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Differential Equations Lecture Work Solutions 251

Differential Equations Lecture Work Solutions 251 - 4 a The...

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4. a. The set of ODEs is dx dt = c du dt = e 3 x The solution of the first is x = x 0 + ct Substituting x in the second ODE du dt = e 3( x 0 + ct ) Now integrate u ( x, t ) = K + e 3 x 0 1 3 c e 3 ct At t = 0 we get f ( x 0 ) = u ( x 0 , 0) = K + e 3 x 0 1 3 c Therefore the constant of integration K is K = f ( x 0 ) + e 3 x 0 1 3 c Substitute this K in the solution u ( x, t ) = f ( x 0 ) + e 3 x 0 1 3 c e 3 x 0 1 3 c e 3 ct Recall that x 0 = x ct thus u ( x, t ) = f ( x ct ) + 1 3 c e 3( x ct ) 1 3 c e 3 x b. The set of ODEs is dx dt = t du dt = 5 The solution of the first is
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