AEM_3e_Chapter_10

# X t e a t c e a t t t e a s f s ds cosh t sinh t sinh

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X ( t ) = e A t C + e A t t t 0 e A s F ( s ) ds = cosh t sinh t sinh t cosh t c 1 c 2 + cosh t sinh t sinh t cosh t t 0 cosh s sinh s sinh s cosh s cosh s sinh s ds = c 1 cosh t + c 2 sinh t c 1 sinh t + c 2 cosh t + cosh t sinh t sinh t cosh t t 0 1 0 ds = c 1 cosh t + c 2 sinh t c 1 sinh t + c 2 cosh t + cosh t sinh t sinh t cosh t s 0 t 0 = c 1 cosh t + c 2 sinh t c 1 sinh t + c 2 cosh t + cosh t sinh t sinh t cosh t t 0 = c 1 cosh t + c 2 sinh t c 1 sinh t + c 2 cosh t + t cosh t t sinh t = c 1 cosh t sinh t + c 2 sinh t cosh t + t cosh t sinh t . 15. From s I A = s 4 3 4 s + 4 we find ( s I A ) 1 = 3 / 2 s 2 1 / 2 s + 2 3 / 4 s 2 3 / 4 s + 2 1 s 2 + 1 s + 2 1 / 2 s 2 + 3 / 2 s + 2 and e A t = 3 2 e 2 t 1 2 e 2 t 3 4 e 2 t 3 4 e 2 t e 2 t + e 2 t 1 2 e 2 t + 3 2 e 2 t . The general solution of the system is then X = e A t C = 3 2 e 2 t 1 2 e 2 t 3 4 e 2 t 3 4 e 2 t e 2 t + e 2 t 1 2 e 2 t + 3 2 e 2 t c 1 c 2 = c 1 3 2 1 e 2 t + c 1 1 2 1 e 2 t + c 2 3 4 1 2 e 2 t + c 2 3 4 3 2 e 2 t = 1 2 c 1 + 1 4 c 2 3 2 e 2 t + 1 2 c 1 3 4 c 2 1 2 e 2 t = c 3 3 2 e 2 t + c 4 1 2 e 2 t . 18. From s I A = s 1 2 s + 2 we find ( s I A ) 1 = s + 1 + 1 ( s + 1) 2 + 1 1 ( s + 1) 2 + 1 2 ( s + 1) 2 + 1 s + 1 1 ( s + 1) 2 + 1 189

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10.5 Matrix Exponential and e A t = e t cos t + e t sin t e t sin t 2 e t sin t e t cos t e t sin t . The general solution of the system is then X = e A t C = e t cos t + e t sin t e t sin t 2 e t sin t e t cos t e t sin t c 1 c 2 = c 1 1 0 e t cos t + c 1 1 2 e t sin t + c 2 0 1 e t cos t + c 2 1 1 e t sin t = c 1 cos t + sin t 2sin t e t + c 2 sin t cos t sin t e t . 21. The eigenvalues are λ 1 = 1 and λ 2 = 3. This leads to the system e t = b 0 b 1 e 3 t = b 0 + 3 b 1 , which has the solution b 0 = 3 4 e t + 1 4 e 3 t and b 1 = 1 4 e t + 1 4 e 3 t . Then e A t = b 0 I + b 1 A = e 3 t 2 e t + 2 e 3 t 0 e t . The general solution of the system is then X = e A t C = e 3 t 2 e t + 2 e 3 t 0 e t c 1 c 2 = c 1 1 0 e 3 t + c 2 2 1 e t + c 2 2 0 e 3 t = c 2 2 1 e t + ( c 1 + 2 c 2 ) 1 0 e 3 t = c 3 2 1 e t + c 4 1 0 e 3 t . 190
CHAPTER 10 REVIEW EXERCISES 24. From equation (3) in the text e D t = 1 0 · · · 0 0 1 · · · 0 . . . . . . . . . . . . 0 0 · · · 1 + λ 1 0 · · · 0 0 λ 2 · · · 0 . . . . . . . . . . . . 0 0 · · · λ n + 1 2! t 2 λ 2 1 0 · · · 0 0 λ 2 2 · · · 0 . . . . . . . . . . . . 0 0 · · · λ 2 n + 1 3! t 3 λ 3 1 0 · · · 0 0 λ 3 2 · · · 0 . . . . . . . . . . . . 0 0 · · · λ 3 n + · · · = 1 + λ 1 t + 1 2! ( λ 1 t ) 2 + · · · 0 · · · 0 0 1 + λ 2 t + 1 2!

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