Differential Equations Lecture Work Solutions 286

Differential Equations Lecture Work Solutions 286 - 3a. utt...

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3a. u tt = c 2 u xx 0 <x< u x (0 ,t )=0 u ( x, 0) = 00 2 12 3 0 x> 3 u t ( x, 0) = 0 u ( x, t )= F ( x ct )+ G ( x + ct ) F ( x 1 2 f ( x ) 1 2 c Z x 0 g ( ξ ) = 1 2 f ( x ) since g 0 0 G ( x 1 2 f ( x 1 2 c Z x 0 g ( ξ ) = 1 2 f ( x ) g 0 0 u ( x, t f ( x + ct f ( x ct ) 2 ,x > c t u x (0 F 0 ( ct G 0 ( ct F 0 ( ct G 0 ( ct ) F 0 ( z G 0 ( z ) Integrate F ( z G ( z K F ( z G ( z ) K F ( x ct G ( ct x ) Kx ct < 0 = 1 2 f ( ct x ) ct < 0 u ( x, t 1 2 f ( x + ct 1 2 f ( ct x ) K To fnd K we look at x =0 u (0 , 0) = 0 From initial condition but u (0 , 0) = 1 2 f (0) + 1 2
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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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