G x sin l nc nc 0 l 2 n x dx sin l 0 f x sin a f x

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Unformatted text preview: L nπ nπc Bn sin x = g (x) . L L VI. Find An and Bn , where L nπ x dx nπ 2L L 0 An = f (x) sin x dx and = L L0 L 2 nπ x dx sin L 0 L nπ g (x) sin x dx L L0 2 nπ L Bn = = x dx. g (x) sin L nπc nπc 0 L 2 nπ x dx sin L 0 f (x) sin (a) f (x) = sin 2π x L and g (x) = 0 nπ 2π 2L x sin x dx = sin An = L0 L L L 2 nπ Bn = x dx = 0 (0) sin nπc 0 L 1, if n = 2 0, if n = 2 Therefore, the solution is u (x, t) = sin (b) f (x) = 0 and g (x) = sin 2π x cos L 2πc t. L 2π x L 2L nπ An = x dx = 0 (0) sin L0 L L nπ 2π 2 sin x sin x dx = Bn = nπc 0 L L L , 2πc 0, Therefore, the solution is u (x, t) = (c) f (x) = sin 2π x L L sin 2πc and g (x) = sin 2 2π x L 2π x sin L 2πc t. L if n = 2 if n = 2 ; By (a) and (b), 1, if n = 2, 0, if n = 2. An = Bn = L , 2πc 0, if n = 2, if n = 2. Therefore, the solution is 2π x L u (x, t) = sin 2π x L (d) f (x) = sin Similarly, cos and g (x) = sin L 2πc t+ sin L 2πc . 3π x L 1, if n = 2, 0, if n = 2, An = 2πc t L Bn = L , 3πc 0, if n = 3, if n = 3. Therefore, the solution is 2π x cos L u (x, t) = sin 2π x L (e) f (x) = sin L 2πc t+ sin L 3πc 3π x sin L 3πc t. L L nπ x dx L and g (x) = 2x 1, if n = 2 0, if n = 2 An = L L 2 nπ 2 x dx = × (2x) sin nπc 0 L nπc L n+1 n+1 L (−1) 4L (−1) 4L2 = × = nπc nπ n2 π 2 c (2x) sin Bn = 0 Therefore, the solution is u (x, t) = sin 2. 2π x cos L nπ nπc 2πc (−1)n+1 4L2 t+ sin x sin t. 2π2c L n L L n=1 ∞ I. Let the solution be u (x, t) = X (x) T (t), then X ′′ (x) + λX (x) = 0, X (0) = X (π ) = 0 · · · SLP T ′...
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This document was uploaded on 01/23/2014.

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