) find theparticular solution that satisfies the initial condition, and (c) use a graphing utility to graph the particular solution.Compare the graph with the sketch in part (a).

Differential Equation Initial Condition

15.

16.

Figure for 15 Figure for 16

Finding a Particular Solution In Exercises 17–22, findthe particular solution that satisfies the initial condition.

Differential Equation Initial Condition

17.

18.

19.

20.

21.

22.

Finding an Integrating Factor In Exercises 23–32, findthe integrating factor that is a function of or alone and useit to find the general solution of the differential equation.

23.

24.

25.

26.

27.

28.

29.

30.

31.

32.

Using an Integrating Factor In Exercises 33–36, use theintegrating factor to find the general solution of the differentialequation.

Integrating Factor Differential Equation

33.

34.

35.

36.

37. Integrating Factor Show that each expression is an integrating factor for the differential equation

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

38. Integrating Factor Show that the differential equation

is exact only when For show that is an integrating factor, where

n � �2a � ba � b

.m � �2b � aa � b

,

xmyna � b,a � b.

�axy2 � by� dx � �bx2y � ax� dy � 0

1x2 � y2

1xy

1y2

1x2

y dx � x dy � 0.

�y3 dx � �xy2 � x2� dy � 0u�x, y� � x�2y�2

��y5 � x2y� dx � �2xy4 � 2x3� dy � 0u�x, y� � x�2y�3

�3y2 � 5x2y� dx � �3xy � 2x3� dy � 0u�x, y� � x2y

�4x2y � 2y2� dx � �3x3 � 4xy� dy � 0u�x, y� � xy2

��2y3 � 1� dx � �3xy2 � x3� dy � 0

2y dx � �x � sin �y � dy � 0

�x2 � 2x � y� dx � 2 dy � 0

y2 dx � �xy � 1� dy � 0

�2x2y � 1� dx � x3 dy � 0

�x � y� dx � tan x dy � 0

�5x2 � y2� dx � 2y dy � 0

�5x2 � y� dx � x dy � 0

�2x3 � y� dx � x dy � 0

y dx � �x � 6y2� dy � 0

yx

y��1� � 8�2xy2 � 4� dx � �2x2y � 6� dy � 0

y�0� � �3�2xy � 9x2� dx � �2y � x2 � 1� dy � 0

y�3� � 1�x2 � y2� dx � 2xy dy � 0

y�0� � �e3x�sin 3y dx � cos 3y dy� � 0

y�0� � 41

x2 � y2 �x dx � y dy� � 0

y�2� � 4y

x � 1 dx � �ln�x � 1� � 2y� dy � 0

x−4 −2 2 4

−4

−2

2

4

y

x−4 −2 2 4

−4

−2

4

2

y

�4, 3�1�x2 � y2

�x dx � y dy� � 0

12

, �

4��2x tan y � 5� dx � �x2 sec2 y� dy � 0

ey cos xy � y dx � �x � tan xy� dy� � 0

1�x � y�2 �y2 dx � x2 dy� � 0

e��x2�y2��x dx � y dy� � 0

1x2 � y2 �x dy � y dx� � 0

2y2exy2 dx � 2xyexy2

dy � 0

�4x3 � 6xy2� dx � �4y3 � 6xy� dy � 0

2 cos�2x � y� dx � cos�2x � y� dy � 0

�3y2 � 10xy2� dx � �6xy � 2 � 10x2y� dy � 0

yex dx � ex dy � 0

�2x � 3y� dx � �2y � 3x� dy � 0

yexy dx � xexy dy � 0

x sin y dx � x cos y dy � 0

�1 � xy� dx � �y � xy� dy � 0

�2x � xy2� dx � �3 � x2y� dy � 0

9781285060293_1601.qxp 10/15/12 5:09 PM Page 1131


Tangent Curves In Exercises 39–42, use a graphing utilityto graph the family of curves tangent to the force field.

39.

40.

41.

42.

Finding an Equation of a Curve In Exercises 43 and 44,find an equation of the curve with the specified slope passingthrough the given point.

Slope Point

43.

44.

Euler’s Method In Exercises 47 and 48, (a) use Euler’sMethod and a graphing utility to graph the particular solutionof the initial value problem over the indicated interval with thespecified value of and initial condition, (b) find the exact solutionof the differential equation analytically, and (c) use a graphingutility to graph the particular solution and compare the resultwith the graph in part (a).

47. 0.05

48. 0.2

49. Euler’s Method Repeat Exercise 47 for and discuss how the accuracy of the result changes.

50. Euler’s Method Repeat Exercise 48 for and discuss how the accuracy of the result changes.

True or False? In Exercises 53–56, determine whether thestatement is true or false. If it is false, explain why or give anexample that shows it is false.

53. The differential equation is exact.

54. If is exact, then is alsoexact.

55. If is exact, then is also exact.

56. The differential equation is exact.

Exact Differential Equation In Exercises 57 and 58, findall values of such that the differential equation is exact.

57.

58.

59. Exact Differential Equation Find all nonzero functionsand such that

is exact.

60. Exact Differential Equation Find all nonzero functionssuch that

is exact.

g�y� ey dx � xy dy � 0

g

g�y� sin x dx � y2 f �x� dy � 0

gf

�ye2xy � 2x� dx � �kxe2xy � 2y� dy � 0

�xy2 � kx2y � x3� dx � �x3 � x2y � y2� dy � 0

k

f �x� dx � g�y� dy � 0

N� dy � 0� f �x� � M � dx � �g�y� �M dx � N dy � 0

xM dx � xN dy � 0M dx � N dy � 0

2xy dx � �y2 � x2� dy � 0

h � 0.5

h � 1

y�0� � 1�0, 5�y� �6x � y2

y�3y � 2x�

y�2� � 1�2, 4�y� ��xy

x2 � y2

InitialConditionhInterval

DifferentialEquation

h

�0, 2�dydx

��2xy

x2 � y2

�2, 1�dydx

�y � x3y � x

F�x, y� � �1 � x2�i � 2xy j

F�x, y� � 4x2y i � 2xy2 �xy2� j

F�x, y� �x

�x2 � y2 i �

y�x2 � y2

j

F�x, y� �y

�x2 � y2 i �

x�x2 � y2

j

In a manufacturing process where represents thecost of producing units, the elasticity of cost is defined as

Find the cost function when the elasticity function is

where

and x 100.

C�100� � 500

E�x� �20x � y

2y � 10x

E�x� �marginal costaverage cost

�C��x�

C�x��x�

xy dydx

.

xy � C�x�

45. Cost

46. HOW DO YOU SEE IT? The graph of the cost function in Exercise 45 is shown below. Usethe figure to estimate the limit of the cost functionas approaches 100 from the right.

50 100 150 200 250 300 350 400 450 500

1000

2000

3000

4000

5000

x

y

x

WRITING ABOUT CONCEPTS51. Testing for Exactness Explain how to determine

whether a differential equation is exact.

52. Finding an Integrating Factor Outline the procedurefor finding an integrating factor for the differential equationM�x, y� dx � N�x, y� dy � 0.

Mircea BEZERGHEANU/Shutterstock.com

9781285060293_1601.qxp 10/15/12 5:10 PM Page 1132

16.2 Second-Order Homogeneous Linear Equations 1133

16.2 Second-Order Homogeneous Linear Equations

Solve a second-order linear differential equation.Solve a higher-order linear differential equation.Use a second-order linear differential equation to solve an applied problem.

Second-Order Linear Differential EquationsIn this section and the next section, you will learn methods for solving higher-order lineardifferential equations.

Homogeneous equations are discussed in this section, and the nonhomogeneouscase is discussed in the next section.

The functions are linearly independent when the only solution ofthe equation

is the trivial one

Otherwise, this set of functions is linearly dependent.

Linearly Independent and Dependent Functions

a. The functions

and

are linearly independent because the only values of and for which

for all are and

b. It can be shown that two functions form a linearly dependent set if and only if oneis a constant multiple of the other. For example,

and

are linearly dependent because

has the nonzero solutions

and C2 � 1.C1 � �3

C1x � C2�3x� � 0

y2�x� � 3xy1�x� � x

C2 � 0.C1 � 0x

C1 sin x � C2x � 0

C2C1

y2�x� � xy1�x� � sin x

C1 � C2 � . . . � Cn � 0.

C1y1 � C2y2 � . . . � Cnyn � 0

y1, y2, . . . , yn

Definition of Linear Differential Equation of Order

Let and be functions of with a common (interval) domain. An equation of the form

is a linear differential equation of order If then the equationis homogeneous; otherwise, it is nonhomogeneous.

f �x� � 0,n.

y�n� � g1�x�y�n�1� � g2�x�y�n�2� � . . . � gn�1�x�y� � gn�x�y � f �x�

xfg1, g2, . . . , gn

n

REMARK Notice that thisuse of the term homogeneousdiffers from that in Section 6.3.

9781285060293_1602.qxp 10/15/12 5:10 PM Page 1133

The next theorem points out the importance of linear independence in constructingthe general solution of a second-order linear homogeneous differential equation withconstant coefficients.

Proof This theorem is proved in only one direction. Letting and be solutions of

you obtain 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 a fullcourse on differential equations.See LarsonCalculus.com for Bruce Edwards’s video of this proof.

Theorem 16.3 states that when you can find two linearly independent solutions,you can 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

So, 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,y�m2,y

y � ay� � by � 0.

m2 � am � b � 0.

y � emxemx

emx�m2 � am � b� � 0.

m2emx � amemx � bemx � 0

y � ay� � by � 0

y � emx

y � m2emx.y� � memx

y � emx.y � ay� � by � 0

y � C1y1 � C2y2

�C1y1 �x� � C2y2 �x�� a�C1y1��x� � C2y2��x�� b�C1y1�x� � C2y2�x�� 0

C2,C1

y2 �x� � ay2� �x� � by2�x� � 0

y1 �x� � ay1� �x� � by1�x� � 0

y � ay� � by � 0

y2y1


THEOREM 16.3 Linear Combinations of Solutions

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

General solution

where and are constants.C2C1

y � C1y1 � C2y2

y � ay� � by � 0,y2y1

9781285060293_1602.qxp 10/15/12 5:10 PM Page 1134

Characteristic Equation: Distinct Real Zeros


Solution In this case, the characteristic equation is

Characteristic equation

So, Therefore,

and

are particular solutions of the differential equation. Furthermore, because these two solutionsare linearly independent, you can apply Theorem 16.3 to conclude that the general solution is

General solution

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

and

which fall into one of three cases.

1. Two distinct real zeros,

2. Two equal real zeros,

3. Two complex conjugate zeros, and

In terms of the differential equation

these three cases correspond to three different types of general solutions.

y � ay� � by � 0

m2 � � � �im1 � � � �i

m1 � m2

m1 � m2

m2 ��a � �a2 � 4b

2m1 �

�a � �a2 � 4b2

m2 � am � b � 0

y � C1e2x � C2e

�2x.

y2 � em2x � e�2xy1 � em1x � e2x

m � ±2.

m2 � 4 � 0.

y � 4y � 0.


THEOREM 16.4 Solutions of

The solutions of

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

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

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

3. Complex Zeros If and are complex zeros ofthe characteristic equation, then the general solution is

y � C1e�x cos �x � C2e

�x sin �x.

m2 � � � �im1 � � � �i

y � C1em1x � C2xem1x � �C1 � C2x�em1x.

m1 � m2

y � C1em1x � C2e

m2x.

m1 � m2

m2 � am � b � 0.

y � ay� � by � 0

y � ay� � by � 0

FOR FURTHER INFORMATIONFor more information on Theorem 16.4, see the article “A Note on a DifferentialEquation” by Russell Euler in the1989 winter issue of the MissouriJournal of Mathematical Sciences.

Exploration

For each differential equation below, find thecharacteristic equation. Solvethe characteristic equationfor and use the values of

to find a general solutionof the differential equation.Using your results, develop ageneral solution of differentialequations with characteristicequations that have distinctreal roots.

(a)

(b) y � 6y� � 8y � 0

y � 9y � 0

mm,

9781285060293_1602.qxp 10/15/12 5:10 PM Page 1135

Characteristic Equation: Complex Zeros

Find the general solution of the differential equation

Solution The characteristic equation

has two complex zeros, as follows.

So, and and the general solution is

The graphs of the basic solutions

and

along with other members of the family of solutions, are shown in Figure 16.5.

In Example 3, note that although the characteristic equation has two complex zeros,the solution of the differential equation is real.

Characteristic Equation: Repeated Zeros


subject to the initial conditions and


has two equal zeros given by So, the general solution is

General solution

Now, because when you have

Furthermore, because when you have

Therefore, the solution is

Particular solution

Try checking this solution in the original differential equation.

y � 2e�2x � 5xe�2x.

5 � C2.

1 � �2�2��1� � C2��2�0��1� � 1� y� � �2C1e

�2x � C2��2xe�2x � e�2x�

x � 0,y� � 1

2 � C1.2 � C1�1� � C2�0��1�

x � 0,y � 2

y � C1e�2x � C2xe�2x.

m � �2.

�m � 2�2 � 0m2 � 4m � 4 � 0

y��0� � 1.y�0� � 2

y � 4y� � 4y � 0

g�x� � e�3x sin �3xf �x� � e�3x cos �3x

y � C1e�3x cos �3x � C2e

�3x sin �3x.

� � �3,� � �3

� �3 ± �3i

� �3 ± ��3

��6 ± 2��3

2

��6 ± ��12

2

m ��6 ± �36 � 48

2

m2 � 6m � 12 � 0

y � 6y� � 12y � 0.


x1 432

3f

gf + g

g − f

y

The basic solutions in Example 3,andare shown in

the graph along with other members ofthe family of solutions. Notice that as

all of these solutions approach 0.Figure 16.5x → ,

g�x� � e�3x sin �3x,f �x� � e�3x cos �3x

9781285060293_1602.qxp 10/15/12 5:10 PM Page 1136

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 zeros of the characteristic equation. Then, based on these zeros,you form a linearly independent collection of solutions. The major difference is thatwith equations of third or higher order, zeros of the characteristic equation may occurmore than twice. When this happens, the linearly independent solutions are formed bymultiplying by increasing powers of as demonstrated in Examples 6 and 7.

Solving a Third-Order Equation

Find the general solution of

Solution The characteristic equation and its zeros are

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

General solution




Because the zero occurs three times, the general solution is

General solution

Solving a Fourth-Order Equation




Because each of the zeros

and

occurs twice, the general solution is

General solutiony � C1 cos x � C2 sin x � C3x cos x � C4x sin x.

m2 � � � �i � 0 � im1 � � � �i � 0 � i

m � ± i.

�m2 � 1�2 � 0

m4 � 2m2 � 1 � 0

y�4� � 2y � y � 0.

y � C1e�x � C2xe�x � C3x

2e�x.

m � �1

m � �1.

�m � 1�3 � 0

m3 � 3m2 � 3m � 1 � 0

y�� 3y � 3y� � y � 0.

y � C1 � C2e�x � C3e

x.

m � 0, 1, �1.

m�m � 1��m � 1� � 0

m3 � m � 0

y�� y� � 0.

x,

nnn


9781285060293_1602.qxp 10/15/12 5:10 PM Page 1137

ApplicationOne of the many applications of linear differential equations is describing the motionof an oscillating spring. According to Hooke’s Law, a spring that is stretched (or compressed)

units from its natural length tends to restore itself to its natural length by a force that is proportional to That is, where is the spring constant and indicates the stiffness of the spring.

A rigid object of mass is attached to the end of a spring and causes a displacement,as shown in Figure 16.6. Assume that the mass of the spring is negligible comparedwith When the object is pulled downward and released, the resulting oscillations area product of two opposing forces—the spring force and the weight ofthe object. Under such conditions, you can use a differential equation to find the position of the object as a function of time According to Newton’s Second Law ofMotion, the force acting on the weight is where is the acceleration.Assuming that the motion is undamped—that is, there are no other external forces acting on the object—it follows that and you have

Undamped motion of a spring

Undamped Motion of a Spring

A 4-pound weight stretches a spring 8 inches from its natural length. The weight ispulled downward an additional 6 inches and released with an initial upward velocity of8 feet per second. Find a formula for the position of the weight as a function of time

Solution By Hooke’s Law, so Moreover, because the weight is

given by it follows that

So, the resulting differential equation for this undamped motion is

The characteristic equation has complex zeros so thegeneral solution is

When seconds, Using this initial condition, you have

To determine note that feet per second when seconds.

Consequently, the position at time is given by

y �12

cos 4�3 t �2�3

3 sin 4�3 t.

t

2�3

3� C2

y��0� � 8 8 � �4�3 12��0� � 4�3 C2�1�

y��t� � �4�3 C1 sin 4�3 t � 4�3 C2 cos 4�3 t

t � 0y� � 8C2,

y�0� �12C1 �

12

.12

� C1�1� � C2�0�

y � 6 inches �12 foot.t � 0

� C1 cos 4�3 t � C2 sin 4�3 t.

y � C1e0 cos 4�3 t � C2e

0 sin 4�3 t

m � 0 ± 4�3i,m2 � 48 � 0

d2ydt2

� 48y � 0.

m �wg

�4

32�

18

.

mg,

wk � 6.4 � k�23�,

t.

d2ydt2

� km

�y � 0.

�ky,m�d2y�dt2� �

a � d2y�dt2F � ma,t.y

mgF�y� � �kym.

m

kF� y� � �ky,y.Fly


m

l = naturallength

y = displacement

A rigid object of mass attached tothe end of the spring causes a displacement of Figure 16.6

y.

m

A common type of spring is acoil spring, also called a helicalspring because the shape of thespring is a helix. A tension coilspring resists being stretched(see photo above and Example 8).A compression coil spring resistsbeing compressed, such as thespring in a car suspension.

ICHIRO/Photodisc/Getty Images

9781285060293_1602.qxp 10/15/12 5:10 PM Page 1138

The object in Figure 16.7 undergoes an additional damping or frictional force that isproportional to its velocity. A case in point would be the damping force resulting fromfriction and movement through a fluid. Considering this damping force

Damping force

the differential equation for the oscillation is

or, in standard linear form,

Damped motion of a springd2ydt2

�pm

dydt� �

km

y � 0.

m d2ydt2

� �ky � p dydt

�p dydt


A damped vibration could be caused byfriction and movement through a liquid.Figure 16.7

Verifying a Solution In Exercises 1–4, verify the solutionof the differential equation. Then use a graphing utility tograph the particular solutions for several different values of and What do you observe?

Solution Differential Equation

1.

2.

3.

4.

Finding a General Solution In Exercises 5–30, find thegeneral solution of the linear differential 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. Finding a Particular Solution Consider the differentialequation and the solution

Find the particular solution satisfying each initial condition.

(a) (b)

(c)

32. Finding a Particular Solution Determine and suchthat is a particular solution of the differentialequation where

Finding a Particular Solution: Initial Conditions InExercises 33–38, find the particular solution of the linear differential equation that satisfies the initial conditions.

33. 34.

35. 36.

37. 38.

Finding a Particular Solution: Boundary Conditions

In Exercises 39–44, find the particular solution of the linear differential equation that satisfies the boundary conditions, ifpossible.

39. 40.

41. 42.

43. 44.

y�0� � 4, y�� 8y�0� � 2, y�1� � �1

y � 6y� � 45y � 04y � 28y� � 49y � 0

y�0� � 3, y�2� � 0y�0� � 3, y�� 5

4y � 20y� � 21y � 0y � 9y � 0

y�0� � 2, y�� 5y�0� � 1, y�1� � 3

4y � y � 0y � 4y� � 3y � 0

y�0� � 3, y��0� � �1y�0� � 2, y��0� � 1

4y � 4y� � y � 0y � 2y� � 3y � 0

y�0� � 2, y��0� � 1y�0� � 0, y��0� � 2

9y � 6y� � y � 0y � 16y � 0

y�0� � 3, y��0� � 3y�0� � 1, y��0� � �4

y � 7y� � 12y � 0y � y� � 30y � 0

y��0� � �5.y � �y � 0,y � C sin�3 t

�C

y�0� � �1, y��0� � 3

y�0� � 0, y��0� � 2y�0� � 2, y��0� � 0

y � C1 cos 10x � C2 sin 10x.y � 100y � 0

y�� 3y � 3y� � y � 0y�� 3y � 7y� � 5y � 0

y�� y � y� � y � 0y�� 6y � 11y� � 6y � 0

y�4� � y � 0y�4� � y � 0

2y � 6y� � 7y � 09y � 12y� � 11y � 0

3y � 4y� � y � 0y � 3y� � y � 0

y � 4y� � 21y � 0y � 2y� � 4y � 0

y � 2y � 0y � 9y � 0

y � 4y � 0y � y � 0

9y � 12y� � 4y � 016y � 8y� � y � 0

y � 10y� � 25y � 0y � 6y� � 9y � 0

16y � 16y� � 3y � 02y � 3y� � 2y � 0

y � 6y� � 5y � 0y � y� � 6y � 0

y � 2y� � 0y � y� � 0

y � 2y� � 10y � 0y � C1e�x cos 3x � C2e

�x sin 3x

y � 4y � 0y � C1 cos 2x � C2 sin 2x

y � 4y � 0y � C1e2x � C2e

�2x

y � 6y� � 9y � 0y � �C1 � C2x�e�3x

C2.C1


WRITING ABOUT CONCEPTS45. Characteristic Equation The solutions of the

differential equation fall into what threecases? What is the relationship of these solutions to the characteristic equation of the differential equation?

46. Linearly Independent Functions Two functionsare said to be linearly independent provided what?

y � ay� � by � 0

9781285060293_1602.qxp 10/15/12 5:10 PM Page 1139


Vibrating Spring In Exercises 49–52, match the differentialequation with the graph of a particular solution. [The graphsare labeled (a), (b), (c), and (d).] The correct match can bemade by comparing the frequency of the oscillations or the rateat which the oscillations are being damped with the appropriatecoefficient in the differential equation.

(a) (b)

(c) (d)

49. 50.

51. 52.

Vibrating Spring In Exercises 53–58, describe the motionof a 32-pound weight suspended on a spring. Assume that theweight stretches the spring foot from its natural position.

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

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

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

56. The weight is pulled foot below the equilibrium position andstarted off with an upward velocity of foot per second.

57. 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.

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

59. Real Zeros The characteristic equation of the differentialequation has two equal real zeros given by

Show that is a solution.

60. Complex Zeros The characteristic equation of the differentialequation

has complex zeros given by and Show that is a solution.

True or False? In Exercises 61–64, determine whether thestatement is true or false. If it is false, explain why or give anexample that shows it is false.

61. is the general solution of

62. is the general solution of

63. is a solution of if and only if

64. It is possible to choose and such that is a solutionof

Wronskian The Wronskian of two differentiable functions and denoted by is defined as the function given bythe determinant

The functions and are linearly independent when thereexists at least one value of for which In Exercises65–68, use the Wronskian to verify the linear independence ofthe two functions.

65. 66.

67. 68.

y2 � x2y2 � eax cos bx, b � 0

y1 � xy1 � eax sin bx

y2 � xeaxy2 � ebx, a � b

y1 � eaxy1 � eax

W� f, g � 0.xgf

W� f, g � � ff�

gg��.

W� f, g ,g,f

y � ay� � by � 0.y � x2exba

a1 � a0 � 0.a0y � 0any�n� � an�1y

�n�1� � . . . � a1y� �y � x

y�4� � 2y � y � 0.y � �C1 � C2x�sin x � �C3 � C4x�cos x

9 � 0.y � 6y� �y � C1e

3x � C2e�3x

y � C1e�x cos �x � C2e

�x sin �xm2 � � � �i.m1 � � � �i

y � ay� � by � 0

y � C1erx � C2xerxm � r.

y � ay� � by � 0

14�v�

12

18

12

12

12

12

23

23

12

23

y � y� �374 y � 0y � 2y� � 10y � 0

y � 25y � 0y � 9y � 0

61

3

3

x

y

5 642 3

3

x

y

6

3

42 31x

y

5 6

3

421x

y

Several shock absorbers areshown at the right.Do you think the motion of the spring in a shock absorber isundamped or damped?

47. Undamped or Damped Motion?

48. HOW DO YOU SEE IT? Give a geometricargument to explain why the graph cannot be asolution of the differential equation. (It is not necessary to solve the differential equation.)

(a) (b)

−3 −2 1 2 3

1

3

2

x

y

−3−3 −2 −1 1 2 3

1

3

2

5

4

x

y

y � �12 y�y � y�

Monty Rakusen/Cultura/Getty Images

9781285060293_1602.qxp 10/15/12 5:10 PM Page 1140

16.3 Second-Order Nonhomogeneous Linear Equations 1141

16.3 Second-Order Nonhomogeneous Linear Equations

Recognize the general solution of a second-order nonhomogeneous linear differential equation.Use the method of undetermined coefficients to solve a second-order nonhomogeneous linear differential equation.Use the method of variation of parameters to solve a second-order nonhomogeneous linear differential equation.

Nonhomogeneous EquationsIn the preceding section, damped oscillations of a spring were represented by the homogeneous 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 subjectto an external periodic force, such as caused by vibrations at the opposite endof the spring, then the motion is called forced, and it is characterized by the nonhomogeneousequation

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 next theorem.

y � yh � yp

y � yp

y � yh

d2ydt2 �

pm

dydt� �

km

y � a sin bt.

a sin bt,

d2ydt2 �

pm

dydt� �

km

y � 0.

THEOREM 16.5 Solution of Nonhomogeneous Linear Equation

Let

be a second-order nonhomogeneous linear differential equation. If is a particular solution of this equation and is the general solution of the corresponding homogeneous equation, then

is the general solution of the nonhomogeneous equation.

y � yh � yp

yh

yp

y � ay� � by � F�x�

SOPHIE GERMAIN (1776–1831)

Many of the early contributors tocalculus were interested in formingmathematical models for vibratingstrings and membranes, oscillatingsprings, and elasticity. One ofthese was the French mathematician Sophie Germain,who in 1816 was awarded a prizeby the French Academy for apaper entitled “Memoir on theVibrations of Elastic Plates.”See LarsonCalculus.com to readmore of this biography.

The Granger Collection, NYC

9781285060293_1603.qxp 10/15/12 5:11 PM Page 1141

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

consists of sums or products of or you can find a particular solutionby the method of undetermined coefficients. The object of this method is to guess

that the solution is a generalized form of Here are some examples.

1. For choose

2. For choose

3. For choose

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

Method of Undetermined Coefficients

Find the general solution of the equation

Solution To find solve the characteristic equation.

or

So, 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

Therefore,

and the general solution is

� C1e�x � C2e

3x �15

cos x �25

sin x.

y � yh � yp

yp �15

cos x �25

sin x

B � �25

.A �15

2A � 4B � 2�4A � 2B � 0

��4A � 2B�cos x � �2A � 4B�sin x � 2 sin x.

�A cos x � B sin x � 2A sin x � 2B cos x � 3A cos x � 3B sin x � 2 sin x

y � 2y� � 3y � 2 sin x

yp � �A cos x � B sin x

yp� � �A sin x � B cos x

yp � A cos x � B sin x

2 sin x.ypyh � C1e�x � C2e

3x.

m � 3 m � �1

�m � 1��m � 3� � 0

m2 � 2m � 3 � 0

yh,

y � 2y� � 3y � 2 sin x.

yp � �Ax � B� � C sin 2x � D cos 2x.F�x� � x � sin 2x,

yp � Axex � Bex.F�x� � 4xex,

yp � Ax2 � Bx � C.F�x� � 3x2,

F�x�.yp

yp

sin �x,xn, emx, cos �x,

y � ay� � by � F�x�

F�x�yp.

yh


9781285060293_1603.qxp 10/15/12 5:11 PM Page 1142

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 was

because this solution would yield 0. In such cases, you should alter your guess by multiplying by the lowest power of that removes the duplication. For this particularproblem, you would guess

Method of Undetermined Coefficients



has solutions and So,

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

Substitution into the differential equation produces

Equating coefficients of like terms yields the system

with solutions and Therefore,


� C1 � C2e2x �

14

x �14

x2 � 2ex.

y � yh � yp

yp � �14

x �14

x2 � 2ex

C � �2.A � B � �14

�C � 2�4B � 1,2B � 2A � 0,

�2B � 2A� � 4Bx � Cex � x � 2ex.

2B � Cex � 2�A � 2Bx � Cex� � x � 2ex

y � 2y� � x � 2ex

yp � 2B � Cex.

yp� � A � 2Bx � Cex

yp � Ax � Bx2 � Cex

xC1,yh

�A � Bx� � Cex.ypF�x� � x � 2ex,

yh � C1 � C2e2x.

m � 2.m � 0

m2 � 2m � 0

y � 2y� � x � 2ex.

yp � Axe�x.

x

y � Ae�x

y � 2y� � 3y � e�x.

y � ay� � by � F�x�.

F�x�

yh � C1e�x � C2e

3x


9781285060293_1603.qxp 10/15/12 5:11 PM Page 1143

In Example 2, the polynomial part of the initial guess for overlapped by a constant term with

and it was necessary to multiply the polynomial part by a power of that removed theoverlap. The next example further illustrates some choices for that eliminate overlapwith Remember that in all cases, the first guess for should match the types offunctions occurring in

Choosing the Form of the Particular Solution

Determine a suitable choice for for each differential equation, given its general solution of the homogeneous equation.

a.

b.

c.

Solution

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

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

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




Solution From Example 6 in Section 16.2, you know that the homogeneous solution is

Because let and obtain and So, by substitutioninto the general solution, you have

So, and which implies that Therefore, the general solution is

� C1e�x � C2xe�x � C3x

2e�x � 3 � x.

y � yh � yp

yp � �3 � x.A � �3,B � 1

�3B � A� � Bx � x.

0 � 3�0� � 3�B� � A � Bx � x

yp � 0.yp� � Byp � A � BxF�x� � x,

yh � C1e�x � C2xe�x � C3x2e�x.

y�� 3y � 3y� � y � x.

yp � Ax2e2x.

x2xe2xyh � C1e2x � C2xe2x

Ae2x.ypF�x� � e2x,

yp � A cos 3x � B sin 3x.

e�x,yhF�x� � 4 sin 3x

yp � Ax2 � Bx3 � Cx4.

x2yh � C1 � C2xA � Bx � Cx2.ypF�x� � x2,

C1e2x � C2xe2xy � 4y� � 4 � e2x

C1e�x cos 3x � C2e

�x sin 3xy � 2y� � 10y � 4 sin 3x

C1 � C2xy � x2

yhy � ay� � by � F �x

yp

F�x�.ypyh.

yp

x

yh � C1 � C2e2x

yp�A � Bx� � Cex


9781285060293_1603.qxp 10/15/12 5:11 PM Page 1144

Variation of ParametersThe method of undetermined coefficients works well when is made up of polynomialsor functions whose successive derivatives have a cyclical pattern. For functions such as

and which do not have such characteristics, it is better to use a more generalmethod called variation of parameters. In this method, you assume that has thesame form as except that the constants in are replaced by variables.

Variation of Parameters



has one repeated solution, So, 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 substitutionin the first equation, you have Finally, integration yields

and

From this result, it follows that a particular solution is


y � C1ex � C2xex �

12

xex � xex ln �x.

yp � �12

xex � �ln �x �xex

u2 �12

� 1x dx �

12

ln x � ln �x.u1 � �� 12

dx � �x2

u1� � �12.

u2� � 1��2x�.

u1� ex � u2� �xex � ex� �

ex

2x.

u1� ex � u2� xex � 0

yp � u1y1 � u2y2 � u1ex � u2xex.

u2u1C2C1

yh � C1y1 � C2y2 � C1ex � C2xex.

m � 1.

�m � 1�2 � 0m2 � 2m � 1 � 0

x > 0.y � 2y� � y �ex

2x,

yhyh,yp

tan x,1�x

F�x�



To find the general solution of the equation use these 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 � yh � yp.u2.u1

u1�y1� � u2�y2� � F�x�u1�y1 � u2�y2 � 0

u2�.u1�

yp � u1y1 � u2y2.

yh � C1y1 � C2y2.

y � ay� � by � F�x�,

9781285060293_1603.qxp 10/15/12 5:11 PM Page 1145



Solution Because the characteristic equation has solutions thehomogeneous solution is

Replacing and by and produces The resulting systemof 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


� C1 cos x � C2 sin x � cos x ln �sec x � tan x�. y � yh � yp

� �cos x ln �sec x � tan x� yp � sin x cos x � cos x ln �sec x � tan x� � sin x cos x

u2 � � sin x dx � �cos x

� sin x � ln �sec x � tan x� u1 � � �cos x � sec x� dx

� cos x � sec x.

�cos2 x � 1

cos x

u1� � �sin2 xcos x

u2� � sin x,

�u1� sin x cos x � u2� cos2 x � sin x.

u1� sin x cos x � u2� sin2 x � 0

cos xsin x

�u1� sin x � u2� cos x � tan x.

u1� cos x � u2� sin x � 0

yp � u1 cos x � u2 sin x.u2u1C2C1

yh � C1 cos x � C2 sin x.

m � ± i,m2 � 1 � 0

y � y � tan x.


Exploration

Notice in Example 5 that the constants of integration were not introducedwhen finding and Show that for

and

the general solution

yields the same result as the solution obtained in the example.

y � yh � yp � C1ex � C2e

x �12

xex � xex ln �x

u2 � ln �x � a2u1 � �x2

� a1

u2.u1

9781285060293_1603.qxp 10/15/12 5:11 PM Page 1146



Verifying a Solution In Exercises 1–4, verify the solutionof the differential equation.


1.

2.

3.

4.

Finding a Particular Solution In Exercises 5–10, find aparticular solution of the differential equation.

5.

6.

7.

8.

9.

10.

Method of Undetermined Coefficients In Exercises 11–18,solve the differential equation by the method of undeterminedcoefficients.

11. 12.

13. 14.

15.

16.

17.

18.

Method of Undetermined Coefficients In Exercises19–24, solve the differential equation by the method of undetermined coefficients that satisfies the initial condition(s).

19. 20.

21. 22.

23. 24.

Method of Variation of Parameters In Exercises 25–30,solve the differential equation by the method of variation ofparameters.

25. 26.

27. 28.

29. 30.

Vibrating Spring In Exercises 35–38, find the particular solution of the differential equation

for the oscillating motion of an object on the end of a spring.Use a graphing utility to graph the solution. In the equation,is the displacement from equilibrium (positive direction isdownward), measured in feet, and is time in seconds (see figure).The constant is the weight ofthe object, is the acceleration dueto gravity, is the magnitude ofthe resistance to the motion,is the spring constant fromHooke’s Law, and is the acceleration imposed on the system.

35.

y�0� �14, y��0� � 0

2432y � 48y �

2432�48 sin 4t�

F�t

kbg

wt

y

wg

y �t � by��t � ky�t �wg

F�t

y � 4y� � 4y �e2x

xy � 2y� � y � ex ln x

y � 4y� � 4y � x2e2xy � 4y � csc 2x

y � y � sec x tan xy � y � sec x

y�

2� �25

y�0� �13

y� � 2y � sin xy� � 4y � xex � xe4x

y�0� � �1, y��0� � 2y�0� � 0, y��0� � �3

y � y� � 2y � 3 cos 2xy � y� � 2 sin x

y�0� � 1, y��0� � 6y�0� � 1, y��0� � 0

y � 4y � 4y � y � x3

y�� 3y� � 2y � 2e�2x

y � 9y � sin 3x

16y � 8y� � y � 4�x � ex�y � 10y� � 25y � 5 � 6ex

y � 9y � 5e3xy � 2y� � 2ex

y � 2y� � 3y � x2 � 1y � 3y� � 2y � 2x

y � 4y� � 5y � ex cos x

y � 2y� � 15y � sin x

y � y� � 3y � e2x

y � 8y� � 16y � e3x

y � y� � 6y � 4

y � 7y� � 12y � 3x � 1

y � y � csc x cot xy � �5 � ln �sin x��cos x � x sin x

y � y � tan xy � 3 sin x � cos x ln �sec x � tan x�y � y � cos xy � �2 �

12x�sin x

y � y � 10e2xy � 2�e2x � cos x�

WRITING ABOUT CONCEPTS31. Choosing Using the method of undetermined

coefficients, determine a suitable choice for for each differential equation. Explain your reasoning. (You do notneed to solve the differential equations.)

(a) (b)

32. Variation of Parameters Describe the steps forsolving a differential equation by the method of variationof parameters.

y � y� � 12y � e4xy � y� � 12y � x2

yp

yp

m

l = naturallength

y = displacement

Spring displacement

In Exercises 33 and 34, use the electrical circuit differentialequation

where is the resistance (in ohms),

is the capacitance (in farads), is the inductance (in henrys),

is the electromotiveforce (in volts), and

is the charge on thecapacitor (in coulombs).Find the charge as a function of time for the electricalcircuit described. Assume that and

33.

34. E�t� � 10 sin 5tR � 20, C � 0.02, L � 1,

E�t� � 12 sin 5tR � 20, C � 0.02, L � 2,

q� �0 � 0.q�0 � 0q

q

E�t

LC

R

d 2qdt2

� �RL�

dqdt

� � 1LC�q � �1

L�E�t

Electrical Circuits

nsj-images/iStockphoto

9781285060293_1603.qxp 10/15/12 5:11 PM Page 1147


36.

37.

38.

39. Vibrating Spring Rewrite in the solution to Exercise 35by using the identity

where

41. Vibrating Spring Refer to the differential equation andthe initial conditions given in Exercise 40.

(a) When there is no resistance to the motion describe the motion.

(b) For what is the ultimate effect of the retarding force?

(c) Is there a real number such that there will be no oscillationsof the spring for Explain your answer.

42. Solving a Differential Equation Solve the differentialequation given that and are solutions of the correspondinghomogeneous equation.

(a)

(b)

True or False? In Exercises 43 and 44, determine whetherthe statement is true or false. If it is false, explain why or givean example that shows it is false.

43. is a particular solution of the differentialequation

44. is a particular solution of the differential equation

y � 6y� � e2x.

yp � �18e2x

y � 3y� � 2y � cos e�x.

yp � �e2x cos e�x

y1 � sin �ln x2�, y2 � cos �ln x2�x2y � xy� � 4y � sin �ln x�y1 � x, y2 � x ln x

x2y � xy� � y � 4x ln x

y2y1

b > M ?M

b > 0,

�b � 0�,

� � arctan a�b.

a cos �t � b sin �t � �a2 � b2 sin��t � ��

yh

y�0� �12, y��0� � �4

432y �

12y� �

252 y � 0

y�0� �14, y��0� � �3

232y � y� � 4y �

232�4 sin 8t�

y�0� �14, y��0� � 0

232y � 4y �

232 �4 sin 8t�

40. HOW DO YOU SEE IT? The figure shows theparticular solution of the differential equation

that satisfies the initial conditions andfor values of the resistance component

in the interval (Note that when the problem is identical to that of Exercise 38.)According to the figure, is the motion damped or undamped when when (You do not need to solve the differential equation.)

y

t

bGenerated by Maple

b = 0

b = 1

b = 12

b > 0?b � 0?

b �12,�0, 1�.b

y��0� � �4y�0� �

12

432

y � by� �252

y � 0

PUTNAM EXAM CHALLENGE45. For all real the real-valued function satisfies

(a) If for all real must for all real Explain.

(b) If for all real must for all real Explain.

This problem was composed by the Committee on the Putnam Prize Competition.© The Mathematical Association of America. All rights reserved.

x?f �x� > 0x,f� �x� > 0

x?f� �x� > 0x,f �x� > 0

y � 2y� � y � 2ex.y � f �x�x,

The fall of a parachutist is described by the second-order lineardifferential equation

where is the weight of the parachutist, is the height at time is the acceleration due to gravity, and is the drag factor of the parachute.

(a) The parachute is opened at 2000 feet, so

At that time, the velocity is

feet per second.

For a 160-pound parachutist, using the differentialequation is

Using the initial conditions, verify that the solution of the differential equation is

(b) Consider a 192-pound parachutist who has a parachute with adrag factor of Using the initial conditions given in part(a), write and solve a differential equation that describes thefall of the parachutist.

k � 9.

y � 1950 � 50e�1.6t � 20t.

�5y � 8y� � 160.

k � 8,

y��0� � �100

y�0� � 2000.

k

gt,y

w

wg

d 2ydt2 � k

dydt

� w

Parachute Jump

Danshutter/Shutterstock.com

9781285060293_1603.qxp 10/15/12 5:11 PM Page 1148

16.4 Series Solutions of Differential Equations 1149

16.4 Series Solutions of Differential Equations

Use a power series to solve a differential equation.Use a Taylor series to find the series solution of a differential equation.

Power Series Solution of a Differential EquationPower series can be used to solve certain types of differential equations. This sectionbegins with the general power series solution method.

Recall from Chapter 9 that a power series represents a function 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 solution methoddemonstrated in the first two examples.

Power Series Solution

Use a power series to solve the differential equation

Solution Assume that is a solution. Then,

Substituting for and you obtain the following series form of the differentialequation. (Note that, from the third step to the fourth, the index of summation is changedto ensure that occurs in both sums.)

Now, by equating coefficients of like terms, you obtain the recursion formula

which implies that

This formula generates the following results.

. . .

. . .

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

� a0e2x.

� a0 �

n�0 �2x�n

n!

y � �

n�0 2na0

n! xn

25a0

5!24a0

4!23a0

3!22a0

22a0a0

a5a4a3a2a1a0

n ≥ 0.an�1 �2an

n � 1,

�n � 1�an�1 � 2an

�

n�0 �n � 1�an�1x

n � �

n�0 2anxn

�

n�1 nanxn�1 � �

n�0 2anxn

�

n�1 nanxn�1 � 2 �

n�0 anxn � 0

y� � 2y � 0

xn

�2y,y�

y� � �

n�1 nanx

n�1.

y � �

n�0anx

n

y� � 2y � 0.

f�, f ,

f

Exploration

In Example 1, the differentialequation could be solvedeasily without using a series.Determine which methodshould be used to solve thedifferential equation

and show that the result isthe same as that obtained inthe example.

y� � 2y � 0

9781285060293_1604.qxp 10/15/12 5:12 PM Page 1149

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.

Power Series Solution

Use a power series to solve the differential equation

Solution Assume that is a solution. Then you have

Substituting for and in the given differential equation, you obtain the followingseries.

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

By equating coefficients, you have

from which you obtain the recursion formula

and the coefficients of the solution series are as follows.

So, you can represent the general solution as the sum of two series—one for the even-powered terms with coefficients in terms of and one for the odd-powered termswith coefficients in terms of

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

a1,a0

� a0 �

k�0 ��1�kx2k

2k�k!� � a1 �

k�0

��1�kx2k�1

3 � 5 � 7 . . . �2k � 1�

y � a01 �x2

2�

x4

2 � 4� . . .� � a1x �

x3

3�

x5

3 � 5� . . .�

a1.a0,

a2k�1 ��1�k a1

3 � 5 � 7 . . . �2k � 1� a2k ��1�k a0

2 � 4 � 6 . . . �2k� ��1�ka0

2k�k!�

��

a7 � �a5

7� �

a1

3 � 5 � 7 a6 � �

a4

6� �

a0

2 � 4 � 6

a5 � �a3

5�

a1

3 � 5 a4 � �

a2

4�

a0

2 � 4

a3 � �a1

3 a2 � �

a0

2

n 0,an�2 � ��n � 1�

�n � 2��n � 1� an � �an

n � 2,

�n � 2��n � 1�an�2 � ��n � 1�an

�

n�0 �n � 2��n � 1�an�2xn � � �

n�0 �n � 1�anxn.

n � 2nx,

�

n�2 n�n � 1�anxn�2 � � �

n�0 �n � 1�anxn

�

n�2n�n � 1�anxn�2 � �

n�0 nanxn � �

n�0 anxn � 0

yy, xy�,

y � �

n�2 n�n � 1�anxn�2.xy� � �

n�1 nanxn,y� � �

n�1 nanxn�1,

y � �

n�0anx

n

y � xy� � y � 0.


9781285060293_1604.qxp 10/15/12 5:12 PM Page 1150

x 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

y 1.0000 1.1057 1.2264 1.3691 1.5432 1.7620 2.0424 2.4062 2.8805 3.4985 4.3000

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 9.10.

Approximation by Taylor Series


Use a Taylor series to find the first six terms of the series solution of

for the initial condition when Then, use this polynomial to approximatevalues of for

Solution Recall from Section 9.10 that, for

Because and you obtain the following.

So, can be approximated by the first six terms of the series solution shown below.

Using this polynomial, you can approximate values for in the interval asshown in the table below.

In addition to approximating values of a function, you can also use a series solution tosketch a graph. In Figure 16.8, the series solutions of using the first two,four, and six terms are shown, along with anapproximation found using a computer algebrasystem. The approximations are nearly the same for values of close to 0. As approaches 1, however, there is a noticeable difference among the approximations. For a series solution that is more accurate near repeat Example 3 using c � 1.

x � 1,

xx

y� � y2 � x

0 � x � 1,y

� 1 � x �12

x2 �43!

x3 �144!

x4 �665!

x5

y � y�0� � y��0�x �y �0�

2! x2 �

y��0�3!

x3 �y�4��0�

4! x4 �

y�5��0�5!

x5

y

y�5��0� � 28 � 32 � 6 � 66 y�5� � 2yy�4� � 8y�y�� 6�y �2

y�4��0� � 8 � 6 � 14 y�4� � 2yy�� 6y�y

y��0� � 2 � 2 � 4 y�� 2yy � 2�y��2

y �0� � 2 � 1 � 1 y � 2yy� � 1

y��0� � 1 y� � y2 � x

y�0� � 1

y� � y2 � x,y�0� � 1

y � y�0� � y��0�x �y �0�

2! x2 �

y��0�3!

x3 � . . . .

c � 0,

0 � x � 1.yx � 0.y � 1

y� � y2 � x

16.4 Series Solutions of Differential Equations 1151

x0.2

6 terms

4 terms

2 terms

1.00.80.60.4

8

6

4

2

y

Figure 16.8

9781285060293_1604.qxp 10/15/12 5:12 PM Page 1151



Equivalent Solution Techniques In Exercises 1–6, verifythat the power series solution of the differential equation isequivalent to the solution found using previously learned solutiontechniques.

1. 2.

3. 4.

5. 6.

Power Series Solution In Exercises 7–10, use powerseries to solve the differential equation and find the interval ofconvergence of the series.

7. 8.

9. 10.

Finding Terms of a Power Series Solution In Exercises 11and 12, find the first three terms of each of the power seriesrepresenting independent solutions of the differential equation.

11. 12.

Approximation by Taylor Series In Exercises 13 and 14,use a Taylor series to find the first terms of the series solutionof the differential equation under the specified initial conditions.Use this polynomial to approximate for the given value of and compare the result with the approximation given byEuler’s Method for

13.

14.

17. Investigation Consider the differential equation

with the initial conditions

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.

Approximation by Taylor Series In Exercises 19–22, usea Taylor series to find the first terms of the series solution ofthe differential equation under the specified initial conditions.Use this polynomial to approximate for the given value of

19.

20.

21.

22.

Verifying that a Series Converges In Exercises 23–26,verify that the series converges to the given function on theindicated interval. (Hint: Use the given differential equation.)

23.

Differential equation:

24.


25.


26.


27. Airy’s Equation Find the first six terms in the series solution of Airy’s equation, y � xy � 0.

�1 � x2�y � xy� � 0

�

n�0

�2n�!x2n�1

�2nn!�2�2n � 1� � arcsin x, ��1, 1�

�x2 � 1�y � 2xy� � 0

�

n�0 ��1�nx2n�1

2n � 1� arctan x, ��1, 1�

y � y � 0

�

n�0 ��1�nx2n

�2n�! � cos x, �� , �

y� � y � 0

�

n�0 xn

n!� ex, �� , �

x �15n � 4,y��0� � 1,y�0� � �2,y � exy� � �sin x�y � 0,

x �13n � 4,y��0� � 2,y�0� � 3,y � x2y� � �cos x�y � 0,

y � 2xy� � y � 0, y�0� � 1, y��0� � 2, n � 8, x �12

y � 2xy � 0, y�0� � 1, y��0� � �3, n � 6, x �14

x.y

n

y��0� � 2.y�0� � 0

y � xy� � 0

y� � 2xy � 0, y�0� � 1, n � 4, x � 1

y� � �2x � 1�y � 0, y�0� � 2, n � 5, x �12

h � 0.1.

xy

n

y � x2y � 0�x2 � 4�y � y � 0

y � xy� � y � 0y � xy� � 0

y� � 2xy � 0y� � 3xy � 0

y � k2y � 0y � 4y � 0

y � k2y � 0y � 9y � 0

y� � ky � 0y� � y � 0

WRITING ABOUT CONCEPTS15. Power Series Solution Method Describe how to

use power series to solve a differential equation.

16. Recursion Formula What is a recursion formula?Give an example.

18. HOW DO YOU SEE IT? Consider the differentialequation

with initial conditions and The figure shows the graph of the solution of the differential equation and the third-degree and fifth-degree polynomial approximations of thesolution. Identify each.

x

y

1 2

3

2

1

y��0� � 6.y�0� � 2

y � 9y � 0

9781285060293_1604.qxp 10/15/12 5:12 PM Page 1152

Review Exercises 1153

Testing for Exactness In Exercises 1 and 2, determinewhether the differential equation is exact. Explain your reasoning.

1.

2.

Solving an Exact Differential Equation In Exercises3–8, determine whether the differential equation is exact. If itis, find the general solution.

3.

4.

5.

6.

7.

8.

Graphical and Analytic Analysis In Exercises 9 and 10,(a) sketch an approximate solution of the differential equationsatisfying the initial condition on the slope field, (b) find theparticular solution that satisfies the initial condition, and (c) usea graphing utility to graph the particular solution. Comparethe graph with the sketch in part (a).

9.

10.

Finding a Particular Solution In Exercises 11 and 12, findthe particular solution that satisfies the initial condition.

11.

12.

Finding an Integrating Factor In Exercises 13–16, findthe integrating factor that is a function of or alone and useit to find the general solution of the differential equation.

13.

14.

15.

16.

Verifying a Solution In Exercises 17 and 18, verify thesolution of the differential equation. Then use a graphing utilityto graph the particular solutions for several different values of

and What do you observe?


17.

18.

Finding a Particular Solution: Initial Conditions InExercises 19–22, find the particular solution of the differentialequation that satisfies the initial conditions. Use a graphing utility to graph the solution.

Differential Equation Initial Conditions

19.

20.

21.

22.

Finding a Particular Solution: Boundary Conditions

In Exercises 23 and 24, find the particular solution of the differential equation that satisfies the boundary conditions.Use a graphing utility to graph the solution.

Differential Equation Boundary Conditions

23.

24.

25. Think About It Is the differential equation

homogeneous? Why or why not?

26. Solutions of a Differential Equation Find all valuesof for which the differential equation

has a general solution of the indicated form.

(a)

(b)

(c) y � C1e�x cos �x � C2e

�x�x

y � C1em1x � C2xem1x

y � C1em3x � C2e

m2x

y � 2ky� � ky � 0

k

y � y� � 5y � sin x

y�0� � 2, y��2� � 1y � y � 0

y�1� � 4, y�2� � 0y � 2y� � 5y � 0

y�0� � 2, y��0� � 1y � 12y� � 36y � 0

y�0� � 2, y��0� � 0y � 2y� � 3y � 0

y�0� � 2, y��0� � �7y � 4y� � 5y � 0

y�0� � 0, y��0� � 3y � y� � 2y � 0

y � 4y � 0y � C1 cos 2x � C2 sin 2x

y � 4y � 0y � C1e2x � C2e

�2x

C2.C1

cos y dx � �2�x � y� sin y � cos y� dy � 0

dx � �3x � e�2y� dy � 0

2xy dx � �y2 � x2� dy � 0

�3x2 � y2� dx � 2xy dy � 0

yx

y�1� � 23x2y2 dx � �2x3y � 3y2� dy � 0,

y�2� � 0�2x � y � 3� dx � �x � 3y � 1� dy � 0,

y

x42−2−4

4

2

−2

−4

y�0� � 1�6xy � y3� dx � �4y � 3x2 � 3xy2� dy � 0,

y

x42−2−4

4

2

−2

−4

y�2� � 2�2x � y� dx � �2y � x� dy � 0,

y sin�xy� dx � �x sin�xy� � y� dy � 0

xy dx �

xy2 dy � 0

�3x2 � 5xy2� dx � �2y3 � 5xy2� dy � 0

�x � y � 5� dx � �x � 3y � 2� dy � 0

�2x � 2y3 � y� dx � �x � 6xy2� dy � 0

�10x � 8y � 2� dx � �8x � 5y � 2� dy � 0

�5x � y� dx � �5y � x� dy � 0

�y � x3 � xy2� dx � x dy � 0

Review Exercises See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.

9781285060293_160R.qxp 10/15/12 5:13 PM Page 1153


Finding a General Solution In Exercises 27–32, find thegeneral solution of the second-order differential equation.

27.

28.

29.

30.

31.

32.

Finding a Particular Solution In Exercises 33–38, findthe particular solution of the differential equation that satisfiesthe initial conditions.

Differential Equation Initial Conditions

33.

34.

35.

36.

37.

38.

Vibrating Spring In Exercises 39 and 40, describe themotion of a 64-pound weight suspended on a spring. Assume

that the weight stretches the spring feet from its natural position.

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

40. 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.

41. Investigation The differential equation

models the motion of a weight suspended on a spring.

(a) Solve the differential equation and use a graphing utility tograph the solution for each of the assigned quantities for

and

(i)

(ii)

(iii)

(iv)

(b) Describe the effect of increasing the resistance to motion

(c) Explain how the motion of the object changes when astiffer spring greater value of is used.

(d) Matching the input and natural frequencies of a system isknown as resonance. In which case of part (a) does thisoccur, and what is the result?

42. True or False? The function

is a particular solution of the differential equation

43. Think About It

(a) Explain how, by observation, you know that a form of aparticular solution of the differential equation

is

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

(c) Compare the algebra required to find particular solutionsin parts (a) and (b) with that required when the form of the particular solution is

44. Think About It Explain how you can find a particular solution of the differential equation

by observation.

Power Series Solution In Exercises 45 and 46, find theseries solution of the differential equation.

45.

46.

Approximation by Taylor Series In Exercises 47 and 48,use a Taylor series to find the first terms of the series solutionof the differential equation under the specified initial conditions.Use this polynomial to approximate for the given value of

47.

48. x �12

n � 6,y��0� � 1,y�0� � 1,y � xy � 0,

x �14

n � 4,y��0� � 0,y�0� � 2,y � y� � exy � 0,

x.y

n

y � 3xy� � 3y � 0

�x � 4�y� � y � 0

y � 4y� � 6y � 30

yp � A cos x � B sin x.

y � 5y � 10 cos x.

yp � A sin x.

y � 3y � 12 sin x

y � 4y� � 5y � sin x � cos x.

yp �14

cos x

k��

b.

b � 1, k � 2, F�t� � 0

b � 0.1, k � 2, F�t� � 0

b � 0, k � 2, F�t� � 24 sin�2�2 t�b � 0, k � 1, F�t� � 24 sin �t

F�t�.k,b,

y�0� �12

, y��0� � 08

32y � by� � ky �

832

F�t�,

18

12

12

43

y�0� � 1, y��0� � 1, y �0� � 1y�� y � 4x2

y�0� � 1, y��0� � 3y � y� � 2y � 1 � xe�x

y�0� � 2, y��0� �103y � 3y� � 6x

y�0� � 6, y��0� � �6y � 4y � cos x

y�0� � 0, y��0� � 0y � 25y � ex

y�0� � 2, y��0� � 0y � y� � 6y � 54

y � 2y� � y �1

x2ex

y � 2y� � y � 2xex

y � 5y� � 4y � x2 � sin 2x

y � y � 2 cos x

y � 2y � e2x � x

y � y � x3 � x

9781285060293_160R.qxp 10/15/12 5:13 PM Page 1154

P.S. Problem Solving 1155

1. Finding a General Solution Find the value of thatmakes the differential equation

exact. Using this value of find the general solution.

2. Using an Integrating Factor The differential equationis not exact, but the integrating factor

makes it exact.

(a) Use this information to find the value of

(b) Using this value of find the general solution.

3. Finding a General Solution Find the general solution ofthe differential equation Show that thegeneral solution can be written in the form

4. Finding a General Solution Find the general solution ofthe differential equation Show that the generalsolution can be written in the form

5. Distinct Real Zeros Given that the characteristic equationof the differential equation has two distinctreal zeros, and where and are real,show that the general solution of the differential equation can bewritten in the form

6. Limit of a Solution Given that and are positive andthat is a solution of the differential equation

show that

7. Trivial and Nontrivial Solutions Consider the differentialequation with boundary conditions and

for some nonzero real number

(a) For show that the differential equation has only thetrivial solution

(b) For show that the differential equation has only the trivial solution

(c) For find the value(s) of for which the solution isnontrivial. Then find the corresponding solution(s).

8. Euler’s Differential Equation Euler’s differentialequation is of the form

when and are constants.

(a) Show that this equation can be transformed into a second-order linear equation with constant coefficients byusing the substitution

(b) Solve

9. Pendulum Consider a pendulum of length that swingsby the force of gravity only.

For small values of the motion of the pendulum canbe approximated by the differential equation

where is the acceleration due to gravity.

(a) Find the general solution of the differential equation andshow that it can be written in the form

(b) Find the particular solution for a pendulum of length 0.25meter when the initial conditions are radianand radian per second. (Use metersper second per second.)

(c) Determine the period of the pendulum.

(d) Determine the maximum value of

(e) How much time from does it take for to be 0 thefirst time? the second time?

(f) What is the angular velocity when the first time?the second time?

10. Deflection of a Beam A horizontal beam with a lengthof 2 meters rests on supports located at the ends of the beam.

The beam is supporting a load of kilograms per meter. Theresulting deflection of the beam at a horizontal distance of meters from the left end can be modeled by

where is a positive constant.

(a) Solve the differential equation to find the deflection as afunction of the horizontal distance

(b) Use a graphing utility to determine the location and valueof the maximum deflection.

x.y

A

A d 2ydx2 � 2Wx �

12

Wx2

xyW

2 meters

� � 0��

�t � 0

�.

g � 9.8��0� � 0.5��0� � 0.1

��t� � A cos� �gL

�t � ��.

g

d 2�

dt 2 �gL

� � 0

� � ��t�,

Lθ

L

x2y � 6xy� � 6y � 0.

x � et.

ba

x > 0x2 y � axy� � by � 0,

aa > 0,

y � 0.a < 0,

y � 0.a � 0,

L.y�L� � 0y�0� � 0y � ay � 0

limx→

y�x� � 0.

y � ay� � by � 0

y�x�ba

y � erx�C1 cosh sx � C2 sinh sx�.

srm2 � r � s,m1 � r � sy � ay� � by � 0

y � C sin��x � ��, 0 � � < 2�.

y � �2y � 0.

y � C1 cosh ax � C2 sinh ax.

y � a2y � 0, a > 0.

k,

k.

1�x2�kx2 � y2� dx � kxy dy � 0

k,

�3x2 � kxy2� dx � �5x2y � ky2� dy � 0

k

P.S. Problem Solving See CalcChat.com for tutorial help andworked-out solutions to odd-numbered exercises.

9781285060293_160R.qxp 10/15/12 5:13 PM Page 1155


Damped Motion In Exercises 11–14, consider a dampedmass-spring system whose motion is described by the differentialequation

The zeros of its characteristic equation are

and

For , the system is overdamped; for , itis critically damped; and for , it is underdamped.

(a) Determine whether the differential equation represents anoverdamped, critically damped, or underdamped system.

(b) Find the particular solution corresponding to the giveninitial conditions.

(c) Use a graphing utility to graph the particular solutionfound in part (b). Explain how the graph illustrates thetype of damping in the system.

11. 12.

13. 14.

15. Airy’s Equation Consider Airy’s equation given inSection 16.4, Exercise 27. Rewrite the equation as

Then use a power series of the form

to find the first eight terms of the solution. Compare yourresult with that of Exercise 27 in Section 16.4.

16. Chebyshev’s Equation Consider Chebyshev’s equation

Polynomial solutions of this differential equation are calledChebyshev polynomials and are denoted by They satisfythe recursion equation

(a) Given that and determine theChebyshev polynomials and

(b) Verify that and are solutionsof the given differential equation.

(c) Show that

and

17. Bessel’s Equation: Order Zero The differential equation is known as Bessel’s equationof order zero.

(a) Use a power series of the form

to find the solution.

(b) Compare your result with that of the function givenin Section 9.8, Exercise 65.

18. Bessel’s Equation: Order One The differential equation

is known as Bessel’s equation of order one.


to find the solution.

(b) Compare your result with that of the function givenin Section 9.8, Exercise 66.

19. Hermite’s Equation Consider Hermite’s equation


to find the solution when [Hint: Choose the arbitrary constants such that the leading term is ]

(b) Polynomial solutions of Hermite’s equation are calledHermite polynomials and are denoted by The generalform for can be written as

where is the greatest integer less than or equal to Use this formula to determine the Hermite polynomials

and

20. Laguerre’s Equation Consider Laguerre’s equation

(a) Polynomial solutions of Laguerre’s equation are calledLaguerre polynomials and are denoted by Use apower series of the form

to show that

Assume that

(b) Determine the Laguerre polynomials and L4�x�.L3�x�,

L2�x�,L1�x�,L0�x�,a0 � 1.

Lk�x� � �k

n�0

��1�nk!xn

�k � n�!�n!�2.

y � �

n�0 an x

n

Lk�x�.

xy � �1 � x�y� � ky � 0.

H4�x�.H3�x�,H2�x�,H0�x�, H1�x�,

k�2.P

Hk�x� � �P

n�0

��1�nk!�2x�k�2n

n!�k � 2n�!

Hk�x�Hk(x).

�2x�k.k � 4.

y � �

n�0 an x

n

y � 2xy� � 2ky � 0.

J1�x�

y � �

n�0an x n

x2y � xy� � �x2 � 1�y � 0

J0�x�

y � �

n�0an xn

x2y � xy� � x2y � 0

T7�x� � 64x7 � 112x5 � 56x3 � 7x.

T6�x� � 32x6 � 48x4 � 18x2 � 1,

T5�x� � 16x5 � 20x3 � 5x,

T4�x�T0�x�, T1�x�, T2�x�, T3�x�,T4�x�.T2�x�, T3�x�,

T1�x� � x,T0�x� � 1

Tn�1�x� � 2xTn�x� � Tn�1�x).

Tk�x�.

�1 � x2�y � xy� � k2y � 0.

y � �

n�0an�x � 1�n

y � �x � 1�y � y � 0.

y�0� � 2, y� �0� � �1y�0� � 2, y� �0� � �20

d 2ydt 2 � 2

dydt

� y � 0d 2ydt 2 � 20

dydt

� 64y � 0

y�0� � 1, y� �0� � 4y�0� � 1, y� �0� � 1

d 2ydt 2 � 2

dydt

� 26y � 0d 2ydt 2 � 8

dydt

� 16y � 0

�2 � �2 < 0�2 � �2 � 0�2 � �2 > 0

m2 � �� 2 � �2.

m1 � �� 2 � �2

d 2ydt 2 � 2�

dydt

� �2y � 0.

9781285060293_160R.qxp 10/15/12 5:13 PM Page 1156

Documents

16 Additional Topics in Differential Equations - Cengage · Use an integrating factor to make a differential equation exact. Exact Differential Equations ... Solution The differential