AEM_3e_Chapter_10

# 30 from x 3 1 1 1 1 1 1 1 1 x t 2 e t 186 104

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30. From X = 3 1 1 1 1 1 1 1 1 X + 0 t 2 e t 186

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10.4 Nonhomogeneous Linear Systems we obtain X c = c 1 1 1 1 e t + c 2 1 1 0 e 2 t + c 3 1 0 1 e 2 t . Then Φ = e t e 2 t e 2 t e t e 2 t 0 e t 0 e 2 t and Φ 1 = e t e t e t e 2 t 0 e 2 t e 2 t e 2 t 0 so that U = Φ 1 F dt = te t + 2 2 e t te 2 t dt = te t e t + 2 t 2 e t 1 2 te 2 t + 1 4 e 2 t and X p = ΦU = 1 2 1 1 2 t + 3 4 1 3 4 + 2 2 0 e t + 2 2 2 te t . 33. Let I = i 1 i 2 so that I = 11 3 3 3 I + 100sin t 0 and I c = c 1 1 3 e 2 t + c 2 3 1 e 12 t . Then Φ = e 2 t 3 e 12 t 3 e 2 t e 12 t , Φ 1 = 1 10 e 2 t 3 10 e 2 t 3 10 e 12 t 1 10 e 12 t , U = Φ 1 F dt = 10 e 2 t sin t 30 e 12 t sin t dt = 2 e 2 t (2sin t cos t ) 6 29 e 12 t (12sin t cos t ) , and I p = ΦU = 332 29 sin t 76 29 cos t 276 29 sin t 168 29 cos t so that I = c 1 1 3 e 2 t + c 2 3 1 e 12 t + I p . If I (0) = 0 0 then c 1 = 2 and c 2 = 6 29 . 36. λ 1 = 1, λ 2 = 4, K 1 = 3 2 , K 2 = 1 1 , P = 3 1 2 1 , P 1 = 1 5 1 1 2 3 , P 1 F = 0 e t ; Y = 1 0 0 4 Y + 0 e t y 1 = c 1 e t , y 2 = 1 3 e t + c 2 e 4 t X = PY = 3 1 2 1 c 1 e t 1 3 e t + c 2 e 4 t = 1 3 e t + 3 c 1 e t + c 2 e 4 t 1 3 e t 2 c 1 e t + c 2 e 4 t = c 1 3 2 e t + c 2 1 1 e 4 t 1 3 1 1 e t 187
EXERCISES 10.5 Matrix Exponential 10.5 Matrix Exponential 3. For A = 1 1 1 1 1 1 2 2 2 we have A 2 = 1 1 1 1 1 1 2 2 2 1 1 1 1 1 1 2 2 2 = 0 0 0 0 0 0 0 0 0 . Thus, A 3 = A 4 = A 5 = · · · = 0 and e A t = I + A t = 1 0 0 0 1 0 0 0 1 + t t t t t t 2 t 2 t 2 t = t + 1 t t t t + 1 t 2 t 2 t 2 t + 1 . 6. From Problem 2, e A t = cosh t sinh t sinh t cosh t , so X = cosh t sinh t sinh t cosh t c 1 c 2 = c 1 cosh t sinh t + c 2 sinh t cosh t . 9. To solve X = 1 0 0 2 X + 3 1 we identify t 0 = 0, F ( t ) = 3 1 , and use the results of Problem 1 and equation (6) in the text. X ( t ) = e A t C + e A t t t 0 e A s F ( s ) ds = e t 0 0 e 2 t c 1 c 2 + e t 0 0 e 2 t t 0 e s 0 0 e 2 s 3 1 ds = c 1 e t c 2 e 2 t + e t 0 0 e 2 t t 0 3 e s e 2 s ds = c 1 e t c 2 e 2 t + e t 0 0 e 2 t 3 e s 1 2 e 2 s t 0 = c 1 e t c 2 e 2 t + e t 0 0 e 2 t 3 e t + 3 1 2 e 2 t 1 2 = c 1 e t c 2 e 2 t + 3 + 3 e t 1 2 1 2 e 2 t = c 3 1 0 e t + c 4 0 1 e 2 t + 3 1 2 . 188

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10.5 Matrix Exponential 12. To solve X = 0 1 1 0 X + cosh t sinh t we identify t 0 = 0, F ( t ) = cosh t sinh t , and use the results of Problem 2 and equation (6) in the text.
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