Differential Equations Lecture Work Solutions 254

Differential Equations Lecture Work Solutions 254 - 5. a....

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5. a. The set of ODEs is dx dt = u 2 du dt =3 u The solution of the Frst ODE requires the yet unknown u thus we tackle the second ODE du u dt Now integrate this ln u ( x, t )=3 t + K or u ( x, t )= Ce 3 t At t =0thesolutionis u ( x 0 , 0) = f ( x 0 C Thus u ( x, t f ( x 0 ) e 3 t Now substitute this solution in the characteristic equation (Frst ODE) dx dt = f ( x 0 ) e 3 t 2 = ( f ( x 0 )) 2 e 6 t Integrating x = ( f ( x 0 )) 2 Z e 6 t dt = 1 6 ( f ( x 0 )) 2 e 6 t + K ±or t =0weget x 0 = 1 6 ( f ( x 0 )) 2 + K Thus K = x 0 + 1 6 ( f ( x 0 )) 2 and the characteristics are x = 1 6 ( f ( x 0 )) 2 e 6 t + x 0 + 1 6 ( f (
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