Differential Equations Lecture Work Solutions 141

Differential Equations Lecture Work Solutions 141 - C = 1 u...

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For n =0 1 2 ˙ u 0 ( t )= 1 2 e t = 1 2 s 0 ( t ) n 6 =0 ˙ u n + n 2 u n =0 Solve the ODES u n = C n e n 2 t u n (0) = 0 n 6 =2 C n =0 u 2 (0) = 1 C 2 =1 u 0 = e t + C 0 u 0 (0) = 0 C 0 1=0
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Unformatted text preview: C = 1 u ( x, t ) = 1 e t + e 4 t cos 2 x 141...
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