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Differential Equations Lecture Work Solutions 77

Differential Equations Lecture Work Solutions 77 - n k n π...

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11. u t = k u xx + e 2 t cos 3 π L q ( x, t ) x u x (0 , t ) = 0 u x ( L, t ) = 0 u ( x, 0) = f ( x ) The boundary conditions imply u ( x, t ) = n =0 b n ( t ) cos n π L x Let q ( x, t ) = n =0 q n ( t ) cos n π L x q 0 ( t ) = e t q 3 ( t ) = e 2 t the rest are zero ! Thus ˙ b n = k n π L 2 b n + q n n = 0 , 1 , · · · n = 0 ˙ b 0 = q 0 = e t b 0 = e t ˙ b 1 + k n π L 2 b 1 = q 1 = 0 ˙ b 2 + k 2 n π L 2 b 2 = 0 homogeneous ˙ b 3 + k 3 n π L 2 b 3 = e 2 t rest are homogeneous. One can solve each equation to obtain all b n . ˙ b
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Unformatted text preview: n + k n π L ± 2 b n = 0 ⇒ b n = C n e − k ( n π L ) 2 t n = 1 , 2 , 4 , 5 , ··· note: n 6 = 3 ˙ b 3 + k 3 π L ± 2 b 3 = e − 2 t Solution of homogeneous is b 3 = C 3 e − k ( 3 π L ) 2 t For particular solution try b 3 = C e − 2 t 77...
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