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Differential Equations Lecture Work Solutions 139

# Differential Equations Lecture Work Solutions 139 - X n =1...

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1c. u x (0 ,t )= t u x ( L, t )= t 2 w = t 2 t 2 L x 2 + tx v t = kv xx 2 t 1 2 L x 2 x + k t 2 t L + x | {z } this gives s n ( t ) v x (0 ,t )= v x ( L, t )=0 λ n = L 2 n =0 , 1 , 2 , ··· X n ( x )=co s L xn =0 , 1 , 2 , ··· v ( x, 0) = x ( L x ) w ( x, 0) | {z } =0 = x ( L x ) s n ( t )= R L 0 n 2 t 1 2 L x 2 + k t 2 t L o cos L xdx R L 0 cos 2 L xdx v n ( t )= v n (0) e k ( L ) 2 t + R t 0 s n ( τ ) e k ( L ) 2 ( t τ ) v ( x, t )= 1 2 v 0 ( t )+
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Unformatted text preview: X n =1 v n ( t ) cos n π L x v n (0) = R L x ( L − x ) cos n π L x dx R L cos 2 n π L x dx u = v + t 2 − t 2 L x 2 + tx 139...
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