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806 Answers to Selected Problems

6

5
2
4
1
3

0 2
0 1 2 3 4 5
t
1
–1
0
0 2 4 6 8 10 12 14
t
FIGURE A.5 Graphs in Problem 9, Section 2.4
FIGURE A.6 Graphs in Problem 15, Section 2.4

1 √ 6582 √
15. (a) y(t) = e −3t 2266cosh( 7t) + √ sinh( 7t)
373 7
1
+ (−28cos(3t) + 72sin(3t))
373
1 √ 2106 √
(b) y(t) = e −3t 29cosh( 7t) + √ sinh( 7t)
373 7
1
+ (−28cos(3t) + 72sin(3t))
373
Graphs are shown in Figure A.6.
1 √ 74 √
17. (a) y(t) = e −t/2 98cos( 11t/2) + √ sin( 11t/2)
15 11
1
+ (−8cos(3t) + 4sin(3t))
15
1 √ 164 √
(b) y(t) = e −t/2 8cos( 11t/2) + √ sin( 11t/2)
15 11
1
+ (−8cos(3t) + 4sin(3t))
15
Graphs are given in Figure A.7.

Section 2.5 Euler’s Differential Equation
2
1. y = c 1 x + c 2 x −3 3. y = c 1 cos(2ln(x)) + c 2 sin(2ln(x))
4
5. y = c 1 x + c 2 x −4 7. y = c 1 x −2 + c 2 x −3
9. y = x −12 (c 1 + c 2 ln(x))
3 −7
7 x 3 x
11. y = +
10 2 10 2
6
2
13. y = x (4 − 3ln(x)) 15. y = 3x − 2x 4
t
2
t

17. Let x = e and Y(t) = y(x ). This transforms the Euler equation x y + Axy + By = 0 to the constant coefficient
equation Y (t) + (A − 1)Y (t) + BY(t) = 0. Solve this and then put t = ln(x).

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October 14, 2010 17:50 THM/NEIL Page-806 27410_25_Ans_p801-866