CHAPTER

15

7. Roots are 2 ± 3. General solution is

15.1 Concepts Review

1. r 2 + a1r + a2 = 0; complex conjugate roots

+ C2 e –

3x

).

(

x

)

9. Auxiliary equation: r 2 + 4 = 0 has roots ±2i.

General solution: y = C1 cos 2 x + C2 sin 2 x

If x = 0 and y = 2, then 2 = C1; if x =

Problem Set 15.1

1. Roots are 2 and 3. General solution is

y = C1e2 x + C2 e3 x .

2. Roots are –6 and 1. General solution is

y = C1e –6 x + C2 e x .

3. Auxiliary equation: r 2 + 6r – 7 = 0,

(r + 7)(r – 1) = 0 has roots –7, 1.

General solution: y = C1e –7 x + C2 e x

y ′ = –7C1e –7 x + C2 e x

If x = 0, y = 0, y ′ = 4, then 0 = C1 + C2 and

4 = –7C1 + C2 , so C1 = –

1

1

and C2 = .

2

2

e x – e –7 x

Therefore, y =

.

2

4. Roots are –2 and 5. General solution is

y = C1e

3x

y = e –3 x C1e 11x + C2 e – 11x .

4. C1 cos x + C2 sin x

–2 x

y = e 2 x (C1e

8. Roots are –3 ± 11. General solution is

2. C1e – x + C2 e x

3. (C1 + C2 x)e

Differential

Equations

+ C2 e . Particular solution is

5x

⎛ 12 ⎞

⎛5⎞

y = ⎜ ⎟ e5 x – ⎜ ⎟ e –2 x .

7

⎝ ⎠

⎝7⎠

5. Repeated root 2. General solution is

y = (C1 + C2 x)e 2 x .

6. Auxiliary equation:

r 2 + 10r + 25 = 0, (r + 5)2 = 0 has one repeated

root −5 .

General solution: y = C1e –5 x + C2 xe –5 x or

y = (C1 + C2 x)e –5 x

Instructor’s Resource Manual

y = 3, then 3 = C2 .

Therefore, y = 2 cos 2 x + 3sin 2 x .

π

and

4

10. Roots are ±3i. General solution is

y = (C1 cos 3 x + C2 sin 3 x). Particular solution is

y = − sin 3 x − 3cos 3x .

11. Roots are –1 ± i. General solution is

y = e – x (C1 cos x + C2 sin x).

12. Auxiliary equation: r 2 + r + 1 = 0 has roots

–1

3

±

i.

2

2

General solution:

⎛ 3⎞

⎛ 3⎞

–1/ 2 ) x

(–1/ 2) x

y = C1e(

cos ⎜⎜

sin ⎜⎜

⎟⎟ x + C2 e

⎟⎟ x

2

⎝

⎠

⎝ 2 ⎠

⎡

⎛ 3⎞

⎛ 3⎞ ⎤

y = e – x / 2 ⎢C1 cos ⎜⎜

⎟⎟ x + C2 sin ⎜⎜

⎟⎟ x ⎥

⎢⎣

⎝ 2 ⎠

⎝ 2 ⎠ ⎥⎦

13. Roots are 0, 0, –4, 1.

General solution is

y = C1 + C2 x + C3e –4 x + C4 e x .

14. Roots are –1, 1, ±i. General solution is

y = C1e – x + C2 e x + C3 cos x + C4 sin x.

15. Auxiliary equation: r 4 + 3r 2 – 4 = 0,

(r + 1)(r – 1)(r 2 + 4) = 0 has roots –1, 1, ±2i.

General solution:

y = C1e – x + C2 e x + C3 cos 2 x + C4 sin 2 x

Section 15.1

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16. Roots are –2, 3, ±i. General solution is y = C1e –2 x + C2 e3 x + C3 cos x + C4 sin x.

17. Roots are –2, 2. General solution is y = C1e –2 x + C2 e2 x .

y = C1 (cosh 2 x – sinh 2 x) + C2 (sinh 2 x + cosh 2 x) = (– C1 + C2 ) sinh 2 x + (C1 + C2 ) cosh 2 x

= D1 sinh 2 x + D2 cosh 2 x

18. eu = cosh u + sinh u and e – u = cosh u – sinh u.

Auxiliary equation: r 2 – 2br – c 2 = 0

Roots of auxiliary equation:

General solution: y = C1e(b +

2b ± 4b 2 + 4c 2

= b ± b2 + c2

2

b2 + c2 ) x

+ C2 e(b –

b2 + c 2 ) x

⎡ ⎛

⎞

⎛

⎞⎤

= ebx ⎢C1 ⎜ cosh ⎛⎜ b 2 + c 2 x ⎞⎟ + sinh ⎛⎜ b 2 + c 2 x ⎞⎟ ⎟ + C2 ⎜ cosh ⎛⎜ b 2 + c 2 x ⎞⎟ – sinh ⎛⎜ b 2 + c 2 x ⎞⎟ ⎟ ⎥

⎝

⎠

⎝

⎠⎠

⎝

⎠

⎝

⎠ ⎠⎦

⎝

⎣ ⎝

⎡

⎤

⎡

⎤

= ebx ⎢( C1 + C2 ) cosh ⎛⎜ b 2 + c 2 x ⎞⎟ + (C1 + C2 ) sin ⎛⎜ b 2 + c 2 x ⎞⎟ ⎥ = ebx ⎢ D1 cosh ⎛⎜ b 2 + c 2 x ⎞⎟ + D2 sinh ⎛⎜ b 2 + c 2 x ⎞⎟ ⎥

⎝

⎠

⎝

⎠⎦

⎝

⎠

⎝

⎠⎦

⎣

⎣

⎛ 1⎞ ⎛ 3⎞

19. Repeated roots ⎜ – ⎟ ± ⎜⎜

⎟ i.

⎝ 2 ⎠ ⎝ 2 ⎟⎠

⎡

⎛ 3⎞

⎛ 3⎞ ⎤

General solution is y = e – x / 2 ⎢ (C1 + C2 x) cos ⎜⎜

⎟⎟ x + (C3 + C4 x) sin ⎜⎜

⎟⎟ x ⎥ .

⎝ 2 ⎠

⎝ 2 ⎠ ⎥⎦

⎣⎢

20. Roots 1 ± i. General solution is

y = e (C1 cos x + C2 sin x)

x

= e x (c sin γ cos x + c cos γ sin x) = ce x sin( x + γ ).

21. (*) x 2 y ′′ + 5 xy ′ + 4 y = 0

Let x = e z . Then z = ln x;

dy dy dz dy 1

y′ =

=

=

;

dx dz dx dz x

y ′′ =

dy ′ d ⎛ dy 1 ⎞ dy –1 1 d 2 y dz

= ⎜

+

⎟=

dx dx ⎝ dz x ⎠ dz x 2 x dz 2 dx

dy –1 1 d 2 y 1

+

dz x 2 x dz 2 x

⎛ dy d 2 y ⎞ ⎛ dy ⎞

⎜ – + 2 ⎟ + ⎜5 ⎟ + 4y = 0

⎜ dz dz ⎟ ⎝ dz ⎠

⎝

⎠

(Substituting y ′ and y ′′ into (*))

=

d2y

dz 2

+4

dy

+ 4y = 0

dz

Auxiliary equation: r 2 + 4r + 4 = 0, (r + 2) 2 = 0

has roots –2, –2.

General solution: y = (C1 + C2 z )e –2 z ,

y = (C1 + C2 ln x)e –2 ln x

y = (C1 + C2 ln x) x –2

892

Section 15.1

22. As done in Problem 21,

⎡ ⎛ dy ⎞

⎛ d 2 y ⎞⎤

⎛ dy ⎞

⎢ – a ⎜ ⎟ + a ⎜ 2 ⎟ ⎥ + b ⎜ ⎟ + cy = 0.

⎜

⎟

⎝ dz ⎠

⎢⎣ ⎝ dz ⎠

⎝ dx ⎠ ⎥⎦

⎛ d2y ⎞

⎛ dy ⎞

Therefore, a ⎜

⎟ + (b – a ) ⎜ ⎟ + cy = 0.

⎜ dz 2 ⎟

⎝ dz ⎠

⎝

⎠

23. We need to show that y ''+ a1 y '+ a2 y = 0 if r1 and

r2 are distinct real roots of the auxiliary equation.

We have,

y ' = C1r1er1 x + C2 r2 er2 x

.

y '' = C1r12 er1 x + C2 r2 2 e r2 x

When put into the differential equation, we

obtain

y ''+ a1 y '+ a2 y = C1r12 er1x + C2 r2 2 e r2 x

(

)

(

)

+ a1 C1r1er1 x + C2 r2 er2 x + a2 C1er1 x + C2 er2 x (*)

The solutions to the auxiliary equation are given

by

1

r1 = −a1 − a12 − 4a2 and

2

1

r2 = −a1 + a12 − 4a2 .

2

Putting these values into (*) and simplifying

yields the desired result: y ''+ a1 y '+ a2 y = 0 .

(

(

)

)

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24. We need to show that y ''+ a1 y '+ a2 y = 0 if α ± β i are complex conjugate roots of the auxiliary equation. We

have,

y ' = eα x ( (α C1 + β C2 ) cos ( β x ) + (α C2 − β C1 ) sin ( β x ) )

.

y '' = eα x α 2C1 − β 2C1 + 2αβ C2 cos ( β x ) + α 2C2 − β 2C2 − 2αβ C1 sin ( β x )

((

)

(

When put into the differential equation, we obtain

((

)

)

)

(

)

y ''+ a1 y '+ a2 y = eα x α 2C1 − β 2C1 + 2αβ C2 cos ( β x ) + α 2C2 − β 2C2 − 2αβ C1 sin ( β x )

(

)

)

+ a1eα x ( (α C1 + β C2 ) cos ( β x ) + (α C2 − β C1 ) sin ( β x ) ) + a2 C1eα x cos ( β x ) + C2 eα x sin ( β x ) (*)

From the solutions to the auxiliary equation, we find that

−a

1

α = 1 and β = − i a12 − 4a2 .

2

2

Putting these values into (*) and simplifying yields the desired result: y ''+ a1 y '+ a2 y = 0 .

25. a.

⎛ b 2 b 4 b6 ⎞

⎛

⎞

(bi )2 (bi )3 (bi ) 4 (bi )5

b 3 b5 b 7

+

+

+

+… = ⎜ 1 –

+

–

+ ⎟ …+ i ⎜ b –

+

–

+…⎟

⎜

⎟

⎜

⎟

2!

3!

4!

5!

2! 4! 6! ⎠

3! 5! 7!

⎝

⎝

⎠

= cos (b) + i sin (b)

ebi = 1 + (bi ) +

b.

e a +bi = ea ebi = e a [cos(b) + i sin(b)]

c.

Dx ⎡e(α + β i ) x ⎤ = Dx [eα x (cos β x + i sin β x)] = α eα x (cos β x + i sin β x) + eα x (– i β sin β x + i β cos β x )

⎣

⎦

= eα x [(α + β i ) cos β x + (α i – β ) sin β x]

(α + β )e(α + β i ) x = (α + β i )[eα x (cos β x + i sin β x)] = eα x [(α + β i ) cos β x + (α i – β ) sin β x]

Therefore, Dx [e(α + β i ) x ] = (α + i β )e(α + β i ) x

26. c1e(α + β i ) x + c2 e(α + β i ) x [c1 and c2 are complex constants.]

= c1eα x [cos β x + i sin β x] + c2 eα x [cos(– β x) + i sin(– β x)] = eα x [(c1 + c2 ) cos β x + (c1 – c2 )i sin β x]

= eα x [C1 cos β x + C2 sin β x], where C1 = c1 + c2 , and C2 = c1 – c2 .

Note: If c1 and c2 are complex conjugates, then C1 and C2 are real.

27. y = 0.5e5.16228 x + 0.5e –1.162278 x

28. y = 3.5 xe –2.5 x + 2e –2.5 x

29. y = 1.29099e –0.25 x sin(0.968246 x)

30. y = e0.333333 x [2.5cos(0.471405 x) – 4.94975sin(0.471405 x)]

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15.2 Concepts Review

1. particular solution to the nonhomogeneous equation; homogeneous equation

2. –6 + C1e –2 x + C2 e3 x

3. y = Ax 2 + Bx + C

1x

4. y = Bxe 3

Problem Set 15.2

1. yh = C1e –3 x + C2 e3 x

⎛ 1⎞

yp = ⎜ – ⎟ x + 0

⎝ 9⎠

⎛ 1⎞

y = ⎜ – ⎟ x + C1e –3 x + C2 e3 x

⎝ 9⎠

2. yh = C1e –3 x + C2 e 2 x

⎛ 1⎞

⎛ 1⎞

⎛ 7 ⎞

y p = ⎜ – ⎟ x2 + ⎜ – ⎟ x + ⎜ – ⎟

3

9

⎝

⎠

⎝

⎠

⎝ 54 ⎠

⎛ 1⎞

⎛1⎞

⎛ 7 ⎞

y = ⎜ – ⎟ x 2 – ⎜ ⎟ x – ⎜ ⎟ + C1e –3 x + C2 e2 x

⎝ 3⎠

⎝9⎠

⎝ 54 ⎠

3. Auxiliary equation: r 2 – 2r + 1 = 0 has roots 1, 1.

yh = (C1 + C2 x)e x

Let y p = Ax 2 + Bx + C ; y ′p = 2 Ax + B;

y ′′p = 2 A .

Then (2 A) – 2(2 Ax + B ) + ( Ax 2 + Bx + C ) = x 2 + x.

Ax 2 + (–4 A + B ) x + (2 A – 2 B + C ) = x 2 + x

Thus, A = 1, –4A + B = 1, 2A – 2B + C = 0, so

A = 1, B = 5, C = 8.

General solution: y = x 2 + 5 x + 8 + (C1 + C2 x)e x

4. yh = C1e – x + C2 ⋅ y p = 2 x 2 + (–4) x

y = 2 x 2 – 4 x + C1e – x + C2

⎛1⎞

5. yh = C1e2 x + C2 e3 x ⋅ y p = ⎜ ⎟ e x ⋅ y

⎝2⎠

1

⎛ ⎞

= ⎜ ⎟ e x + C1e 2 x + C2 e3 x

⎝2⎠

894

Section 15.2

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6. Auxiliary equation: r 2 + 6r + 9 = 0, (r + 3)2 = 0 has roots –3, –3.

yh = (C1 + C2 x )e –3 x

Let y p = Be – x ; y ′p = – Be – x ; y ′′p = Be – x .

Then ( Be – x ) + 6(– Be – x ) + 9( Be – x ) = 2e – x ; 4 Be – x = 2e – x ; B =

1

2

⎛1⎞

General solution: y = ⎜ ⎟ e – x + (C1 + C2 x)e –3 x

⎝2⎠

7. yh = C1e –3 x + C2 e – x

⎛ 1⎞

y p = ⎜ – ⎟ xe –3 x

⎝ 2⎠

⎛ 1 ⎞ –3 x

y = ⎜ – ⎟ xe

+ C1e –3 x + C2 e – x

⎝ 2⎠

8. yh = e – x (C1 cos x + C2 sin x)

⎛3⎞

y p = ⎜ ⎟ e –2 x

⎝2⎠

⎛3⎞

y = ⎜ ⎟ e –2 x + e – x (C1 cos x + C2 sin x)

⎝2⎠

9. Auxiliary equation: r 2 – r – 2 = 0,

(r + 1)(r – 2) = 0 has roots –1, 2.

yh = C1e – x + C2 e2 x

Let y p = B cos x + C sin x; y ′p = – B sin x + C cos x; y ′′p = – B cos x – C sin x.

Then (− B cos x − C sin x) − (− B sin x + C cos x)

−2( B cos x + C sin x) = 2sin x.

1

–3

(−3B − C ) cos x + ( B − 3C ) sin x = 2sin x , so − 3B − C = 0 so –3B – C = 0 and B − 3C = 2 ; B = ; C = .

5

5

⎛1⎞

⎛3⎞

General solution: ⎜ ⎟ cos x – ⎜ ⎟ sin x + C1e2 x + C2 e – x

5

⎝ ⎠

⎝5⎠

10. yh = C1e –4 x + C2

⎛ 1⎞

⎛ 4⎞

y p = ⎜ – ⎟ cos x + ⎜ ⎟ sin x

⎝ 17 ⎠

⎝ 17 ⎠

1

⎛

⎞

⎛ 4⎞

y = ⎜ – ⎟ cos x + ⎜ ⎟ sin x + C1e –4 x + C2

⎝ 17 ⎠

⎝ 17 ⎠

11. yh = C1 cos 2 x + C2 sin 2 x

⎛1⎞

y p = (0) x cos 2 x + ⎜ ⎟ x sin 2 x

⎝2⎠

⎛1⎞

y = ⎜ ⎟ x sin 2 x + C1 cos 2 x + C2 sin 2 x

⎝2⎠

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12. Auxiliary equation: r 2 + 9 = 0 has roots ±3i, so yh = C1 cos 3x + C2 sin 3 x.

Let y p = Bx cos 3 x + Cx sin 3 x; y ′p = (–3bx + C ) sin 3 x + ( B + 3Cx) cos 3x ;

y ′′p = (–9 Bx + 6C ) cos 3 x + (–9Cx – 6 B ) sin 3x .

Then substituting into the original equation and simplifying, obtain 6C cos 3 x - 6 B sin 3x = sin 3x , so C = 0 and

1

B=– .

6

⎛ 1⎞

General solution: y = ⎜ – ⎟ x cos 3x + C1 cos 3x + C2 sin 3x

⎝ 6⎠

13. yh = C1 cos 3 x + C2 sin 3x

⎛1⎞

⎛1⎞

y p = (0) cos x + ⎜ ⎟ sin x + ⎜ ⎟ e2 x

8

⎝ ⎠

⎝ 13 ⎠

⎛1⎞

⎛1⎞

y = ⎜ ⎟ sin x + ⎜ ⎟ e2 x + C1 cos 3x + C2 sin 3 x

⎝8⎠

⎝ 13 ⎠

14. yh = C1e – x + C2

⎛1⎞

⎛3⎞

y p = ⎜ ⎟ e x + ⎜ ⎟ x 2 + (–3) x

⎝2⎠

⎝2⎠

⎛1⎞

⎛3⎞

y = ⎜ ⎟ e x + ⎜ ⎟ x 2 – 3 x + C1e – x + C2

⎝2⎠

⎝2⎠

15. Auxiliary equation: r 2 – 5r + 6 = 0 has roots 2 and 3, so yh = C1e2 x + C2 e3 x .

Let y p = Be x ; y ′p = Be x ; y ′′p = Be x .

Then ( Be x ) – 5( Be x ) + 6( Be x ) = 2e x ; 2 Be x = 2e x ; B = 1 .

General solution: y = e x + C1e2 x + C2 e3 x

y ′ = e x + 2C1e2 x + 3C2 e3 x

If x = 0, y = 1, y ′ = 0, then 1 = 1 + C1 + C2 and 0 = 1 + 2C1 + 3C2 ; C1 = 1, C2 = –1.

Therefore, y = e x + e2 x – e3 x .

16. yh = C1e –2 x + C2 e2 x

⎛ 4⎞

y p = (0) cos x + ⎜ – ⎟ sin x

⎝ 5⎠

⎛ 4⎞

y = ⎜ – ⎟ sin x + C1e –2 x + C2 e2 x

⎝ 5⎠

⎛ 4⎞

⎛9⎞

⎛ 11 ⎞

y = ⎜ – ⎟ sin x + ⎜ ⎟ e –2 x + ⎜ ⎟ e2 x satisfies the conditions.

⎝ 5⎠

⎝5⎠

⎝5⎠

17. yh = C1e x + C2 e2 x

⎛1⎞

y p = ⎜ ⎟ (10 x + 19)

⎝4⎠

⎛1⎞

y = ⎜ ⎟ (10 x + 19) + C1e x + C2 e2 x

⎝4⎠

896

Section 15.2

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18. Auxiliary equation: r 2 – 4 = 0 has roots 2, –2, so yh = C1e2 x + C2 e –2 x .

Let y p = v1e2 x + v2 e –2 x , subject to v1′e2 x + v2′ e –2 x = 0, and v1′ (2e2 x ) + v2′ (–2e –2 x ) = e2 x .

Then v1′ (4e2 x ) = e2 x and v2′ (–4e –2 x ) = e2 x ; v1′ =

e4 x

1

x

and v2′ = – e4 x / 4 ; v1 = and v2 = –

.

16

4

4

xe2 x e2 x

+ C1e2 x + C2 e –2 x

–

4

16

General solution: y =

19. yh = C1 cos x + C2 sin x

y p = – cos ln sin x – cos x – x sin x

y = – cos x ln sin x – x sin x + C3 cos x + C2 sin x (combined cos x terms)

20. yh = C1 cos x + C2 sin x

y p = – sin x ln csc x + cot x

y = – sin x ln csc x + cot x + C1 cos x + C2 sin x

21. Auxiliary equation: r 2 – 3r + 2 = 0 has roots 1, 2, so yh = C1e x + C2 e2 x .

Let y p = v1e x + v2 e2 x subject to v1′e x + v2′ e2 x = 0, and v1′ (e x ) + v2′ (2e2 x ) = e x (ex + 1) –1.

Then v1′ =

–ex

e (e + 1)

x

x

so v1 = ∫

–e x

e (e + 1)

x

x

dx = ∫

–1

du

u (u + 1)

1 ⎞

ex +1

⎛ –1

⎛ u +1⎞

–x

= ∫⎜ +

⎟ du = – ln u + ln(u + 1) = ln ⎜

⎟ = ln x = ln(1 + e )

+

u

u

1

u

⎝

⎠

⎝

⎠

e

v2′ =

ex

so v2 = – e – x + ln(1 + e – x )

e (e + 1)

(similar to finding v1 )

2x

x

General solution: y = e x ln(1 + e – x ) – e x + e2 x ln(1 + e – x ) + C1e x + C2 e2 x

y = (e x + e2 x ) ln(1 + e – x ) + D1e x + D2 e2 x

22. yh = C1e2 x + C2 e3 x ; y p = e x

y = e x + C1e2 x + C2 e3 x

23. L( y p ) = (v1u1 + v2u2 )′′ + b(v1u1 + v2u2 )′ + c(v1u1 + v2 u2 )

= (v1′u1 + v1u1′ + v2′ u2 + v2u2′ ) + b(v1′u1 + v1u1′ + v2′ u2 + v2 u2′ ) + c(v1u1 + v2 u2 )

= (v1′′u1 + v1′u1′ + v1′u1′ + v1u1′′ + v2′′u2 + v2′ u2′ + v2′ u2′ + v2u2′′ ) + b(v1′u1 + v1u1′ + v2′ u2 + v2u2′ ) + c(v1u1 + v2 u2 )

= v1 (u1′′ + bu1′ + cu1 ) + v2 (u2′′ + bu2′ + cu2 ) + b(v1′u1 + v2′ u2 ) + (v1′′u1 + v1′u1′ + v2′′u2 + v2′ u2 ) + (v1′u1′ + v2′ u2′ )

= v1 (u1′′ + bu1′ + cu1 ) + v2 (u2′′ + bu2′ + cu2 ) + b(v1′u1 + v2′ u2 ) + (v1′u1 + v2′ u2 )′ + (v1′u1′ + v2′ u2′ )

= v1 (0) + v2 (0) + b(0) + (0) + k ( x) = k ( x)

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24. Auxiliary equation: r 2 + 4 = 0 has roots ±2i.

yh = C1 cos 2 x + C2 sin 2 x

Now write sin 3 x in a form involving sin βx’s or

cos βx’s.

3

1

sin 3 x = sin x – sin 3x

4

4

(C.R.C. Standard Mathematical Tables, or derive it using half-angle and product identities.)

Let y p = A sin x + B cos x + C sin 3 x + D cos 3 x ;

y ′p = A cos x – B sin x + 3C cos 3x – 3D sin 3 x ;

y ′′p = – A sin x – B cos x – 9C sin 3 x – 9 D cos 3 x .

Then

3

1

y ′′p + 4 y p = 3 A sin x + 3B cos x – 5C sin 3 x – 5 D cos 3 x = sin x – sin 3x, so

4

4

1

1

A = , B = 0, C = , D = 0.

20

4

1

1

General solution: y = sin x + sin 3 x + C1 cos 2 x + C2 sin 2 x

4

20

15.3 Concepts Review

1. 3; π

2. π ; decreases

3. 0

4. electric circuit

Problem Set 15.3

1. k = 250, m = 10, B 2 = k / m = 250 /10 = 25, B = 5

(the problem gives the mass as m = 10 kg )

Thus, y '' = −25 y. The general solution is y = C1 cos 5t + C2 sin 5t. Apply the initial condition to get y = 0.1cos 5t.

The period is

2π

seconds.

5

2. k = 100 lb/ft, w = 1 lb, g = 32 ft/s2, y0 =

1

ft,

12

⎛1⎞

B = 40 2 . Then y = ⎜ ⎟ cos(40 2)t.

⎝ 12 ⎠

1

ft = 1 in.

Amplitude is

12

2π

≈ 0.1111 s.

Period is

40 2

3. y = 0.1cos 5t = 0 whenever 5t =

π

2

+ π k or t =

π

10

+

π

5

k.

⎛π π ⎞

⎛π π ⎞

⎛π

⎞

y ' ⎜ + k ⎟ = 0.5 sin 5 ⎜ + k ⎟ = 0.5 sin ⎜ + π k ⎟ = 0.5 meters per second

10

5

10

5

2

⎝

⎠

⎝

⎠

⎝

⎠

898

Section 15.3

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⎛1⎞

4. 10 = k ⎜ ⎟ , so k = 30 lb/ft, w = 20 lb,

⎝3⎠

g = 32 ft/s2, y0 = –1 ft, v0 = 2 ft/s, B = 4 3

Then y = C1 cos(4 3t ) + C2 sin(4 3t ).

⎛ 3t ⎞

y = cos(4 3t ) + ⎜⎜

⎟⎟ sin(4 3t ) satisfies the initial conditions.

⎝ 6 ⎠

5. k = 20 lb/ft; w = 10 lb; y0 = 1 ft; q =

1

s-lb/ft, B = 8, E = 0.32

10

E 2 – 4 B 2 < 0, so there is damped motion. Roots of auxiliary equation are approximately –0.16 ± 8i.

General solution is y ≈ e –0.16t (C1 cos8t + C2 sin 8t ). y ≈ e –0.16t (cos8t + 0.02sin 8t ) satisfies the initial conditions.

6. k = 20 lb/ft; w = 10 lb; y0 = 1 ft; q = 4 s-lb/ft

(20)(32)

(4)(32)

= 8; E =

= 12.8; E 2 – 4 B 2 < 0, so damped motion.

10

10

B=

Roots of auxiliary equation are

– E ± E 2 – 4B2

= –6.4 ± 4.8i.

2

General solution is y = e –6.4t (C1 cos 4.8t + C2 sin 4.8t ).

y ′ = e –6.4t (–4.8C1 sin 4.8t + 4.8C2 cos 4.8t ) – 6.4e –6.4t (C1 cos 4.8t + C2 sin 4.8t )

4

If t = 0, y = 1, y ′ = 0, then 1 = C1 and 0 = 4.8C2 – 6.4C1 , so C1 = 1 and C2 = .

3

4

⎡

⎤

⎛

⎞

Therefore, y = e –6.4t ⎢cos 4.8t + ⎜ ⎟ sin 4.8t ⎥ .

⎝3⎠

⎣

⎦

7. Original amplitude is 1 ft. Considering the contribution of the sine term to be negligible due to the 0.02 coefficient,

the amplitude is approximately e –0.16t .

e –0.16t ≈ 0.1 if t ≈ 14.39 , so amplitude will be about one-tenth of original in about 14.4 s.

8. C1 = 1 and C2 = –0.105, so y = e –0.16t (cos8t + 0.105sin 8t ).

9. LQ′′ + RQ′ +

Q

= E (t ); 106 Q ′ + 106 Q = 1; Q′ + Q = 10 –6

C

Integrating factor: et

D[Qet ] = 10 –6 et ; Qet = 10 –6 et + C ;

Q = 10 –6 + Ce – t

If t = 0, Q = 0, then C = –10 –6.

Therefore, Q(t ) = 10 –6 – 10 –6 e – t = 10 –6 (1 – e – t ).

10. Same as Problem 9, except C = 4 – 10 –6 , so Q(t ) = 10 –6 + (4 – 10 –6 )e – t .

Then I (t ) = Q ′(t ) = –(4 – 10 –6 )e – t .

11.

Q

[2(10 –6 )]

= 120sin 377t

a.

Q(t) = 0.00024 sin 377t

b.

I (t ) = Q ′(t ) = 0.09048cos 377t

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12. LQ′′ + RQ ′ +

Q

Q

= E; 10 –2 Q′′ +

= 20; Q′′ + 109 Q = 2000

C

10 –7

The auxiliary equation, r 2 + 109 = 0, has roots ±109 / 2 i.

Qh = C1 cos109 / 2 t + C2 sin109 / 2 t

Q p = 2000(10 –9 ) = 2(10 –6 ) is a particular solution (by inspection).

General solution: Q(t ) = 2(10 –6 ) + C1 cos109 / 2 t + C2 sin109 / 2 t

Then I (t ) = Q′(t ) = –109 / 2 C1 sin109 / 2 t + 109 / 2 C2 cos109 / 2 t.

If t = 0, Q = 0, I = 0, then 0 = 2(10 –6 ) + C1 and 0 = C2 .

Therefore, I (t ) = –109 / 2 (–2[10 –6 ]) sin109 / 2 t = 2(10 –3 / 2 ) sin109 / 2 t.

13. 3.5Q′′ + 1000Q +

Q

= 120sin 377t

[2(10 –6 )]

(Values are approximated to 6 significant figures for the remainder of the problem.)

Q′′ + 285.714Q ′ + 142857Q = 34.2857 sin 377t

Roots of the auxiliary equation are

–142.857 ± 349.927i.

Qh = e –142.857t (C1 cos 349.927t + C2 sin 349.927t )

Q p = –3.18288(10 –4 ) cos 377t + 2.15119(10 –6 ) sin 377t

Then, Q = –3.18288(10 –4 ) cos 377t + 2.15119(10 –6 ) sin 377t + Qh .

I = Q ′ = 0.119995sin 377t + 0.000810998cos 377t + Qh′

0.000888 cos 377t is small and Qh′ → 0 as t → ∞, so the steady-state current is I ≈ 0.12sin 377t .

14. a. Roots of the auxiliary equation are ±Bi.

yh = C1 cos Bt + C2 sin Bt.

⎡

⎤

c

yp = ⎢

⎥ sin At

2

2

⎣⎢ ( B – A ) ⎦⎥

The desired result follows.

b.

⎛ c ⎞

yp = ⎜ –

⎟ t cos Bt , so

⎝ 2B ⎠

⎛ c ⎞

y = C1 cos Bt + C2 sin Bt – ⎜

⎟ t cos Bt.

⎝ 2B ⎠

c.

900

Due to the t factor in the last term, it

increases without bound.

Section 15.3

15. A sin( β t + γ ) = A(sin β t cos γ + cos β t sin γ )

= ( A cos γ ) sin β t + ( A sin γ ) cos β t

= C1 sin β t + C2 cos β t , where C1 = A cos γ and

C2 = A sin γ .

[Note that

C12 + C22 = A2 cos 2 γ + A2 sin 2 γ = A2 .)

16. The first two terms have period

2π

and the last

B

2π

. Then the sum of the three terms

A

⎛ 2π ⎞

⎛ 2π ⎞

is periodic if m ⎜ ⎟ = n ⎜ ⎟ for some integers

B

⎝ ⎠

⎝ B ⎠

B m

m, n; equivalently, if = , a rational number.

A n

has period

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17. The magnitudes of the tangential components of

the forces acting on the pendulum bob must be

equal.

15.4 Chapter Review

Concepts Test

1. False:

y 2 is not linear in y.

2. True:

y and y ′′ are linear in y and y ′′,

respectively.

3. True:

y ′ = sec 2 x + sec x tan x

2 y ′ – y 2 = (2sec 2 x + 2sec x tan x)

Therefore, – m

d 2s

s = Lθ, so

dt 2

d 2s

dt 2

=L

Therefore, – mL

–(tan 2 x + 2sec x tan x + sec2 x)

= mg sin θ .

d 2θ

dt 2

d 2θ

dt 2

= sec2 x – tan 2 x = 1

.

= mg sin θ .

4. False:

It should involve 6.

5. True:

D 2 adheres to the conditions for

linear operators.

D 2 (kf ) = kD 2 ( f )

d θ

2

g

Hence,

= – sin θ .

2

L

dt

18. a.

D2 ( f + g ) = D2 f + D2 g

Since the roots of the auxiliary equation are

g

⎛g⎞

± i , the solution of θ ′′(t ) + ⎜ ⎟θ = 0 is

L

⎝L⎠

g

g

t + C2 sin

t , which can be

L

L

⎛ g

⎞

t + γ ⎟⎟

written as θ = C ⎜⎜

⎝ L

⎠

(by Problem 15).

The period of this function is

6. False:

Replacing y by C1u1 ( x) + C2u2 ( x)

would yield, on the left side,

C1 f ( x) + C2 f ( x) = (C1 + C2 ) f ( x)

which is f(x) only if C1 + C2 = 1 or

f(x) = 0.

7. True:

–1 is a repeated root, with multiplicity

3, of the auxiliary equation.

8. True:

L(u1 – u2 ) = L(u1 ) – L(u2 )

= f ( x) – f ( x) = 0

9. False:

That is the form of yh . y p should

θ = C1 cos

2π

g

L

= 2π

L

G

= 2π

LR 2

L

= 2πR

.

GM

GM

2πR1

p

Therefore, 1 =

p2 2πR

2

L1

GM

L2

GM

have the form Bx cos 3x + Cx sin 3x.

=

R1 L1

R2 L2

.

b. To keep perfect time at both places, require

R 80.85

p1 = p2 . Then 1 = 2

, so

3960 81

R2 ≈ 3963.67.

The height of the mountain is about

3963.67 - 3960 = 3.67 mi (about 19,387 ft).

10. True:

See Problem 15, Section 15.3.

Sample Test Problems

1. u ′ + 3u = e x . Integrating factor is e3 x .

D[ue3 x ] = e 4 x

⎛1⎞

y = ⎜ ⎟ e x + C1e –3 x

⎝4⎠

⎛1⎞

y ′ = ⎜ ⎟ e x + C1e –3 x

⎝4⎠

⎛1⎞

y = ⎜ ⎟ e x + C3e –3 x + C2

⎝4⎠

2. Roots are –1, 1. y = C1e – x + C2 e x

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3. (Second order homogeneous)

12. (Fourth-order homogeneous)

The auxiliary equation, r 2 – 3r + 2 = 0, has roots

The auxiliary equation, r 4 – 3r 2 – 10 = 0 or

1, 2. The general solution is y = C1e x + C2 e2 x .

(r 2 – 5)(r 2 + 2) = 0, has roots – 5, 5, ± 2i.

General solution:

y ′ = C1e x + 2C2 e2 x

If x = 0, y = 0, y ′ = 3, then 0 = C1 + C2 and

3 = C1 + 2C2 , so C1 = –3, C2 = 3.

Therefore, y = –3e x + 3e2 x .

3

4. Repeated root – . y = (C1 + C2 x)e(–3 / 2) x

2

5. yh = C1e – x + C2 e x (Problem 2)

y = C1e

+ C2 e –

5x

+ C3 cos 2 x + C4 sin 2 x

13. Repeated roots ± 2

y = (C1 + C2 x)e –

2x

+ (C3 + C4 x)e

14. a.

Q′(t ) = 3 – 0.02Q

b.

Q′(t ) + 0.02Q = 3

2x

Integrating factor is e0.02t

y p = –1 + C1e – x + C2 e x

D[Qe0.02t ] = 3e0.02t

Q(t ) = 150 + Ce –0.02t

6. (Second-order nonhomogeneous) The auxiliary

equation, r + 4r + 4 = 0, has roots –2, –2.

2

Q(t ) = 150 – 30e –0.02t goes through (0, 120).

yh = C1e –2 x + C2 xe –2 x = (C1 + C2 x)e –2 x

Let y p = Be x ; y ′p = Be x ; y ′′p = Be x .

1

( Be x ) + 4( Be x ) + 4( Be x ) = 3e x , so B = .

3

General solution: y =

5x

ex

+ (C1 + C2 x)e –2 x

3

7. yh = (C1 + C2 x)e –2 x (Problem 12)

⎛1⎞

y p = ⎜ ⎟ x 2 e –2 x

⎝2⎠

⎡⎛ 1 ⎞

⎤

y = ⎢⎜ ⎟ x 2 + C1 + C2 x ⎥ e –2 x

2

⎝

⎠

⎣

⎦

8. Roots are ±2i.

y = C1 cos 2 x + C2 sin 2 x

y = sin 2x satisfies the conditions.

9. (Second-order homogeneous)

The auxiliary equation, r + 6r + 25 = 0, has

roots –3 ± 4i. General solution:

2

y = e –3 x (C1 cos 4 x + C2 sin 4 x)

10. Roots are ±i. yh = C1 cos x + C2 sin x

c.

Q → 150 g, as t → ∞ .

15. (Simple harmonic motion)

k = 5; w = 10; y0 = –1

(5)(32)

=4

10

Then the equation of motion is y = –cos 4t.

2π π

= .

The amplitude is –1 = 1; the period is

4 2

B=

16. It is at equilibrium when y = 0 or –cos 4t = 0, or

π 3π

t= ,

, ….

8 8

y ′(t ) = 4sin 4t , so at equilibrium y ′ = ±4 = 4.

17. Q′′ + 2Q ′ + 2Q = 1

Roots are –1 ± i.

1

Qh = e – t (C1 cos t + C2 sin t ) and Q p = ;

2

1

Q = e – t (C1 cos t + C2 sin t ) +

2

–

t

I (t ) = Q ′(t ) = – e [(C1 – C2 ) cos t + (C1 + C2 ) sin t ]

I (t ) = e – t sin t satisfies the initial conditions.

y p = x cos x – sin x + sin x ln cos x

y = x cos x − sin x ln cos x + C1 cos x + C3 sin x

(combining the sine terms)

11. Roots are –4, 0, 2. y = C1e –4 x + C2 + C3e2 x

902

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15

7. Roots are 2 ± 3. General solution is

15.1 Concepts Review

1. r 2 + a1r + a2 = 0; complex conjugate roots

+ C2 e –

3x

).

(

x

)

9. Auxiliary equation: r 2 + 4 = 0 has roots ±2i.

General solution: y = C1 cos 2 x + C2 sin 2 x

If x = 0 and y = 2, then 2 = C1; if x =

Problem Set 15.1

1. Roots are 2 and 3. General solution is

y = C1e2 x + C2 e3 x .

2. Roots are –6 and 1. General solution is

y = C1e –6 x + C2 e x .

3. Auxiliary equation: r 2 + 6r – 7 = 0,

(r + 7)(r – 1) = 0 has roots –7, 1.

General solution: y = C1e –7 x + C2 e x

y ′ = –7C1e –7 x + C2 e x

If x = 0, y = 0, y ′ = 4, then 0 = C1 + C2 and

4 = –7C1 + C2 , so C1 = –

1

1

and C2 = .

2

2

e x – e –7 x

Therefore, y =

.

2

4. Roots are –2 and 5. General solution is

y = C1e

3x

y = e –3 x C1e 11x + C2 e – 11x .

4. C1 cos x + C2 sin x

–2 x

y = e 2 x (C1e

8. Roots are –3 ± 11. General solution is

2. C1e – x + C2 e x

3. (C1 + C2 x)e

Differential

Equations

+ C2 e . Particular solution is

5x

⎛ 12 ⎞

⎛5⎞

y = ⎜ ⎟ e5 x – ⎜ ⎟ e –2 x .

7

⎝ ⎠

⎝7⎠

5. Repeated root 2. General solution is

y = (C1 + C2 x)e 2 x .

6. Auxiliary equation:

r 2 + 10r + 25 = 0, (r + 5)2 = 0 has one repeated

root −5 .

General solution: y = C1e –5 x + C2 xe –5 x or

y = (C1 + C2 x)e –5 x

Instructor’s Resource Manual

y = 3, then 3 = C2 .

Therefore, y = 2 cos 2 x + 3sin 2 x .

π

and

4

10. Roots are ±3i. General solution is

y = (C1 cos 3 x + C2 sin 3 x). Particular solution is

y = − sin 3 x − 3cos 3x .

11. Roots are –1 ± i. General solution is

y = e – x (C1 cos x + C2 sin x).

12. Auxiliary equation: r 2 + r + 1 = 0 has roots

–1

3

±

i.

2

2

General solution:

⎛ 3⎞

⎛ 3⎞

–1/ 2 ) x

(–1/ 2) x

y = C1e(

cos ⎜⎜

sin ⎜⎜

⎟⎟ x + C2 e

⎟⎟ x

2

⎝

⎠

⎝ 2 ⎠

⎡

⎛ 3⎞

⎛ 3⎞ ⎤

y = e – x / 2 ⎢C1 cos ⎜⎜

⎟⎟ x + C2 sin ⎜⎜

⎟⎟ x ⎥

⎢⎣

⎝ 2 ⎠

⎝ 2 ⎠ ⎥⎦

13. Roots are 0, 0, –4, 1.

General solution is

y = C1 + C2 x + C3e –4 x + C4 e x .

14. Roots are –1, 1, ±i. General solution is

y = C1e – x + C2 e x + C3 cos x + C4 sin x.

15. Auxiliary equation: r 4 + 3r 2 – 4 = 0,

(r + 1)(r – 1)(r 2 + 4) = 0 has roots –1, 1, ±2i.

General solution:

y = C1e – x + C2 e x + C3 cos 2 x + C4 sin 2 x

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16. Roots are –2, 3, ±i. General solution is y = C1e –2 x + C2 e3 x + C3 cos x + C4 sin x.

17. Roots are –2, 2. General solution is y = C1e –2 x + C2 e2 x .

y = C1 (cosh 2 x – sinh 2 x) + C2 (sinh 2 x + cosh 2 x) = (– C1 + C2 ) sinh 2 x + (C1 + C2 ) cosh 2 x

= D1 sinh 2 x + D2 cosh 2 x

18. eu = cosh u + sinh u and e – u = cosh u – sinh u.

Auxiliary equation: r 2 – 2br – c 2 = 0

Roots of auxiliary equation:

General solution: y = C1e(b +

2b ± 4b 2 + 4c 2

= b ± b2 + c2

2

b2 + c2 ) x

+ C2 e(b –

b2 + c 2 ) x

⎡ ⎛

⎞

⎛

⎞⎤

= ebx ⎢C1 ⎜ cosh ⎛⎜ b 2 + c 2 x ⎞⎟ + sinh ⎛⎜ b 2 + c 2 x ⎞⎟ ⎟ + C2 ⎜ cosh ⎛⎜ b 2 + c 2 x ⎞⎟ – sinh ⎛⎜ b 2 + c 2 x ⎞⎟ ⎟ ⎥

⎝

⎠

⎝

⎠⎠

⎝

⎠

⎝

⎠ ⎠⎦

⎝

⎣ ⎝

⎡

⎤

⎡

⎤

= ebx ⎢( C1 + C2 ) cosh ⎛⎜ b 2 + c 2 x ⎞⎟ + (C1 + C2 ) sin ⎛⎜ b 2 + c 2 x ⎞⎟ ⎥ = ebx ⎢ D1 cosh ⎛⎜ b 2 + c 2 x ⎞⎟ + D2 sinh ⎛⎜ b 2 + c 2 x ⎞⎟ ⎥

⎝

⎠

⎝

⎠⎦

⎝

⎠

⎝

⎠⎦

⎣

⎣

⎛ 1⎞ ⎛ 3⎞

19. Repeated roots ⎜ – ⎟ ± ⎜⎜

⎟ i.

⎝ 2 ⎠ ⎝ 2 ⎟⎠

⎡

⎛ 3⎞

⎛ 3⎞ ⎤

General solution is y = e – x / 2 ⎢ (C1 + C2 x) cos ⎜⎜

⎟⎟ x + (C3 + C4 x) sin ⎜⎜

⎟⎟ x ⎥ .

⎝ 2 ⎠

⎝ 2 ⎠ ⎥⎦

⎣⎢

20. Roots 1 ± i. General solution is

y = e (C1 cos x + C2 sin x)

x

= e x (c sin γ cos x + c cos γ sin x) = ce x sin( x + γ ).

21. (*) x 2 y ′′ + 5 xy ′ + 4 y = 0

Let x = e z . Then z = ln x;

dy dy dz dy 1

y′ =

=

=

;

dx dz dx dz x

y ′′ =

dy ′ d ⎛ dy 1 ⎞ dy –1 1 d 2 y dz

= ⎜

+

⎟=

dx dx ⎝ dz x ⎠ dz x 2 x dz 2 dx

dy –1 1 d 2 y 1

+

dz x 2 x dz 2 x

⎛ dy d 2 y ⎞ ⎛ dy ⎞

⎜ – + 2 ⎟ + ⎜5 ⎟ + 4y = 0

⎜ dz dz ⎟ ⎝ dz ⎠

⎝

⎠

(Substituting y ′ and y ′′ into (*))

=

d2y

dz 2

+4

dy

+ 4y = 0

dz

Auxiliary equation: r 2 + 4r + 4 = 0, (r + 2) 2 = 0

has roots –2, –2.

General solution: y = (C1 + C2 z )e –2 z ,

y = (C1 + C2 ln x)e –2 ln x

y = (C1 + C2 ln x) x –2

892

Section 15.1

22. As done in Problem 21,

⎡ ⎛ dy ⎞

⎛ d 2 y ⎞⎤

⎛ dy ⎞

⎢ – a ⎜ ⎟ + a ⎜ 2 ⎟ ⎥ + b ⎜ ⎟ + cy = 0.

⎜

⎟

⎝ dz ⎠

⎢⎣ ⎝ dz ⎠

⎝ dx ⎠ ⎥⎦

⎛ d2y ⎞

⎛ dy ⎞

Therefore, a ⎜

⎟ + (b – a ) ⎜ ⎟ + cy = 0.

⎜ dz 2 ⎟

⎝ dz ⎠

⎝

⎠

23. We need to show that y ''+ a1 y '+ a2 y = 0 if r1 and

r2 are distinct real roots of the auxiliary equation.

We have,

y ' = C1r1er1 x + C2 r2 er2 x

.

y '' = C1r12 er1 x + C2 r2 2 e r2 x

When put into the differential equation, we

obtain

y ''+ a1 y '+ a2 y = C1r12 er1x + C2 r2 2 e r2 x

(

)

(

)

+ a1 C1r1er1 x + C2 r2 er2 x + a2 C1er1 x + C2 er2 x (*)

The solutions to the auxiliary equation are given

by

1

r1 = −a1 − a12 − 4a2 and

2

1

r2 = −a1 + a12 − 4a2 .

2

Putting these values into (*) and simplifying

yields the desired result: y ''+ a1 y '+ a2 y = 0 .

(

(

)

)

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24. We need to show that y ''+ a1 y '+ a2 y = 0 if α ± β i are complex conjugate roots of the auxiliary equation. We

have,

y ' = eα x ( (α C1 + β C2 ) cos ( β x ) + (α C2 − β C1 ) sin ( β x ) )

.

y '' = eα x α 2C1 − β 2C1 + 2αβ C2 cos ( β x ) + α 2C2 − β 2C2 − 2αβ C1 sin ( β x )

((

)

(

When put into the differential equation, we obtain

((

)

)

)

(

)

y ''+ a1 y '+ a2 y = eα x α 2C1 − β 2C1 + 2αβ C2 cos ( β x ) + α 2C2 − β 2C2 − 2αβ C1 sin ( β x )

(

)

)

+ a1eα x ( (α C1 + β C2 ) cos ( β x ) + (α C2 − β C1 ) sin ( β x ) ) + a2 C1eα x cos ( β x ) + C2 eα x sin ( β x ) (*)

From the solutions to the auxiliary equation, we find that

−a

1

α = 1 and β = − i a12 − 4a2 .

2

2

Putting these values into (*) and simplifying yields the desired result: y ''+ a1 y '+ a2 y = 0 .

25. a.

⎛ b 2 b 4 b6 ⎞

⎛

⎞

(bi )2 (bi )3 (bi ) 4 (bi )5

b 3 b5 b 7

+

+

+

+… = ⎜ 1 –

+

–

+ ⎟ …+ i ⎜ b –

+

–

+…⎟

⎜

⎟

⎜

⎟

2!

3!

4!

5!

2! 4! 6! ⎠

3! 5! 7!

⎝

⎝

⎠

= cos (b) + i sin (b)

ebi = 1 + (bi ) +

b.

e a +bi = ea ebi = e a [cos(b) + i sin(b)]

c.

Dx ⎡e(α + β i ) x ⎤ = Dx [eα x (cos β x + i sin β x)] = α eα x (cos β x + i sin β x) + eα x (– i β sin β x + i β cos β x )

⎣

⎦

= eα x [(α + β i ) cos β x + (α i – β ) sin β x]

(α + β )e(α + β i ) x = (α + β i )[eα x (cos β x + i sin β x)] = eα x [(α + β i ) cos β x + (α i – β ) sin β x]

Therefore, Dx [e(α + β i ) x ] = (α + i β )e(α + β i ) x

26. c1e(α + β i ) x + c2 e(α + β i ) x [c1 and c2 are complex constants.]

= c1eα x [cos β x + i sin β x] + c2 eα x [cos(– β x) + i sin(– β x)] = eα x [(c1 + c2 ) cos β x + (c1 – c2 )i sin β x]

= eα x [C1 cos β x + C2 sin β x], where C1 = c1 + c2 , and C2 = c1 – c2 .

Note: If c1 and c2 are complex conjugates, then C1 and C2 are real.

27. y = 0.5e5.16228 x + 0.5e –1.162278 x

28. y = 3.5 xe –2.5 x + 2e –2.5 x

29. y = 1.29099e –0.25 x sin(0.968246 x)

30. y = e0.333333 x [2.5cos(0.471405 x) – 4.94975sin(0.471405 x)]

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15.2 Concepts Review

1. particular solution to the nonhomogeneous equation; homogeneous equation

2. –6 + C1e –2 x + C2 e3 x

3. y = Ax 2 + Bx + C

1x

4. y = Bxe 3

Problem Set 15.2

1. yh = C1e –3 x + C2 e3 x

⎛ 1⎞

yp = ⎜ – ⎟ x + 0

⎝ 9⎠

⎛ 1⎞

y = ⎜ – ⎟ x + C1e –3 x + C2 e3 x

⎝ 9⎠

2. yh = C1e –3 x + C2 e 2 x

⎛ 1⎞

⎛ 1⎞

⎛ 7 ⎞

y p = ⎜ – ⎟ x2 + ⎜ – ⎟ x + ⎜ – ⎟

3

9

⎝

⎠

⎝

⎠

⎝ 54 ⎠

⎛ 1⎞

⎛1⎞

⎛ 7 ⎞

y = ⎜ – ⎟ x 2 – ⎜ ⎟ x – ⎜ ⎟ + C1e –3 x + C2 e2 x

⎝ 3⎠

⎝9⎠

⎝ 54 ⎠

3. Auxiliary equation: r 2 – 2r + 1 = 0 has roots 1, 1.

yh = (C1 + C2 x)e x

Let y p = Ax 2 + Bx + C ; y ′p = 2 Ax + B;

y ′′p = 2 A .

Then (2 A) – 2(2 Ax + B ) + ( Ax 2 + Bx + C ) = x 2 + x.

Ax 2 + (–4 A + B ) x + (2 A – 2 B + C ) = x 2 + x

Thus, A = 1, –4A + B = 1, 2A – 2B + C = 0, so

A = 1, B = 5, C = 8.

General solution: y = x 2 + 5 x + 8 + (C1 + C2 x)e x

4. yh = C1e – x + C2 ⋅ y p = 2 x 2 + (–4) x

y = 2 x 2 – 4 x + C1e – x + C2

⎛1⎞

5. yh = C1e2 x + C2 e3 x ⋅ y p = ⎜ ⎟ e x ⋅ y

⎝2⎠

1

⎛ ⎞

= ⎜ ⎟ e x + C1e 2 x + C2 e3 x

⎝2⎠

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6. Auxiliary equation: r 2 + 6r + 9 = 0, (r + 3)2 = 0 has roots –3, –3.

yh = (C1 + C2 x )e –3 x

Let y p = Be – x ; y ′p = – Be – x ; y ′′p = Be – x .

Then ( Be – x ) + 6(– Be – x ) + 9( Be – x ) = 2e – x ; 4 Be – x = 2e – x ; B =

1

2

⎛1⎞

General solution: y = ⎜ ⎟ e – x + (C1 + C2 x)e –3 x

⎝2⎠

7. yh = C1e –3 x + C2 e – x

⎛ 1⎞

y p = ⎜ – ⎟ xe –3 x

⎝ 2⎠

⎛ 1 ⎞ –3 x

y = ⎜ – ⎟ xe

+ C1e –3 x + C2 e – x

⎝ 2⎠

8. yh = e – x (C1 cos x + C2 sin x)

⎛3⎞

y p = ⎜ ⎟ e –2 x

⎝2⎠

⎛3⎞

y = ⎜ ⎟ e –2 x + e – x (C1 cos x + C2 sin x)

⎝2⎠

9. Auxiliary equation: r 2 – r – 2 = 0,

(r + 1)(r – 2) = 0 has roots –1, 2.

yh = C1e – x + C2 e2 x

Let y p = B cos x + C sin x; y ′p = – B sin x + C cos x; y ′′p = – B cos x – C sin x.

Then (− B cos x − C sin x) − (− B sin x + C cos x)

−2( B cos x + C sin x) = 2sin x.

1

–3

(−3B − C ) cos x + ( B − 3C ) sin x = 2sin x , so − 3B − C = 0 so –3B – C = 0 and B − 3C = 2 ; B = ; C = .

5

5

⎛1⎞

⎛3⎞

General solution: ⎜ ⎟ cos x – ⎜ ⎟ sin x + C1e2 x + C2 e – x

5

⎝ ⎠

⎝5⎠

10. yh = C1e –4 x + C2

⎛ 1⎞

⎛ 4⎞

y p = ⎜ – ⎟ cos x + ⎜ ⎟ sin x

⎝ 17 ⎠

⎝ 17 ⎠

1

⎛

⎞

⎛ 4⎞

y = ⎜ – ⎟ cos x + ⎜ ⎟ sin x + C1e –4 x + C2

⎝ 17 ⎠

⎝ 17 ⎠

11. yh = C1 cos 2 x + C2 sin 2 x

⎛1⎞

y p = (0) x cos 2 x + ⎜ ⎟ x sin 2 x

⎝2⎠

⎛1⎞

y = ⎜ ⎟ x sin 2 x + C1 cos 2 x + C2 sin 2 x

⎝2⎠

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12. Auxiliary equation: r 2 + 9 = 0 has roots ±3i, so yh = C1 cos 3x + C2 sin 3 x.

Let y p = Bx cos 3 x + Cx sin 3 x; y ′p = (–3bx + C ) sin 3 x + ( B + 3Cx) cos 3x ;

y ′′p = (–9 Bx + 6C ) cos 3 x + (–9Cx – 6 B ) sin 3x .

Then substituting into the original equation and simplifying, obtain 6C cos 3 x - 6 B sin 3x = sin 3x , so C = 0 and

1

B=– .

6

⎛ 1⎞

General solution: y = ⎜ – ⎟ x cos 3x + C1 cos 3x + C2 sin 3x

⎝ 6⎠

13. yh = C1 cos 3 x + C2 sin 3x

⎛1⎞

⎛1⎞

y p = (0) cos x + ⎜ ⎟ sin x + ⎜ ⎟ e2 x

8

⎝ ⎠

⎝ 13 ⎠

⎛1⎞

⎛1⎞

y = ⎜ ⎟ sin x + ⎜ ⎟ e2 x + C1 cos 3x + C2 sin 3 x

⎝8⎠

⎝ 13 ⎠

14. yh = C1e – x + C2

⎛1⎞

⎛3⎞

y p = ⎜ ⎟ e x + ⎜ ⎟ x 2 + (–3) x

⎝2⎠

⎝2⎠

⎛1⎞

⎛3⎞

y = ⎜ ⎟ e x + ⎜ ⎟ x 2 – 3 x + C1e – x + C2

⎝2⎠

⎝2⎠

15. Auxiliary equation: r 2 – 5r + 6 = 0 has roots 2 and 3, so yh = C1e2 x + C2 e3 x .

Let y p = Be x ; y ′p = Be x ; y ′′p = Be x .

Then ( Be x ) – 5( Be x ) + 6( Be x ) = 2e x ; 2 Be x = 2e x ; B = 1 .

General solution: y = e x + C1e2 x + C2 e3 x

y ′ = e x + 2C1e2 x + 3C2 e3 x

If x = 0, y = 1, y ′ = 0, then 1 = 1 + C1 + C2 and 0 = 1 + 2C1 + 3C2 ; C1 = 1, C2 = –1.

Therefore, y = e x + e2 x – e3 x .

16. yh = C1e –2 x + C2 e2 x

⎛ 4⎞

y p = (0) cos x + ⎜ – ⎟ sin x

⎝ 5⎠

⎛ 4⎞

y = ⎜ – ⎟ sin x + C1e –2 x + C2 e2 x

⎝ 5⎠

⎛ 4⎞

⎛9⎞

⎛ 11 ⎞

y = ⎜ – ⎟ sin x + ⎜ ⎟ e –2 x + ⎜ ⎟ e2 x satisfies the conditions.

⎝ 5⎠

⎝5⎠

⎝5⎠

17. yh = C1e x + C2 e2 x

⎛1⎞

y p = ⎜ ⎟ (10 x + 19)

⎝4⎠

⎛1⎞

y = ⎜ ⎟ (10 x + 19) + C1e x + C2 e2 x

⎝4⎠

896

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18. Auxiliary equation: r 2 – 4 = 0 has roots 2, –2, so yh = C1e2 x + C2 e –2 x .

Let y p = v1e2 x + v2 e –2 x , subject to v1′e2 x + v2′ e –2 x = 0, and v1′ (2e2 x ) + v2′ (–2e –2 x ) = e2 x .

Then v1′ (4e2 x ) = e2 x and v2′ (–4e –2 x ) = e2 x ; v1′ =

e4 x

1

x

and v2′ = – e4 x / 4 ; v1 = and v2 = –

.

16

4

4

xe2 x e2 x

+ C1e2 x + C2 e –2 x

–

4

16

General solution: y =

19. yh = C1 cos x + C2 sin x

y p = – cos ln sin x – cos x – x sin x

y = – cos x ln sin x – x sin x + C3 cos x + C2 sin x (combined cos x terms)

20. yh = C1 cos x + C2 sin x

y p = – sin x ln csc x + cot x

y = – sin x ln csc x + cot x + C1 cos x + C2 sin x

21. Auxiliary equation: r 2 – 3r + 2 = 0 has roots 1, 2, so yh = C1e x + C2 e2 x .

Let y p = v1e x + v2 e2 x subject to v1′e x + v2′ e2 x = 0, and v1′ (e x ) + v2′ (2e2 x ) = e x (ex + 1) –1.

Then v1′ =

–ex

e (e + 1)

x

x

so v1 = ∫

–e x

e (e + 1)

x

x

dx = ∫

–1

du

u (u + 1)

1 ⎞

ex +1

⎛ –1

⎛ u +1⎞

–x

= ∫⎜ +

⎟ du = – ln u + ln(u + 1) = ln ⎜

⎟ = ln x = ln(1 + e )

+

u

u

1

u

⎝

⎠

⎝

⎠

e

v2′ =

ex

so v2 = – e – x + ln(1 + e – x )

e (e + 1)

(similar to finding v1 )

2x

x

General solution: y = e x ln(1 + e – x ) – e x + e2 x ln(1 + e – x ) + C1e x + C2 e2 x

y = (e x + e2 x ) ln(1 + e – x ) + D1e x + D2 e2 x

22. yh = C1e2 x + C2 e3 x ; y p = e x

y = e x + C1e2 x + C2 e3 x

23. L( y p ) = (v1u1 + v2u2 )′′ + b(v1u1 + v2u2 )′ + c(v1u1 + v2 u2 )

= (v1′u1 + v1u1′ + v2′ u2 + v2u2′ ) + b(v1′u1 + v1u1′ + v2′ u2 + v2 u2′ ) + c(v1u1 + v2 u2 )

= (v1′′u1 + v1′u1′ + v1′u1′ + v1u1′′ + v2′′u2 + v2′ u2′ + v2′ u2′ + v2u2′′ ) + b(v1′u1 + v1u1′ + v2′ u2 + v2u2′ ) + c(v1u1 + v2 u2 )

= v1 (u1′′ + bu1′ + cu1 ) + v2 (u2′′ + bu2′ + cu2 ) + b(v1′u1 + v2′ u2 ) + (v1′′u1 + v1′u1′ + v2′′u2 + v2′ u2 ) + (v1′u1′ + v2′ u2′ )

= v1 (u1′′ + bu1′ + cu1 ) + v2 (u2′′ + bu2′ + cu2 ) + b(v1′u1 + v2′ u2 ) + (v1′u1 + v2′ u2 )′ + (v1′u1′ + v2′ u2′ )

= v1 (0) + v2 (0) + b(0) + (0) + k ( x) = k ( x)

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24. Auxiliary equation: r 2 + 4 = 0 has roots ±2i.

yh = C1 cos 2 x + C2 sin 2 x

Now write sin 3 x in a form involving sin βx’s or

cos βx’s.

3

1

sin 3 x = sin x – sin 3x

4

4

(C.R.C. Standard Mathematical Tables, or derive it using half-angle and product identities.)

Let y p = A sin x + B cos x + C sin 3 x + D cos 3 x ;

y ′p = A cos x – B sin x + 3C cos 3x – 3D sin 3 x ;

y ′′p = – A sin x – B cos x – 9C sin 3 x – 9 D cos 3 x .

Then

3

1

y ′′p + 4 y p = 3 A sin x + 3B cos x – 5C sin 3 x – 5 D cos 3 x = sin x – sin 3x, so

4

4

1

1

A = , B = 0, C = , D = 0.

20

4

1

1

General solution: y = sin x + sin 3 x + C1 cos 2 x + C2 sin 2 x

4

20

15.3 Concepts Review

1. 3; π

2. π ; decreases

3. 0

4. electric circuit

Problem Set 15.3

1. k = 250, m = 10, B 2 = k / m = 250 /10 = 25, B = 5

(the problem gives the mass as m = 10 kg )

Thus, y '' = −25 y. The general solution is y = C1 cos 5t + C2 sin 5t. Apply the initial condition to get y = 0.1cos 5t.

The period is

2π

seconds.

5

2. k = 100 lb/ft, w = 1 lb, g = 32 ft/s2, y0 =

1

ft,

12

⎛1⎞

B = 40 2 . Then y = ⎜ ⎟ cos(40 2)t.

⎝ 12 ⎠

1

ft = 1 in.

Amplitude is

12

2π

≈ 0.1111 s.

Period is

40 2

3. y = 0.1cos 5t = 0 whenever 5t =

π

2

+ π k or t =

π

10

+

π

5

k.

⎛π π ⎞

⎛π π ⎞

⎛π

⎞

y ' ⎜ + k ⎟ = 0.5 sin 5 ⎜ + k ⎟ = 0.5 sin ⎜ + π k ⎟ = 0.5 meters per second

10

5

10

5

2

⎝

⎠

⎝

⎠

⎝

⎠

898

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⎛1⎞

4. 10 = k ⎜ ⎟ , so k = 30 lb/ft, w = 20 lb,

⎝3⎠

g = 32 ft/s2, y0 = –1 ft, v0 = 2 ft/s, B = 4 3

Then y = C1 cos(4 3t ) + C2 sin(4 3t ).

⎛ 3t ⎞

y = cos(4 3t ) + ⎜⎜

⎟⎟ sin(4 3t ) satisfies the initial conditions.

⎝ 6 ⎠

5. k = 20 lb/ft; w = 10 lb; y0 = 1 ft; q =

1

s-lb/ft, B = 8, E = 0.32

10

E 2 – 4 B 2 < 0, so there is damped motion. Roots of auxiliary equation are approximately –0.16 ± 8i.

General solution is y ≈ e –0.16t (C1 cos8t + C2 sin 8t ). y ≈ e –0.16t (cos8t + 0.02sin 8t ) satisfies the initial conditions.

6. k = 20 lb/ft; w = 10 lb; y0 = 1 ft; q = 4 s-lb/ft

(20)(32)

(4)(32)

= 8; E =

= 12.8; E 2 – 4 B 2 < 0, so damped motion.

10

10

B=

Roots of auxiliary equation are

– E ± E 2 – 4B2

= –6.4 ± 4.8i.

2

General solution is y = e –6.4t (C1 cos 4.8t + C2 sin 4.8t ).

y ′ = e –6.4t (–4.8C1 sin 4.8t + 4.8C2 cos 4.8t ) – 6.4e –6.4t (C1 cos 4.8t + C2 sin 4.8t )

4

If t = 0, y = 1, y ′ = 0, then 1 = C1 and 0 = 4.8C2 – 6.4C1 , so C1 = 1 and C2 = .

3

4

⎡

⎤

⎛

⎞

Therefore, y = e –6.4t ⎢cos 4.8t + ⎜ ⎟ sin 4.8t ⎥ .

⎝3⎠

⎣

⎦

7. Original amplitude is 1 ft. Considering the contribution of the sine term to be negligible due to the 0.02 coefficient,

the amplitude is approximately e –0.16t .

e –0.16t ≈ 0.1 if t ≈ 14.39 , so amplitude will be about one-tenth of original in about 14.4 s.

8. C1 = 1 and C2 = –0.105, so y = e –0.16t (cos8t + 0.105sin 8t ).

9. LQ′′ + RQ′ +

Q

= E (t ); 106 Q ′ + 106 Q = 1; Q′ + Q = 10 –6

C

Integrating factor: et

D[Qet ] = 10 –6 et ; Qet = 10 –6 et + C ;

Q = 10 –6 + Ce – t

If t = 0, Q = 0, then C = –10 –6.

Therefore, Q(t ) = 10 –6 – 10 –6 e – t = 10 –6 (1 – e – t ).

10. Same as Problem 9, except C = 4 – 10 –6 , so Q(t ) = 10 –6 + (4 – 10 –6 )e – t .

Then I (t ) = Q ′(t ) = –(4 – 10 –6 )e – t .

11.

Q

[2(10 –6 )]

= 120sin 377t

a.

Q(t) = 0.00024 sin 377t

b.

I (t ) = Q ′(t ) = 0.09048cos 377t

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12. LQ′′ + RQ ′ +

Q

Q

= E; 10 –2 Q′′ +

= 20; Q′′ + 109 Q = 2000

C

10 –7

The auxiliary equation, r 2 + 109 = 0, has roots ±109 / 2 i.

Qh = C1 cos109 / 2 t + C2 sin109 / 2 t

Q p = 2000(10 –9 ) = 2(10 –6 ) is a particular solution (by inspection).

General solution: Q(t ) = 2(10 –6 ) + C1 cos109 / 2 t + C2 sin109 / 2 t

Then I (t ) = Q′(t ) = –109 / 2 C1 sin109 / 2 t + 109 / 2 C2 cos109 / 2 t.

If t = 0, Q = 0, I = 0, then 0 = 2(10 –6 ) + C1 and 0 = C2 .

Therefore, I (t ) = –109 / 2 (–2[10 –6 ]) sin109 / 2 t = 2(10 –3 / 2 ) sin109 / 2 t.

13. 3.5Q′′ + 1000Q +

Q

= 120sin 377t

[2(10 –6 )]

(Values are approximated to 6 significant figures for the remainder of the problem.)

Q′′ + 285.714Q ′ + 142857Q = 34.2857 sin 377t

Roots of the auxiliary equation are

–142.857 ± 349.927i.

Qh = e –142.857t (C1 cos 349.927t + C2 sin 349.927t )

Q p = –3.18288(10 –4 ) cos 377t + 2.15119(10 –6 ) sin 377t

Then, Q = –3.18288(10 –4 ) cos 377t + 2.15119(10 –6 ) sin 377t + Qh .

I = Q ′ = 0.119995sin 377t + 0.000810998cos 377t + Qh′

0.000888 cos 377t is small and Qh′ → 0 as t → ∞, so the steady-state current is I ≈ 0.12sin 377t .

14. a. Roots of the auxiliary equation are ±Bi.

yh = C1 cos Bt + C2 sin Bt.

⎡

⎤

c

yp = ⎢

⎥ sin At

2

2

⎣⎢ ( B – A ) ⎦⎥

The desired result follows.

b.

⎛ c ⎞

yp = ⎜ –

⎟ t cos Bt , so

⎝ 2B ⎠

⎛ c ⎞

y = C1 cos Bt + C2 sin Bt – ⎜

⎟ t cos Bt.

⎝ 2B ⎠

c.

900

Due to the t factor in the last term, it

increases without bound.

Section 15.3

15. A sin( β t + γ ) = A(sin β t cos γ + cos β t sin γ )

= ( A cos γ ) sin β t + ( A sin γ ) cos β t

= C1 sin β t + C2 cos β t , where C1 = A cos γ and

C2 = A sin γ .

[Note that

C12 + C22 = A2 cos 2 γ + A2 sin 2 γ = A2 .)

16. The first two terms have period

2π

and the last

B

2π

. Then the sum of the three terms

A

⎛ 2π ⎞

⎛ 2π ⎞

is periodic if m ⎜ ⎟ = n ⎜ ⎟ for some integers

B

⎝ ⎠

⎝ B ⎠

B m

m, n; equivalently, if = , a rational number.

A n

has period

Instructor’s Resource Manual

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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

17. The magnitudes of the tangential components of

the forces acting on the pendulum bob must be

equal.

15.4 Chapter Review

Concepts Test

1. False:

y 2 is not linear in y.

2. True:

y and y ′′ are linear in y and y ′′,

respectively.

3. True:

y ′ = sec 2 x + sec x tan x

2 y ′ – y 2 = (2sec 2 x + 2sec x tan x)

Therefore, – m

d 2s

s = Lθ, so

dt 2

d 2s

dt 2

=L

Therefore, – mL

–(tan 2 x + 2sec x tan x + sec2 x)

= mg sin θ .

d 2θ

dt 2

d 2θ

dt 2

= sec2 x – tan 2 x = 1

.

= mg sin θ .

4. False:

It should involve 6.

5. True:

D 2 adheres to the conditions for

linear operators.

D 2 (kf ) = kD 2 ( f )

d θ

2

g

Hence,

= – sin θ .

2

L

dt

18. a.

D2 ( f + g ) = D2 f + D2 g

Since the roots of the auxiliary equation are

g

⎛g⎞

± i , the solution of θ ′′(t ) + ⎜ ⎟θ = 0 is

L

⎝L⎠

g

g

t + C2 sin

t , which can be

L

L

⎛ g

⎞

t + γ ⎟⎟

written as θ = C ⎜⎜

⎝ L

⎠

(by Problem 15).

The period of this function is

6. False:

Replacing y by C1u1 ( x) + C2u2 ( x)

would yield, on the left side,

C1 f ( x) + C2 f ( x) = (C1 + C2 ) f ( x)

which is f(x) only if C1 + C2 = 1 or

f(x) = 0.

7. True:

–1 is a repeated root, with multiplicity

3, of the auxiliary equation.

8. True:

L(u1 – u2 ) = L(u1 ) – L(u2 )

= f ( x) – f ( x) = 0

9. False:

That is the form of yh . y p should

θ = C1 cos

2π

g

L

= 2π

L

G

= 2π

LR 2

L

= 2πR

.

GM

GM

2πR1

p

Therefore, 1 =

p2 2πR

2

L1

GM

L2

GM

have the form Bx cos 3x + Cx sin 3x.

=

R1 L1

R2 L2

.

b. To keep perfect time at both places, require

R 80.85

p1 = p2 . Then 1 = 2

, so

3960 81

R2 ≈ 3963.67.

The height of the mountain is about

3963.67 - 3960 = 3.67 mi (about 19,387 ft).

10. True:

See Problem 15, Section 15.3.

Sample Test Problems

1. u ′ + 3u = e x . Integrating factor is e3 x .

D[ue3 x ] = e 4 x

⎛1⎞

y = ⎜ ⎟ e x + C1e –3 x

⎝4⎠

⎛1⎞

y ′ = ⎜ ⎟ e x + C1e –3 x

⎝4⎠

⎛1⎞

y = ⎜ ⎟ e x + C3e –3 x + C2

⎝4⎠

2. Roots are –1, 1. y = C1e – x + C2 e x

Instructor’s Resource Manual

Section 15.4

901

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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

3. (Second order homogeneous)

12. (Fourth-order homogeneous)

The auxiliary equation, r 2 – 3r + 2 = 0, has roots

The auxiliary equation, r 4 – 3r 2 – 10 = 0 or

1, 2. The general solution is y = C1e x + C2 e2 x .

(r 2 – 5)(r 2 + 2) = 0, has roots – 5, 5, ± 2i.

General solution:

y ′ = C1e x + 2C2 e2 x

If x = 0, y = 0, y ′ = 3, then 0 = C1 + C2 and

3 = C1 + 2C2 , so C1 = –3, C2 = 3.

Therefore, y = –3e x + 3e2 x .

3

4. Repeated root – . y = (C1 + C2 x)e(–3 / 2) x

2

5. yh = C1e – x + C2 e x (Problem 2)

y = C1e

+ C2 e –

5x

+ C3 cos 2 x + C4 sin 2 x

13. Repeated roots ± 2

y = (C1 + C2 x)e –

2x

+ (C3 + C4 x)e

14. a.

Q′(t ) = 3 – 0.02Q

b.

Q′(t ) + 0.02Q = 3

2x

Integrating factor is e0.02t

y p = –1 + C1e – x + C2 e x

D[Qe0.02t ] = 3e0.02t

Q(t ) = 150 + Ce –0.02t

6. (Second-order nonhomogeneous) The auxiliary

equation, r + 4r + 4 = 0, has roots –2, –2.

2

Q(t ) = 150 – 30e –0.02t goes through (0, 120).

yh = C1e –2 x + C2 xe –2 x = (C1 + C2 x)e –2 x

Let y p = Be x ; y ′p = Be x ; y ′′p = Be x .

1

( Be x ) + 4( Be x ) + 4( Be x ) = 3e x , so B = .

3

General solution: y =

5x

ex

+ (C1 + C2 x)e –2 x

3

7. yh = (C1 + C2 x)e –2 x (Problem 12)

⎛1⎞

y p = ⎜ ⎟ x 2 e –2 x

⎝2⎠

⎡⎛ 1 ⎞

⎤

y = ⎢⎜ ⎟ x 2 + C1 + C2 x ⎥ e –2 x

2

⎝

⎠

⎣

⎦

8. Roots are ±2i.

y = C1 cos 2 x + C2 sin 2 x

y = sin 2x satisfies the conditions.

9. (Second-order homogeneous)

The auxiliary equation, r + 6r + 25 = 0, has

roots –3 ± 4i. General solution:

2

y = e –3 x (C1 cos 4 x + C2 sin 4 x)

10. Roots are ±i. yh = C1 cos x + C2 sin x

c.

Q → 150 g, as t → ∞ .

15. (Simple harmonic motion)

k = 5; w = 10; y0 = –1

(5)(32)

=4

10

Then the equation of motion is y = –cos 4t.

2π π

= .

The amplitude is –1 = 1; the period is

4 2

B=

16. It is at equilibrium when y = 0 or –cos 4t = 0, or

π 3π

t= ,

, ….

8 8

y ′(t ) = 4sin 4t , so at equilibrium y ′ = ±4 = 4.

17. Q′′ + 2Q ′ + 2Q = 1

Roots are –1 ± i.

1

Qh = e – t (C1 cos t + C2 sin t ) and Q p = ;

2

1

Q = e – t (C1 cos t + C2 sin t ) +

2

–

t

I (t ) = Q ′(t ) = – e [(C1 – C2 ) cos t + (C1 + C2 ) sin t ]

I (t ) = e – t sin t satisfies the initial conditions.

y p = x cos x – sin x + sin x ln cos x

y = x cos x − sin x ln cos x + C1 cos x + C3 sin x

(combining the sine terms)

11. Roots are –4, 0, 2. y = C1e –4 x + C2 + C3e2 x

902

Section 15.4

Instructor’s Resource Manual

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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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