Differential Equations Lecture Work Solutions 159

# Differential Equations Lecture Work Solutions 159 - π L x...

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5. u tt u xx =1 u ( x, 0) = f ( x ) u t ( x, 0) = g ( x ) u (0 ,t )=1 u x ( L, t )= B ( t ) w ( x, t )= xB ( t )+1 ; w t = x ˙ B ( t ) w tt = x ¨ B ( t ) w xx =0 w ( x, 0) = xB (0) + 1 w t ( x, 0) = x ˙ B (0) v tt v xx =1 x ¨ B ( t )+0 S ( x, t ) v ( x, 0) = f ( x ) xB (0) 1 F ( x ) v t ( x, 0) = g ( x ) x ˙ B (0) G ( x ) v (0 ,t )= v t ( L, t )=0 λ n = ( n 1 / 2) π L ! 2 φ n =s in ( n 1 / 2) π L xn =1 , 2 , ··· v ( x, t )= X n =1 v n ( t )s in ( n 1 / 2) π L x S ( x, t )= X n =1 s n ( t )s in ( n 1 / 2) π L x s n ( t )= R L 0 S ( x, t )s in ( n 1 / 2) π L xdx R L 0 sin 2 ( n 1 / 2) π L xdx F ( x )= X n =1 v n (0) sin ( n 1 / 2) π L x v n (0) = R L 0 F ( x )s in ( n 1 / 2) π L xdx R L 0 sin 2 ( n 1 / 2) π L xdx G ( x )= X n =1 ˙ v n (0) sin ( n 1 / 2) π L x ˙ v n (0) = R L 0 G ( x )sin (
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Unformatted text preview: π L x dx R L sin 2 ( n − 1 / 2) π L x dx ¨ v n + ( n − 1 / 2) π L 2 v n = s n ( t ) ⇒ v n = c 1 cos ± λ n t + c 2 sin ± λ n t + Z t s n ( τ ) sin √ λ n ( t − τ ) √ λ n dτ v n (0) = c 1 ˙ v n (0) = c 2 √ λ n continue as in 3b. 159...
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## This note was uploaded on 12/22/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.

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