lec7_print

# 2 i sin θ 13 we get ψ t e 2 t cos 7 t 2 7 sin 7 t

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2 i = Sin ( θ ) (13) we get Ψ 0 ( t ) = e - 2 t cos ( 7 t ) + 2 7 sin ( 7 t ) Ψ 1 ( t ) = e - 2 t 7 sin ( 7 t ) Thus e At = Ψ 0 ( t ) I + Ψ 1 ( t ) A = e - 2 t cos ( 7 t ) + 3 7 sin ( 7 t ) 2 7 sin ( 7 t ) - 8 7 sin ( 7 t ) cos ( 7 t ) - 3 7 sin ( 7 t ) Srinivas Palanki (USA) Solution of Linear Differential Equations 14 / 15

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Illustrative Examples Analytical Calculation of e At when n = 2 Suppose the eigenvalues of the matrix A are complex . λ 1 = R + i Ω λ 2 = R - i Ω (14) Then, Ψ 0 ( t ) = cos t ) - R Ω Sin t ) e Rt Ψ 1 ( t ) = 1 Ω Sin t ) e Rt (15) e At = Ψ 0 ( t ) I + Ψ 1 ( t ) A (16) Srinivas Palanki (USA) Solution of Linear Differential Equations 15 / 15
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