Differential Equations Lecture Work Solutions 153

Differential Equations Lecture Work Solutions 153 - 2 n L x...

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4a. u tt c 2 u xx = xt u ( x, 0) = sin x u t ( x, 0) = 0 u (0 ,t )=1 u ( L, t )= t w ( x, t )= t 1 L x +1; w t = x L w tt =0 v tt c 2 v xx = xt v (0 ,t )= v ( L, t )=0 λ n = L 2 φ n =s in L xn =1 , 2 , ··· v ( x, 0) = sin x x L +1 v t ( x, 0) = 0 x L v ( x, t )= X n =1 v n ( t )s in L x xt = X n =1 s n ( t )s in L x s n ( t )= R L 0 xt sin L xdx R L 0 sin 2 L xdx sin x + x L 1= X n =1 v n (0) sin L x v n (0) = R L 0 (sin x + x L 1) sin L xdx R L 0 sin 2 L xdx x L = X n =1 ˙ v n (0) sin L x ˙ v n (0) = R L 0 x L sin L xdx R L
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Unformatted text preview: 2 n L x dx v n + n L 2 c 2 v n = s n ( t ) v n = c 1 cos c n t + c 2 sin c n t + Z t s n ( ) sin c n ( t ) c n d v n (0) = c 1 v n (0) = c 2 c n continue as in 3b. 153...
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