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Unformatted text preview: tribution by expanding it in a Fourier sine series. f ( x ) = x, < x < 40 , ( period = 80) (10) f ( x ) = X n =1 b n sin nx 40 (11) Solving for b n b n = 1 20 Z 40 x sin nx 40 dx (12) = 1 20 1600 sin nx 40 n 2 240 x cos nx 40 n (13) = 80 n 1 , n = odd1 , n = even (14) = 80(1) n +1 n (15) the initial temperature dist. is therefore given by, u ( x, 0) = 80 X n =1 (1) n +1 n sin nx 40 (16) Applying this initial temp. dist. to (9) gives C n , C n = 80(1) n +1 n (17) The solution to the given heat conduction problem is therefore, u ( x, t ) = 80 X n =1 (1) n +1 n en 2 2 t/ 16000 sin nx 40 (18) 1...
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 Spring '07
 TERRELL,R

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