lec7_print

# When ? 1 6 ? 2 Ïˆ t ? 2 ? 1 ? 2 e ? 1 t ? 1 ? 2 ? 1 e

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When λ 1 6 = λ 2 Ψ 0 ( t ) = - λ 2 λ 1 - λ 2 e λ 1 t - λ 1 λ 2 - λ 1 e λ 2 t Ψ 1 ( t ) = 1 λ 1 - λ 2 e λ 1 t + 1 λ 2 - λ 1 e λ 2 t (9) e At = Ψ 0 ( t ) I + Ψ 1 ( t ) A (10) Srinivas Palanki (USA) Solution of Linear Differential Equations 8 / 15

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Analytical Calculation of e At When λ 1 = λ 2 Ψ 0 ( t ) = e λ 1 t - λ 1 te λ 1 t Ψ 1 ( t ) = te λ 1 t (11) e At = Ψ 0 ( t ) I + Ψ 1 ( t ) A (12) Srinivas Palanki (USA) Solution of Linear Differential Equations 9 / 15
Illustrative Examples Example 1 Calculate e At for A = 1 2 4 3 Eigenvalues of A : det ( λ I - A ) = det λ - 1 - 2 - 4 λ - 3 = λ 2 - 4 λ - 5 = ( λ - 5)( λ + 1) Thus λ 1 = 5 , λ 2 = - 1. Srinivas Palanki (USA) Solution of Linear Differential Equations 10 / 15

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Illustrative Examples Substituting in the formulae for Ψ 0 ( t ) and Ψ 1 ( t ) Ψ 0 = 1 6 ( e 5 t + 5 e - t ) Ψ 1 = 1 6 ( e 5 t - e - t ) Thus, e At = Ψ 0 ( t ) I + Ψ 1 ( t ) A = 1 3 e 5 t + 2 3 e - t 1 3 e 5 t - 1 3 e - t 2 3 e 5 t - 2 3 e - t 2 3 e 5 t + 1 3 e - t Srinivas Palanki (USA) Solution of Linear Differential Equations 11 / 15
Illustrative Examples Example 2 Calculate e At for A = 1 2 - 8 - 5 Eigenvalues of A : det ( λ I - A ) = det λ - 1 - 2 8 λ + 5 = λ 2 + 4 λ + 11 Thus λ 1 = - 2 + i 7 λ 2 = - 2 - i 7 Srinivas Palanki (USA) Solution of Linear Differential Equations 12 / 15

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Illustrative Examples Substituting in the formulae for Ψ 0 ( t ) and Ψ 1 ( t ) Ψ 0 = (2 + i 7) e ( - 2+ i 7) t - (2 - i 7) e ( - 2 - i 7) t 2 i 7 Ψ 1 = e ( - 2+ i 7) t - e ( - 2 - i 7) t 2 i 7 Srinivas Palanki (USA) Solution of Linear Differential Equations 13 / 15
Illustrative Examples Using the formula e i θ + e - i θ 2 = Cos ( θ ) e i θ - e - i θ

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