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3672055903

# 3672055903 - Solution 1(a 22 22 T = sin(nx e 4 n t[4n 2 2 =...

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Solution: 1. (a) t n t n e x n n n e x n t T 2 2 2 2 4 2 2 2 2 4 ) sin( 4 ] 4 [ ) sin( π π π π π π = = (b) t n t n t n t n e x n n e x n n x x T x x T e x n n e n x n x T 2 2 2 2 2 2 2 2 4 2 2 4 2 2 4 4 ) sin( ] ) cos( [ ) ( ) cos( ] [ ) cos( π π π π π π π π π π π π = = = = = (c) 0 ) sin( 4 ) sin( 4 4 2 2 2 2 4 2 2 4 2 2 2 2 = + = t n t n e x n n e x n n x T t T π π π π π π 2. (a) Forward difference: t t x T t t x T t T + / )] , ( ) , ( [ / At point x i (i=2,3,4): t t x T t t x T t T i i x x i + = / )] , ( ) , ( [ Thus t T / are x 1 =0 x 2 =0.2 x 3 =0.4 x 4 =0.6 x 5 =0.8 x 6 =1.0 t=0 -8 -8 -8

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(b) Forward difference: 2 2 2 ) /( )] , ( ) , ( 2 ) , 2 ( [ / x t x T t x x T t x x T x T + + + At point i=2,3,4: 2 1 2 0 2 2 ) /( )] , ( ) , ( 2 ) , ( [ x t x T t x T t x T x T i i i t + + + = At t=0, 2 2 / x T are x 1 =0 x 2 =0.2 x 3 =0.4 x 4 =0.6 x 5 =0.8 x 6 =1.0 t=0 -8 -8 -8 (c) Because t t x T i = ) 0 , ( equals 2 2 ) 0 , ( x t x T i = if i=2,3,4, therefore the temperatures at t=0 and at x 2 , x 3 , x 4 satisfy the heat equation 2 2 x T t T =
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