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Differential Equations Lecture Work Solutions 137

# Differential Equations Lecture Work Solutions 137 - n − 1...

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1. u t = ku xx + x u ( x, 0) = x ( L x ) a. u x (0 , t ) = 1 w = x + t L u ( L, t ) = t Solve v t = kv xx 1 + x (see 1a last section) v x (0 , t ) = 0 v ( x, 0) = x ( L x ) ( x L ) = ( x + 1) ( t x ) v ( L, t ) = 0 eigenvalues: n 1 2 π L 2 n = 1 , 2 , · · · eigenfunctions : cos ( n 1 2 ) π L x n = 1 , 2 , · · · v n =1 v n ( t ) cos n 1 2 π L x 1 + x = n =1 s n cos n 1 2 π L x s n = L 0 ( 1 + x ) cos n 1 2 π L x dx L 0 cos 2 n 1 2 π L x dx n =1 ˙ v n ( t ) cos n 1 2 π L x = k n =1 n 1 2 π L 2 v n cos n 1 2 π L x + n =1 s n cos n 1 2 π L x Compare coeﬃcients ˙ v n ( t ) + k n 1 2 π L 2 v n = s n v n = v n (0) e [ ( n 1 2 )
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Unformatted text preview: n − 1 2 ) π L ] 2 kt + s n Z t e − [ ( n − 1 2 ) π L ] 2 k ( t − τ ) dτ | {z } see (8 . 2 . 39) v n (0) = coeﬃcients of expansion of (1 + x ) ( L − x ) v n (0) = R L (1 + x ) ( L − x ) cos n − 1 2 π L x dx R L cos 2 n − 1 2 π L x dx u = v + w 137...
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