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9: Triple Integrals in Spherical Coordinates

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97817_17_ch17_p1186-1194.qk_97817_17_ch17_p1186-1194 11/9/10 10:29 AM Page 1187

SECTION 17.3

APPLICATIONS OF SECOND-ORDER DIFFERENTIAL EQUATIONS

1187

We can also transfer other ideas from one situation to the other. For instance, the steady

state solution discussed in Note 1 makes sense in the spring system. And the phenomenon

of resonance in the spring system can be usefully carried over to electric circuits as electrical resonance.

17.3

Exercises

10. As in Exercise 9, consider a spring with mass m, spring con-

1. A spring has natural length 0.75 m and a 5-kg mass. A force

stant k, and damping constant c ෇ 0, and let ␻ ෇ sk͞m .

If an external force F͑t͒ ෇ F0 cos ␻ t is applied (the applied

frequency equals the natural frequency), use the method of

undetermined coefficients to show that the motion of the

mass is given by

of 25 N is needed to keep the spring stretched to a length of

1 m. If the spring is stretched to a length of 1.1 m and then

released with velocity 0, find the position of the mass after

t seconds.

2. A spring with an 8-kg mass is kept stretched 0.4 m beyond

its natural length by a force of 32 N. The spring starts at its

equilibrium position and is given an initial velocity of 1 m͞s.

Find the position of the mass at any time t.

x͑t͒ ෇ c1 cos ␻ t ϩ c2 sin ␻ t ϩ

11. Show that if ␻ 0

␻, but ␻͞␻ 0 is a rational number, then the

motion described by Equation 6 is periodic.

3. A spring with a mass of 2 kg has damping constant 14, and

a force of 6 N is required to keep the spring stretched 0.5 m

beyond its natural length. The spring is stretched 1 m beyond

its natural length and then released with zero velocity. Find

the position of the mass at any time t.

12. Consider a spring subject to a frictional or damping force.

(a) In the critically damped case, the motion is given by

x ෇ c1 ert ϩ c2 tert. Show that the graph of x crosses the

t-axis whenever c1 and c2 have opposite signs.

(b) In the overdamped case, the motion is given by

x ෇ c1e r t ϩ c2 e r t, where r1 Ͼ r2. Determine a condition

on the relative magnitudes of c1 and c2 under which the

graph of x crosses the t-axis at a positive value of t.

4. A force of 13 N is needed to keep a spring with a 2-kg mass

;

stretched 0.25 m beyond its natural length. The damping constant of the spring is c ෇ 8.

(a) If the mass starts at the equilibrium position with a

velocity of 0.5 m͞s, find its position at time t.

(b) Graph the position function of the mass.

1

2

13. A series circuit consists of a resistor with R ෇ 20 ⍀, an

5. For the spring in Exercise 3, find the mass that would

inductor with L ෇ 1 H, a capacitor with C ෇ 0.002 F, and a

12-V battery. If the initial charge and current are both 0, find

the charge and current at time t.

produce critical damping.

6. For the spring in Exercise 4, find the damping constant that

would produce critical damping.

14. A series circuit contains a resistor with R ෇ 24 ⍀, an induc-

; 7. A spring has a mass of 1 kg and its spring constant is

k ෇ 100. The spring is released at a point 0.1 m above its

equilibrium position. Graph the position function for the following values of the damping constant c: 10, 15, 20, 25, 30.

What type of damping occurs in each case?

F0

t sin ␻ t

2m␻

;

tor with L ෇ 2 H, a capacitor with C ෇ 0.005 F, and a 12-V

battery. The initial charge is Q ෇ 0.001 C and the initial current is 0.

(a) Find the charge and current at time t.

(b) Graph the charge and current functions.

; 8. A spring has a mass of 1 kg and its damping constant is

c ෇ 10. The spring starts from its equilibrium position with a

velocity of 1 m͞s. Graph the position function for the following values of the spring constant k: 10, 20, 25, 30, 40. What

type of damping occurs in each case?

15. The battery in Exercise 13 is replaced by a generator pro-

ducing a voltage of E͑t͒ ෇ 12 sin 10t. Find the charge at

time t.

16. The battery in Exercise 14 is replaced by a generator produc-

9. Suppose a spring has mass m and spring constant k and let

␻ ෇ sk͞m . Suppose that the damping constant is so small

that the damping force is negligible. If an external force

F͑t͒ ෇ F0 cos ␻ 0 t is applied, where ␻ 0 ␻, use the method

of undetermined coefficients to show that the motion of the

mass is described by Equation 6.

;

Graphing calculator or computer required

;

ing a voltage of E͑t͒ ෇ 12 sin 10t.

(a) Find the charge at time t.

(b) Graph the charge function.

17. Verify that the solution to Equation 1 can be written in the

form x͑t͒ ෇ A cos͑␻ t ϩ ␦͒.

1. Homework Hints available at stewartcalculus.com

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

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1188

CHAPTER 17

SECOND-ORDER DIFFERENTIAL EQUATIONS

18. The figure shows a pendulum with length L and the angle ␪

(b) What is the maximum angle from the vertical?

(c) What is the period of the pendulum (that is, the time to

complete one back-and-forth swing)?

(d) When will the pendulum first be vertical?

(e) What is the angular velocity when the pendulum is

vertical?

from the vertical to the pendulum. It can be shown that ␪, as a

function of time, satisfies the nonlinear differential equation

d 2␪

t

ϩ sin ␪ ෇ 0

dt 2

L

where t is the acceleration due to gravity. For small values of

␪ we can use the linear approximation sin ␪ Ϸ ␪ and then the

differential equation becomes linear.

(a) Find the equation of motion of a pendulum with length 1 m

if ␪ is initially 0.2 rad and the initial angular velocity is

17.4

ă

L

Series Solutions

Many differential equations can’t be solved explicitly in terms of finite combinations of

simple familiar functions. This is true even for a simple-looking equation like

yЉ Ϫ 2 xyЈ ϩ y ෇ 0

1

But it is important to be able to solve equations such as Equation 1 because they arise from

physical problems and, in particular, in connection with the Schrödinger equation in quantum mechanics. In such a case we use the method of power series; that is, we look for a solution of the form

y ෇ f ͑x͒ ෇

ϱ

͚cx

n

n

෇ c0 ϩ c1 x ϩ c2 x 2 ϩ c3 x 3 ϩ и и и

n෇0

The method is to substitute this expression into the differential equation and determine the

values of the coefficients c0 , c1, c2 , . . . . This technique resembles the method of undetermined coefficients discussed in Section 17.2.

Before using power series to solve Equation 1, we illustrate the method on the simpler

equation yЉ ϩ y ෇ 0 in Example 1. It’s true that we already know how to solve this equation by the techniques of Section 17.1, but it’s easier to understand the power series method

when it is applied to this simpler equation.

v

EXAMPLE 1 Use power series to solve the equation yЉ ϩ y ෇ 0.

SOLUTION We assume there is a solution of the form

2

y ෇ c0 ϩ c1 x ϩ c2 x 2 ϩ c3 x 3 ϩ и и и ෇

ϱ

͚cx

n

n

n෇0

We can differentiate power series term by term, so

yЈ ෇ c1 ϩ 2c2 x ϩ 3c3 x 2 ϩ и и и ෇

ϱ

͚ nc

n

x nϪ1

n෇1

3

yЉ ෇ 2c2 ϩ 2 и 3c3 x ϩ и и и ෇

ϱ

͚ n͑n Ϫ 1͒c x

n

nϪ2

n෇2

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

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SECTION 17.4

SERIES SOLUTIONS

1189

In order to compare the expressions for y and yЉ more easily, we rewrite yЉ as follows:

By writing out the first few terms of 4 , you can

see that it is the same as 3 . To obtain 4 , we

replaced n by n ϩ 2 and began the summation

yЉ ෇

4

ϱ

͚ ͑n ϩ 2͒͑n ϩ 1͒c

nϩ2

xn

n෇0

Substituting the expressions in Equations 2 and 4 into the differential equation, we

obtain

ϱ

͚ ͑n ϩ 2͒͑n ϩ 1͒c

xn ϩ

nϩ2

n෇0

ϱ

͚cx

n

n

෇0

n෇0

or

ϱ

5

͚ ͓͑n ϩ 2͒͑n ϩ 1͒c

nϩ2

ϩ cn ͔x n ෇ 0

n෇0

If two power series are equal, then the corresponding coefficients must be equal. Therefore the coefficients of x n in Equation 5 must be 0:

͑n ϩ 2͒͑n ϩ 1͒cnϩ2 ϩ cn ෇ 0

6

cnϩ2 ෇ Ϫ

cn

͑n ϩ 1͒͑n ϩ 2͒

n ෇ 0, 1, 2, 3, . . .

Equation 6 is called a recursion relation. If c0 and c1 are known, this equation allows

us to determine the remaining coefficients recursively by putting n ෇ 0, 1, 2, 3, . . . in

succession.

Put n ෇ 0:

c2 ෇ Ϫ

c0

1ؒ2

Put n ෇ 1:

c3 ෇ Ϫ

c1

2ؒ3

Put n ෇ 2:

c4 ෇ Ϫ

c2

c0

c0

3ؒ4

1ؒ2ؒ3ؒ4

4!

Put n ෇ 3:

c5 ෇ Ϫ

c3

c1

c1

4ؒ5

2ؒ3ؒ4ؒ5

5!

Put n ෇ 4:

c6 ෇ Ϫ

c4

c0

c0

෇Ϫ

෇Ϫ

5ؒ6

4! 5 ؒ 6

6!

Put n ෇ 5:

c7 ෇ Ϫ

c5

c1

c1

෇Ϫ

෇Ϫ

6ؒ7

5! 6 ؒ 7

7!

By now we see the pattern:

For the even coefficients, c2n ෇ ͑Ϫ1͒n

c0

͑2n͒!

For the odd coefficients, c2nϩ1 ෇ ͑Ϫ1͒n

c1

͑2n ϩ 1͒!

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

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1190

SECOND-ORDER DIFFERENTIAL EQUATIONS

CHAPTER 17

Putting these values back into Equation 2, we write the solution as

y ෇ c0 ϩ c1 x ϩ c2 x 2 ϩ c3 x 3 ϩ c4 x 4 ϩ c5 x 5 ϩ и и и

ͩ

෇ c0 1 Ϫ

෇ c0

ͪ

x2

x4

x6

x 2n

ϩ

Ϫ

ϩ и и и ϩ ͑Ϫ1͒n

ϩ иии

2!

4!

6!

͑2n͒!

ͩ

ϱ

͚ ͑Ϫ1͒

n

n෇0

ͪ

x3

x5

x7

x 2nϩ1

ϩ

Ϫ

ϩ и и и ϩ ͑Ϫ1͒n

ϩ иии

3!

5!

7!

͑2n ϩ 1͒!

ϩ c1 x Ϫ

ϱ

x 2n

x 2nϩ1

ϩ c1 ͚ ͑Ϫ1͒n

͑2n͒!

͑2n ϩ 1͒!

n෇0

Notice that there are two arbitrary constants, c0 and c1.

NOTE 1 We recognize the series obtained in Example 1 as being the Maclaurin series

for cos x and sin x. (See Equations 11.10.16 and 11.10.15.) Therefore we could write the

solution as

y͑x͒ ෇ c0 cos x ϩ c1 sin x

But we are not usually able to express power series solutions of differential equations in

terms of known functions.

v

EXAMPLE 2 Solve yЉ Ϫ 2 xyЈ ϩ y ෇ 0.

SOLUTION We assume there is a solution of the form

y෇

ϱ

͚c

n

xn

n෇0

yЈ ෇

Then

ϱ

͚ nc

n

x nϪ1

n෇1

yЉ ෇

and

ϱ

͚ n͑n Ϫ 1͒c x

n

nϪ2

n෇2

ϱ

͚ ͑n ϩ 2͒͑n ϩ 1͒c

nϩ2

xn

n෇0

as in Example 1. Substituting in the differential equation, we get

ϱ

͚ ͑n ϩ 2͒͑n ϩ 1͒c

nϩ2

x n Ϫ 2x

n෇0

nϩ2

xn Ϫ

n෇0

ϱ

͚ 2nc

n෇1

n

xn ෇

n

ϱ

n

n෇1

n

͚ ͓͑n ϩ 2͒͑n ϩ 1͒c

xn

n෇0

nϩ2

ϱ

͚c

n

xn ෇ 0

n෇0

͚ 2nc

ϱ

ϱ

͚ 2nc

x nϪ1 ϩ

n෇1

ϱ

͚ ͑n ϩ 2͒͑n ϩ 1͒c

ϱ

͚ nc

xn ϩ

ϱ

͚c

n

xn ෇ 0

n෇0

Ϫ ͑2n Ϫ 1͒cn ͔x n ෇ 0

n෇0

This equation is true if the coefficient of x n is 0:

͑n ϩ 2͒͑n ϩ 1͒cnϩ2 Ϫ ͑2n Ϫ 1͒cn ෇ 0

7

cnϩ2 ෇

2n Ϫ 1

cn

͑n ϩ 1͒͑n ϩ 2͒

n ෇ 0, 1, 2, 3, . . .

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

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SECTION 17.4

SERIES SOLUTIONS

1191

We solve this recursion relation by putting n ෇ 0, 1, 2, 3, . . . successively in Equation 7:

Put n ෇ 0:

c2 ෇

Ϫ1

c0

1ؒ2

Put n ෇ 1:

c3 ෇

1

c1

2ؒ3

Put n ෇ 2:

c4 ෇

3

3

3

c2 ෇ Ϫ

c 0 ෇ Ϫ c0

3ؒ4

1ؒ2ؒ3ؒ4

4!

Put n ෇ 3:

c5 ෇

5

1ؒ5

1ؒ5

c3 ෇

c1 ෇

c1

4ؒ5

2ؒ3ؒ4ؒ5

5!

Put n ෇ 4:

c6 ෇

7

3ؒ7

3ؒ7

c4 ෇ Ϫ

c0 ෇ Ϫ

c0

5ؒ6

4! 5 ؒ 6

6!

Put n ෇ 5:

c7 ෇

9

1ؒ5ؒ9

1ؒ5ؒ9

c5 ෇

c1 ෇

c1

6ؒ7

5! 6 ؒ 7

7!

Put n ෇ 6:

c8 ෇

11

3 ؒ 7 ؒ 11

c6 ෇ Ϫ

c0

7ؒ8

8!

Put n ෇ 7:

c9 ෇

13

1 ؒ 5 ؒ 9 ؒ 13

c7 ෇

c1

8ؒ9

9!

In general, the even coefficients are given by

c2n ෇ Ϫ

3 ؒ 7 ؒ 11 ؒ и и и ؒ ͑4n Ϫ 5͒

c0

͑2n͒!

and the odd coefficients are given by

c2nϩ1 ෇

1 ؒ 5 ؒ 9 ؒ и и и ؒ ͑4n Ϫ 3͒

c1

͑2n ϩ 1͒!

The solution is

y ෇ c0 ϩ c1 x ϩ c2 x 2 ϩ c3 x 3 ϩ c4 x 4 ϩ и и и

ͩ

෇ c0 1 Ϫ

ͩ

ϩ c1 x ϩ

or

8

ͪ

1 2

3 4

3ؒ7 6

3 ؒ 7 ؒ 11 8

x Ϫ

x Ϫ

x Ϫ

x Ϫ иии

2!

4!

6!

8!

ͩ

y ෇ c0 1 Ϫ

ͪ

1 3

1ؒ5 5

1ؒ5ؒ9 7

1 ؒ 5 ؒ 9 ؒ 13 9

x ϩ

x ϩ

x ϩ

x ϩ иии

3!

5!

7!

9!

ϱ

1 2

3 ؒ 7 ؒ и и и ؒ ͑4n Ϫ 5͒ 2n

x Ϫ ͚

x

2!

͑2n͒!

n෇2

ͩ

ϩ c1 x ϩ

ϱ

͚

n෇1

ͪ

1 ؒ 5 ؒ 9 ؒ и и и ؒ ͑4n Ϫ 3͒ 2nϩ1

x

͑2n ϩ 1͒!

ͪ

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

97817_17_ch17_p1186-1194.qk_97817_17_ch17_p1186-1194 11/9/10 10:29 AM Page 1192

1192

CHAPTER 17

SECOND-ORDER DIFFERENTIAL EQUATIONS

NOTE 2 In Example 2 we had to assume that the differential equation had a series solution. But now we could verify directly that the function given by Equation 8 is indeed a

solution.

NOTE 3 Unlike the situation of Example 1, the power series that arise in the solution

of Example 2 do not define elementary functions. The functions

and

y1͑x͒ ෇ 1 Ϫ

ϱ

1 2

3 ؒ 7 ؒ и и и ؒ ͑4n Ϫ 5͒ 2n

x Ϫ ͚

x

2!

͑2n͒!

n෇2

y2͑x͒ ෇ x ϩ

͚

ϱ

n෇1

2

are perfectly good functions but they can’t be expressed in terms of familiar functions. We

can use these power series expressions for y1 and y2 to compute approximate values of the

functions and even to graph them. Figure 1 shows the first few partial sums T0 , T2 , T4 , . . .

(Taylor polynomials) for y1͑x͒, and we see how they converge to y1 . In this way we can

graph both y1 and y2 in Figure 2.

2

_2

1 ؒ 5 ؒ 9 ؒ и и и ؒ ͑4n Ϫ 3͒ 2nϩ1

x

͑2n ϩ 1͒!

T¡¸

NOTE 4 If we were asked to solve the initial-value problem

_8

yЉ Ϫ 2 xyЈ ϩ y ෇ 0

y͑0͒ ෇ 0

yЈ͑0͒ ෇ 1

FIGURE 1

we would observe from Theorem 11.10.5 that

15

c0 ෇ y͑0͒ ෇ 0

_2.5

2.5

This would simplify the calculations in Example 2, since all of the even coefficients would

be 0. The solution to the initial-value problem is

_15

y͑x͒ ෇ x ϩ

FIGURE 2

17.4

ϱ

͚

n෇1

1 ؒ 5 ؒ 9 ؒ и и и ؒ ͑4n Ϫ 3͒ 2nϩ1

x

͑2n ϩ 1͒!

Exercises

11. yЉ ϩ x 2 yЈ ϩ x y ෇ 0,

1–11 Use power series to solve the differential equation.

1. yЈ Ϫ y ෇ 0

2. yЈ ෇ x y

3. yЈ ෇ x y

4. ͑x Ϫ 3͒yЈ ϩ 2y ෇ 0

5. yЉ ϩ x yЈ ϩ y ෇ 0

6. yЉ ෇ y

2

x 2 yЉ ϩ x yЈ ϩ x 2 y ෇ 0

8. yЉ ෇ x y

9. yЉ Ϫ x yЈ Ϫ y ෇ 0,

10. yЉ ϩ x 2 y ෇ 0,

y͑0͒ ෇ 1,

y͑0͒ ෇ 1,

yЈ͑0͒ ෇ 0

yЈ͑0͒ ෇ 0

Graphing calculator or computer required

y͑0͒ ෇ 0,

yЈ͑0͒ ෇ 1

12. The solution of the initial-value problem

7. ͑x Ϫ 1͒ yЉ ϩ yЈ ෇ 0

;

c1 ෇ yЈ͑0͒ ෇ 1

;

y͑0͒ ෇ 1

yЈ͑0͒ ෇ 0

is called a Bessel function of order 0.

(a) Solve the initial-value problem to find a power series

expansion for the Bessel function.

(b) Graph several Taylor polynomials until you reach one that

looks like a good approximation to the Bessel function on

the interval ͓Ϫ5, 5͔.

1. Homework Hints available at stewartcalculus.com

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

97817_17_ch17_p1186-1194.qk_97817_17_ch17_p1186-1194 11/9/10 10:29 AM Page 1193

CHAPTER 17

REVIEW

1193

Review

17

Concept Check

1. (a) Write the general form of a second-order homogeneous

linear differential equation with constant coefficients.

(b) Write the auxiliary equation.

(c) How do you use the roots of the auxiliary equation to solve

the differential equation? Write the form of the solution for

each of the three cases that can occur.

2. (a) What is an initial-value problem for a second-order differ-

ential equation?

(b) What is a boundary-value problem for such an equation?

(b) What is the complementary equation? How does it help

solve the original differential equation?

(c) Explain how the method of undetermined coefficients

works.

(d) Explain how the method of variation of parameters works.

4. Discuss two applications of second-order linear differential

equations.

5. How do you use power series to solve a differential equation?

3. (a) Write the general form of a second-order nonhomogeneous

linear differential equation with constant coefficients.

True-False Quiz

Determine whether the statement is true or false. If it is true, explain why.

If it is false, explain why or give an example that disproves the statement.

3. The general solution of yЉ Ϫ y ෇ 0 can be written as

y ෇ c1 cosh x ϩ c2 sinh x

1. If y1 and y2 are solutions of yЉ ϩ y ෇ 0, then y1 ϩ y2 is also

a solution of the equation.

2. If y1 and y2 are solutions of yЉ ϩ 6yЈ ϩ 5y ෇ x, then

4. The equation yЉ Ϫ y ෇ e x has a particular solution of the form

yp ෇ Ae x

c1 y1 ϩ c2 y2 is also a solution of the equation.

Exercises

1–10 Solve the differential equation.

11–14 Solve the initial-value problem.

1. 4yЉ Ϫ y ෇ 0

11. yЉ ϩ 6yЈ ෇ 0,

2. yЉ Ϫ 2yЈ ϩ 10y ෇ 0

12. yЉ Ϫ 6yЈ ϩ 25y ෇ 0,

3. yЉ ϩ 3y ෇ 0

13. yЉ Ϫ 5yЈ ϩ 4y ෇ 0,

14. 9yЉ ϩ y ෇ 3x ϩ e Ϫx,

4. 4yЉ ϩ 4yЈ ϩ y ෇ 0

5.

d 2y

dy

Ϫ4

ϩ 5y ෇ e 2x

2

dx

dx

6.

d 2y

dy

ϩ

Ϫ 2y ෇ x 2

dx 2

dx

7.

d 2y

dy

Ϫ2

ϩ y ෇ x cos x

dx 2

dx

8.

d 2y

ϩ 4 y ෇ sin 2 x

dx 2

y͑0͒ ෇ 0,

y͑0͒ ෇ 1,

yЈ͑0͒ ෇ 1

yЈ͑0͒ ෇ 1

yЈ͑0͒ ෇ 2

15–16 Solve the boundary-value problem, if possible.

15. yЉ ϩ 4yЈ ϩ 29y ෇ 0,

y͑0͒ ෇ 1,

y͑␲ ͒ ෇ Ϫ1

16. yЉ ϩ 4yЈ ϩ 29y ෇ 0,

y͑0͒ ෇ 1,

y͑␲ ͒ ෇ Ϫe Ϫ2␲

17. Use power series to solve the initial-value problem

y͑0͒ ෇ 0

yЈ͑0͒ ෇ 1

18. Use power series to solve the equation

d y

dy

Ϫ

Ϫ 6y ෇ 1 ϩ eϪ2x

dx 2

dx

d 2y

ϩ y ෇ csc x,

10.

dx 2

yЈ͑1͒ ෇ 12

y͑0͒ ෇ 2,

yЉ ϩ x yЈ ϩ y ෇ 0

2

9.

y͑1͒ ෇ 3,

0 Ͻ x Ͻ ␲ ͞2

yЉ Ϫ x yЈ Ϫ 2y ෇ 0

19. A series circuit contains a resistor with R ෇ 40 ⍀, an inductor

with L ෇ 2 H, a capacitor with C ෇ 0.0025 F, and a 12-V battery. The initial charge is Q ෇ 0.01 C and the initial current

is 0. Find the charge at time t.

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

97817_17_ch17_p1186-1194.qk_97817_17_ch17_p1186-1194 11/9/10 10:29 AM Page 1194

1194

CHAPTER 17

SECOND-ORDER DIFFERENTIAL EQUATIONS

20. A spring with a mass of 2 kg has damping constant 16, and a

force of 12.8 N keeps the spring stretched 0.2 m beyond its

natural length. Find the position of the mass at time t if it

starts at the equilibrium position with a velocity of 2.4 m͞s.

21. Assume that the earth is a solid sphere of uniform density with

mass M and radius R ෇ 3960 mi. For a particle of mass m

within the earth at a distance r from the earth’s center, the

gravitational force attracting the particle to the center is

Fr ෇

ϪGMr m

r2

where G is the gravitational constant and Mr is the mass of the

earth within the sphere of radius r.

ϪGMm

r.

R3

(b) Suppose a hole is drilled through the earth along a diameter. Show that if a particle of mass m is dropped from rest

at the surface, into the hole, then the distance y ෇ y͑t͒ of

the particle from the center of the earth at time t is given by

(a) Show that Fr ෇

yЉ͑t͒ ෇ Ϫk 2 y͑t͒

where k 2 ෇ GM͞R 3 ෇ t͞R.

(c) Conclude from part (b) that the particle undergoes simple

harmonic motion. Find the period T.

(d) With what speed does the particle pass through the center

of the earth?

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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. ### Tài liệu bạn tìm kiếm đã sẵn sàng tải về

9: Triple Integrals in Spherical Coordinates

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