Page 830 - Advanced_Engineering_Mathematics o'neil

810 Answers to Selected Problems

1 1 2
x 2 (t) = − cos(2t) + cos(3t)
18 10 45

1 1 2
− − cos(2(t − 2)) + cos(3(t − 2)) H(t − 2)
18 10 45
17. The equations of motion are

m 1 y = k(y 2 − y 1 ),m 2 y = k(y 1 − y 2 )
1
2
with initial conditions y 1 (0) = y (0) = y (0) = 0, y 2 (0) = d. Transform these to find that Y 1 (s) and Y 2 (s) have

2
1
quadratic factors
m 1 + m 2
2
s + k
m 1 m 2
√
in their denominators, implying that the motion has frequency ω = (m 1 + m 2 )k/m 1 m 2 , hence period

m 1 m 2
2π .
m 1 + m 2
19. With E(t) = 5δ(t − 1),
1 −(t−1) −(t−1)/6
i 1 (t) = e + 3e H(t − 1)
10
1 −(t−1) −(t−1)/6
i 2 (t) = −e + e H(t − 1)
10
21. x 1 (t)= e −3t/50 + 9e −t/100 + 3(e −(t−3)/100 − e −3(t−3)/50 )H(t − 3)
x 2 (t)=−e −3t/50 + 6e −t/100 + (3e −(t−3)/50 + 2e −(t−3)/100 )H(t − 3)
Section 3.7 Polynomial Coefficients

1. With u = 1/t, obtain −z (u) − 2z = 2where z(u) = y(t(u)).Then y(t) =−1 + ce −2/t .
3. y = 7t 2
2 −t
5. y = ct e
7. y = 4
3 2
9. y = t
2
11. With W(s) being inverted, use is made of the formula for the Laplace transform of n!/s n+1 for n any nonnegative
integer.
CHAPTER FOUR SERIES SOLUTIONS
Section 4.1 Power Series Solutions
1
1. a 0 is arbitrary; a 1 = 1, 2a 2 − a 0 =−1, and a n = a n−2 for n = 3,4,···;
n
1 1 1
3
7
5
y(x) =a 0 + x + x + x + x +···
3 3 · 5 3 · 5 · 7
1 1 1

6
4
2
(a 0 − 1) x + x + x +···
2 2 · 4 2 · 4 · 6
3. a 0 is arbitrary; a 1 + a 0 = 0, 2a 2 + a 1 = 1,
1
a n+1 = (a n−2 − a n ) for n = 2,3,··· ,
n + 1
and

1 1 7
3
4
2
y(x) =a 0 1 − x + x + x − x +···
2! 3! 4!
1 1 1 11 31
6
5
3
4
2
+ x − x + x + x − x +···
2! 3! 4! 5! 6!
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October 14, 2010 17:50 THM/NEIL Page-810 27410_25_Ans_p801-866