diff eqn Differential equations

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A20 APPENDIX C Differential Equations

General Solution of a Differential Equation

A differential equation is an equation involving a differentiable function and one

or more of its derivatives. For instance,

Differential equation

is a differential equation. A function is a solution of a differential equa-

tion if the equation is satisfied when y and its derivatives are replaced by and

its derivatives. For instance,

Solution of differential equation

is a solution of the differential equation shown above. To see this, substitute for y

and in the original equation.

Substitute for y and

In the same way, you can show that and are

also solutions of the differential equation. In fact, each function given by

General solution

where C is a real number, is a solution of the equation. This family of solutions

is called the general solution of the differential equation.

EXAMPLE 1 Checking Solutions

Show that (a) and (b) are solutions of the differential equation

Solution

(a) Because and it follows that

So, is a solution.

(b) Because and it follows that

So, is also a solution. y Ce x

y y Ce x Ce x

0.

y Ce x, y Ce x

y Ce x

y y Ce x Ce x

0.

y Ce x, y Ce x

y y 0.

y Ce x y Ce x

y Ce2 x

y 12e2 x y 2e2 x, y 3e2 x,

y . y 2 y 2e2 x 2e2 x 0

y 2e2 x

y e2 x

f x y f x

y 2 y 0

C Differential Equations

C.1 Solutions of Differential Equations

Find general solutions of differential equations. • Find particular solutions of differential equations.



Particular Solutions and Initial Conditions

A particular solution of a differential equation is any solution that is obtained

by assigning specific values to the constants in the general equation.*

Geometrically, the general solution of a differential equation is a family of

graphs called solution curves. For instance, the general solution of the differen-tial equation is

General solution

Figure A.7 shows several solution curves of this differential equation.

Particular solutions of a differential equation are obtained from initial

conditions placed on the unknown function and its derivatives. For instance,

in Figure A.7, suppose you want to find the particular solution whose graph

passes through the point This initial condition can be written as

when Initial condition

Substituting these values into the general solution produces which

implies that So, the particular solution isParticular solution

EXAMPLE 2 Finding a Particular Solution

Verify that

General solution

is a solution of the differential equation for any value of C . Then

find the particular solution determined by the initial condition

when Initial condition

Solution The derivative of is Substituting into the differen-tial equation produces

Thus, is a solution for any value of C . To find the particular solution,

substitute and into the general solution to obtain

or

This implies that the particular solution is

Particular solution

*Some differential equations have solutions other than those given by their general solutions. These

are called singular solutions. In this brief discussion of differential equations, singular solutions will

not be discussed.

y 2

27

x3.

C 2

27.2 C 33

y 2 x 3

y Cx3

0.

xy 3 y x3Cx2 3Cx3

y 3Cx

2

. y Cx3

x 3. y 2

xy 3 y 0

y Cx3

y 3 x2.C

3.

3 C 12,

x 1. y 3

1, 3.

y Cx2.

xy 2 y 0

APPENDIX C Differential Equations A21

3

2

3

2

1

3 2 32

y

x

2 xC y) ) ,1 3

FIGURE A.7



EXAMPLE 3 Finding a Particular Solution

You are working in the marketing department of a company that is producing a

new cereal product to be sold nationally. You determine that a maximum of 10

million units of the product could be sold in a year. You hypothesize that the rateof growth of the sales x (in millions of units) is proportional to the difference

between the maximum sales and the current sales. As a differential equation, this

hypothesis can be written as

The general solution of this differential equation is

General solution

where t is the time in years. After 1 year, 250,000 units have been sold. Sketch

the graph of the sales function over a 10-year period.

Solution Because the product is new, you can assume that when So,

you have two initial conditions.

when First initial condition

when Second initial condition

Substituting the first initial condition into the general solution produces

which implies that Substituting the second initial condition into the

general solution produces

which implies that So, the particular solution is

Particular solution

The table shows the annual sales during the first 10 years, and the graph of the

solution is shown in Figure A.8.

In the first three examples in this section, each solution was given in

explicit form, such as Sometimes you will encounter solutions for

which it is more convenient to write the solution in implicit form, as illustrated

in Example 4.

y f x.

x 10 10e0.0253t .

k ln4039 0.0253.

0.25 10 10ek (1)

C 10.

0 10 Cek (0)

t 1 x 0.25

t 0 x 0

t 0. x 0

x

10

Cekt

0 ≤ x ≤ 10.dx

dt k 10 x,


Sales Projection

Sales(inmillionsofunits)

Time (in years)

3

2

1

x

t

1 2 3 4 5 6 7 8 9 10

x e= 10 10− −0.0253t

FIGURE A.8

t 1 2 3 4 5 6 7 8 9 10

x 0.25 0.49 0.73 0.96 1.19 1.41 1.62 1.83 2.04 2.24

Rate of change

of x

is propor-tional to

the differencebetween10 and x.



EXAMPLE 4 Sketching Graphs of Solutions

Verify that

General solution

is a solution of the differential equation Then sketch the particular

solutions represented by and

Solution To verify the given solution, differentiate each side with respect to x.

Given general solution

Differentiate with respect to x.

Divide each side by 2.

Because the third equation is the given differential equation, you can conclude

that

is a solution. The particular solutions represented by andare shown in Figure A.9.C ±4

C ±1,C 0,

2 y2 x2

C

2 yy x 0

4 yy 2 x 0

2 y2 x2

C

C ±4.C 0, C ±1,

2 yy x 0.

2 y2 x2

C


22

2

3

2

2

1

2

3

1

y

x

x

y

x

y

x

x

C 0

1C

1C

C 4

C 4

y y

TAKE ANOTHER LOOK

Writing a Differential Equation

Write a differential equation that has the family of circles

as a general solution.

x 2 y 2 C

FIGURE A.9 Graphs of Five Particular Solutions



The following warm-up exercises involve skills that were covered in earlier sections.

You will use these skills in the exercise set for this section.

In Exercises 1–4, find the first and second derivatives of the function.

1.

2.

3.

4.

In Exercises 5–8, use implicit differentiation to find

5.

6.

7.

8.

In Exercises 9 and 10, solve for k .

9.

10. 14.75 25 25e2k

0.5 9 9ek

3 xy x2 y 2 10

xy2 3

2 x y3 4 y

x2 y 2

2 x

dy dx .

y 3e x2

y 3e2 x

y 2 x3 8 x 4

y 3 x2 2 x 1


EXER CI SES C.1

In Exercises 1–10, verify that the function is a solution of the

differential equation.

Solution Differential Equation

1.

2.3.

4.

5.

6.

7.

8.

9.

10.

In Exercises 11–28, verify that the function is a solution of the

differential equation for any value of C .


11.

12.

13.

14.

15.

16.

17.

18.

19.

20. y y 0 y C 1 C 2e x

xy y x3 x 4 y x2 2 x

C

x

y y x 2 x y x ln x2 2 x32 Cx

xy 3 x 2 y 0 y Cx2 3 x

y y 10 0 y Cet 10

3dy

dt y 7 0 y Cet 3

7

dy

dx 4 y y Ce4 x

dy

dx 4 y y Ce4 x

dy

dx

x

4 x2 y 4 x2

C

dy

dx

1

x2 y

1

x C

y 3 x2 y 6 xy 0 y e x3

y y 2 y 0 y 2e2 x

xy 2 y 0 y 1

x

x2 y 2 y 0 y x2

y 2

x y 0 y 4 x2

y 3

x y 0 y 2 x3

y 2 xy 0 y 3e x2

y 2 y 0 y e2 x

y 6 x

2

1 y 2 x3

x 1

y 3 x2 y x3 5

WA R M -UP C.1




21.

22.

23.

24.

25.

26.

27.

28.

In Exercises 29–32, use implicit differentiation to verify that the

equation is a solution of the differential equation for any value

of C .


29.

30.

31.

32.

In Exercises 33–36, determine whether the function is a solution

of the differential equation

33.

34.

35.

36.

In Exercises 37–40, determine whether the function is a solution

of the differential equation

37.

38.

39.

40.

In Exercises 41–48, verify that the general solution satisfies the

differential equation.Then find the particular solution that satis-

fies the initial condition.

41. General solution:

Differential equation:

Initial condition: when






Initial condition: and when


Differential equation:Initial condition: and when



Initial condition: and when







when


Differential equation:Initial condition: and when

In Exercises 49–52, the general solution of the differential

equation is given. Use a graphing utility to graph the particular

solutions that correspond to the indicated values of C .

General Solution Differential Equation C-values

49. 1, 2, 4

50. 0,

51.

52.

In Exercises 53–60, use integration to find the general solution of

the differential equation.

53.

54.

55.

56.

57.

58.

59.

60. dy

dx xe x

dy

dx x x 3

dydx

x1 x2

dy

dx

1

x2 1

dy

dx

x 2

x

dy

dx

x 3

x

dy

dx

1

1 x

dy

dx 3 x2

0, ±1, ±2 y y 0 y Ce x

0, ±1, ±2 x 2 y 2 y 0 y C x 22

±1, ±44 yy x 04 y 2 x2

C

xy 2 y 0 y Cx2

x 0 y 1 y 2 y 4 y 4 y x2e2 x

y C 1 C 2 x 112 x

4e2 x

x 3 y 0

x 0 y 4

9 y 12 y 4 y 0

y e2 x3C 1 C 2 x

x 1 y 2

y 2 x 1 y 0

y Ce x x 2

x 0 y 6 y 5

y y 12 y 0

y C 1e4 x C 2e

3 x

x 2 y 4 y 0 x2

y

3 xy

3 y

0

y C 1 x C 2 x3

x 1 y 0.5 y 5

xy y 0

x > 0 y C 1 C 2 ln x,

x 1 y 2

2 x 3 yy 0

2 x2 3 y 2

C

x 0 y 3

y 2 y 0

y Ce2 x

y x ln x

y xe x

y 4e x

29 xe2 x

y 29 xe2 x

y 3 y 2 y 0.

y 4e2 x

y 4 x

y 5 ln x

y e2 x

y 4 16 y 0.

y3 y x2 y 2

0 x2 y 2

C

x2 y 2 x y 0 x2 xy C

x y y x y 0 y 2 2 xy x2

C

y 2 xy

x2 y 2

x2 y 2

Cy

x y xy 0 y xln x C

x y 1 y 4 0 y x ln x Cx 4

y 2 x 1 y 0 y Ce x x2

y 2 xy xy2 y 2

1 Ce x2

2 xy y x3 x y x3

5 x C x

y

ay

x bx3

y

bx4

4 a Cxa

y 3 y 4 y 0 y C 1e4 x C 2e

x

2 y 3 y 2 y 0 y C 1e x2

C 2e2 x





In Exercises 61–64,you are shown the graphs of some of the solu-

tions of the differential equation. Find the particular solution

whose graph passes through the indicated point.

61. 62.

63. 64.

65. Biology The limiting capacity of the habitat of a wildlife

herd is 750. The growth rate of the herd is propor-

tional to the unutilized opportunity for growth, as described

by the differential equation


When the population of the herd is 100. After 2

years, the population has grown to 160.

(a) Write the population function N as a function of t .

(b) Use a graphing utility to graph the population function.

(c) What is the population of the herd after 4 years?

66. Investment The rate of growth of an investment is

proportional to the amount in the investment at any time t .

That is,

The initial investment is $1000, and after 10 years the bal-

ance is $3320.12. The general solution is

What is the particular solution?

67. Marketing You are working in the marketing depart-

ment of a computer software company. Your marketing

team determines that a maximum of 30,000 units of a new

product can be sold in a year. You hypothesize that the rate

of growth of the sales x is proportional to the difference

between the maximum sales and the current sales. That is,


where t is the time in years. During the first year, 2000

units are sold. Complete the table showing the numbers of

units sold in subsequent years.

68. Marketing In Exercise 67, suppose that the maximum

annual sales are 50,000 units. How does this change the

sales shown in the table?

69. Safety Assume that the rate of change in the number of

miles s of road cleared per hour by a snowplow is inverse-

ly proportional to the depth h of the snow. This rate of

change is described by the differential equation

Show that

is a solution of this differential equation.

70. Show that is a solution of the differen-

tial equation

where k is a constant.

71. The function is a solution of the differential

equation

Is it possible to determine C or k from the informationgiven? If so, find its value.

True or False? In Exercises 72 and 73, determine whether

the statement is true or false. If it is false, explain why or give an

example that shows it is false.

72. A differential equation can have more than one solution.

73. If is a solution of a differential equation, then

is also a solution. y f x C

y f x

dy

dx 0.07 y.

y Cekx

y a b y a 1

k dy

dt

y a Cek 1bt

s 25 13

ln 3

lnh

2

ds

dh

k

h.

x 30,000 Cekt

dx

dt k 30,000 x.

A Cekt .

dA

dt kA.

t 0,

N 750 Cekt .

dN dt k 750 N .

dN dt

1

2

1

2

x

(2, 1)

y

1 2 3123

4

5

6

x

(0, 3)

y

2 xy y 0 y y 0

y 2 2Cx y Ce x

4

3

2

4

3

433 x

)4,3(

y

7

4

3

2

1

4

3

2

1

654 x

)4,4(

y

yy 2 x 02 xy

3 y 0

2 x2 y 2

C y 2 Cx3

Year, t 2 4 6 8 10

Units, x

diff eqn Differential equations

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