# Solution manual calculus 8th edition varberg, purcell, rigdon ch15

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

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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

891

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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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Section 15.1

893

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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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Section 15.2

895

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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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Section 15.2

897

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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

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

≈ 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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Section 15.3

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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

and the last
B

. 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

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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Section 15.4

901

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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

Section 15.4

Instructor’s Resource Manual