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Differential Equations Lecture Work Solutions 138

Differential Equations Lecture Work Solutions 138 - R L x L...

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1b. u (0 , t ) = u x ( L, t ) = 1 w = x + 1 v t = k v xx + x v (0 , t ) = 0 v x ( L, t ) = 1 v ( x, 0) = x ( L x ) ( x + 1) eigenvalues: n 1 2 π L 2 n = 1 , 2 , · · · eigenfunctions: sin n 1 2 π L x n = 1 , 2 , · · · v = n =1 v n ( t ) sin n 1 2 π L x x = n =1 s n sin n 1 2 π L x s n = L 0 x sin n 1 2 π L x dx L 0 sin 2 n 1 2 π L x dx ˙ v n + k n 1 2 π L 2 v n = s n v n ( t ) = v n (0) e [ ( n 1 2 ) π L ] 2 kt + s n 1 e [ ( n 1 2 ) π L ] 2 t [ ( n 1 2 ) π L ] 2 v n (0) = L 0 [
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Unformatted text preview: R L [ x ( L − x ) − ( x + 1)] sin n − 1 2 π L x dx R L sin 2 n − 1 2 π L x dx ↑ Coefficients of expansion of initial condition for v u = v + w 138...
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