Differential Equations Lecture Work Solutions 251

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

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Unformatted text preview: 4. a. The set of ODEs is du = e−3x dt dx =c dt The solution of the first is x = x0 + ct Substituting x in the second ODE du = e−3(x0 +ct) dt Now integrate u(x, t) = K + e−3x0 1 −3ct e −3c At t = 0 we get f (x0 ) = u(x0 , 0) = K + e−3x0 1 −3c Therefore the constant of integration K is K = f (x0 ) + e−3x0 1 3c Substitute this K in the solution u(x, t) = f (x0 ) + e−3x0 1 1 −3ct − e−3x0 e 3c 3c Recall that x0 = x − ct thus u(x, t) = f (x − ct) + b. The set of ODEs is 1 −3x 1 −3(x−ct) e e − 3c 3c dx =t dt du =5 dt The solution of the first is x = x0 + 12 t 2 Now integrate the second ODE u(x, t) = 5t + K At t = 0 the solution is u(x0 , 0) = f (x0 ) = K plug t = 0 in the solution u Thus when substituting for x0 in the solution u(x, t) = 5t + f x − 251 12 t 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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