Differential Equations Lecture Work Solutions 165

Differential Equations Lecture Work Solutions 165 - 2 2 u =...

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2. 2 u = e 2 y sin x u (0 ,y )=0 u ( π, y )=0 sin nx u ( x, 0) = 0 u ( x, L )= f ( x ) Use a Fourier sine series in x u ( x, y )= X n =1 u n ( y )s in nx S ( x, y ) is already in a Fourier sine series in x with the coefficients s 1 ( y )= e 2 y and all the other coefficients are zero. X n =1 n 2 u n sin nx | {z } u xx u n sin nx | {z } u yy = X n =1 s n ( x )s in nx ¨ u n ( y ) n 2 u n ( y )=0 fo r n 6 =1 ¨ u 1 ( y ) u 1 ( y )= e 2 y Boundary conditions are coming from u ( x, 0) = 0 u ( x, L )= f ( x ) X n =1 u n (0) sin nx =0 u n (0) = 0 X n =1 u n ( L )s in nx = f ( x ) u n ( L )= 1 L Z L 0 f ( x )s in nx dx The solution of the ODEs is u 1 ( y )= 1 3 e 2 y | {z }
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