Page 849 - Advanced_Engineering_Mathematics o'neil

Answers to Selected Problems 829

t
29. If A is hermitian then A = A ,so

t
t
(AA ) = A(A )
= A(A) = (A)A.

t
31. If S =−S,theneach s jj =−s jj . Write s jj = a jj + ib jj .Then a jj =−a jj ,so a jj = 0and s jj is pure imaginary.
CHAPTER TEN SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS
Section 10.1 Linear Systems
The following solutions give the fundamental matrix (t) and the solution of the initial value problem.
2t 6t 2t 6t
−e 3e −3e + 3e
1. (t) = 2t 6t ; X(t) = 2t 6t
e e 3e + e
√ √
(1+2 3)t (1−2 3)t
4e 4e
3. (t) = √ √ √ √
(−1 + 3)e (1+2 3)t (−1 − 3)e (1−2 3)t
√ √ √
(1+2 3)t (1−2 3)t
√
(1 + 5 3/3)e + (1 − 5 3/3)e
√
√
X(t) = √ (1+2 3)t √ (1−2 3)t
(−1 + 3/6)e + (1 + 3/6)e
⎛ t −3t ⎞ ⎛ t −3t ⎞
e 0 e 10e − 9e
t
5. (t) = ⎝ 0 e t 3e −3t ⎠ ; X(t) = ⎝ 24e − 27e −3t ⎠
t
e −t e t e −3t 14e − 9e −3t
Section 10.2 Solution of X = AX for Constant A

3t 3t
7e 0 7c 1 e
1. (t) = 3t −4t ; X(t) = (t)C = 3t −4t
5e e 5c 1 e + c 2 e

1 e 2t c 1 + c 2 e 2t
3. (t) = 2t ; X(t) = 2t
−1 e −c 1 + c 2 e
⎛ 3t −4t ⎞ ⎛ 3t −4t ⎞
1 2e −e c 1 + 2c 2 e − c 3 e
3t
5. (t) = ⎝ 6 3e 3t 2e −4t ⎠ ; X(t) = ⎝ 6c 1 + 3c 2 e + 2c 2 e −4t ⎠
3t
−13 −2e 3t e −4t −13c 1 − 2c 2 e + c 3 e −4t
4t −3t 4t −3t
2e e 6e − 5e
7. (t) = 4t −3t ; X(t) = 4t −3t
−3e 2e −9e − 10e
⎛ 2t 3t ⎞ ⎛ 2t 3t ⎞
0 e 3e 4e − 3e
3t
2t
3t
9. (t) = ⎝ 1 e 2t e ⎠ ; X(t) = ⎝ 2 + 4e − e ⎠
1 0 e 3t 2 − e 3t
2t 2t
2e cos(2t) 2e sin(2t)
11. (t) = 2t 2t
e sin(2t) −e cos(2t)

t
t
5e cos(t) 5e sin(t)
13. (t) = t t
e [2cos(t) + sin(t)] e [2sin(t) − cos(t)]
⎛ −t −t ⎞
0 e cos(2t) e sin(2t)
−t
−t
15. (t) = ⎝ 0 e [cos(2t) − 2sin(2t)] e [sin(2t) + 2cos(2t)] ⎠
−t
−t
e −2t 3e cos(2t) 3e sin(2t)
3t 3t
e 2te
17. (t) = 3t
0 e
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October 14, 2010 17:50 THM/NEIL Page-829 27410_25_Ans_p801-866