Calculo

CHAPTER

15

Differential Equations
7. Roots are 2 ± 3. General solution is
y = e 2 x (C1e
3x

15.1 Concepts Review
1. r 2 + a1r + a2 = 0; complex conjugate roots 2. C1e – x + C2 e x 3. (C1 + C2 x)e
x

+ C2 e –

3x

).

8. Roots are –3 ± 11. General solution is
y = e –3 x C1e 11x + C2 e – 11x .

(

)

4. C1 cos x + C2 sin 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 = y = 3, then 3 = C2 . Therefore, y = 2 cos 2 x + 3sin 2 x .
π and 4

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 .

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).

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 = –
e x – e –7 x Therefore, y = . 2

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 ⎜ ⎜ 2 ⎟ x + C2 e ⎟ ⎜ 2 ⎟x ⎟ ⎝ ⎠ ⎝ ⎠
⎡ ⎛ 3⎞ ⎛ 3⎞ ⎤ y = e – x / 2 ⎢C1 cos ⎜ ⎜ 2 ⎟ x + C2 sin ⎜ 2 ⎟ x ⎥ ⎟ ⎜ ⎟ ⎢ ⎝ ⎠ ⎝ ⎠ ⎥ ⎣ ⎦

1 1 and C2 = . 2 2

4. Roots are –2 and 5. General solution is
y = C1e
–2 x

+ C2 e . Particular solution is
5x

⎛ 12 ⎞ ⎛5⎞ y = ⎜ ⎟ e5 x – ⎜ ⎟ e –2 x . 7⎠ ⎝ ⎝7⎠13. Roots are 0, 0, –4, 1. General solution is
y = C1 + C2 x + C3e –4 x + C4 e x .

5. Repeated root 2. General solution is
y = (C1 + C2 x)e 2 x .

14. Roots are –1, 1, ±i. General solution is
y = C1e – x + C2 e x + C3 cos x + C4 sin x.

6. Auxiliary equation:
r 2 + 10r + 25 = 0, (r + 5)2 = 0 has one repeated root −5 .

15. Auxiliary equation: r 4 + 3r 2 – 4 = 0,
(r + 1)(r – 1)(r2 + 4) = 0 has roots –1, 1, ±2i. General solution: y = C1e – x + C2 e x + C3 cos 2 x + C4 sin 2 x

General solution: y = C1e –5 x + C2 xe –5 x or
y = (C1 + C2 x)e –5 x

Instructor’s Resource Manual

Section 15.1

891

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion ofthis material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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 x18. eu = cosh u + sinh u and e – u = cosh u – sinh u. Auxiliary equation: r 2 – 2br – c 2 = 0
2b ± 4b 2 + 4c 2 = b ± b2 + c2 2
b2 + c2 ) x

Roots of auxiliary equation: General solution: y = C1e(b +

+ 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 ⎜ ⎜ 2 ⎟ ⎜ 2 ⎟ x⎥ . ⎟ ⎢ ⎝ ⎠ ⎝ ⎠ ⎥ ⎣ ⎦

20. Roots 1 ± i. Generalsolution 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 ′′ =
=

22. As done in Problem 21, ⎡ ⎛ dy ⎞ ⎛ d 2 y ⎞⎤ ⎛ dy ⎞ ⎢ – a ⎜ ⎟ + a ⎜ 2 ⎟ ⎥ + b ⎜ ⎟ + cy = 0. ⎜ dx ⎟ ⎥ ⎝ dz ⎠ ⎢ ⎝ dz ⎠ ⎝ ⎠⎦ ⎣ ⎛ d2y ⎞ ⎛ dy ⎞ Therefore, a ⎜ ⎟ + (b – a ) ⎜...