Differential Equations Lecture Work Solutions 259

Differential Equations Lecture Work Solutions 259 - Now...

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Now from the solution u we have f ( ξ )= ue ct so x ( t )= 1 c ue ct e ct 1 + ξ x ( t )= u c 1 e ct + ξ ξ = x ( t )+ u c 1 e ct so we get u ( x, t )= f x + u c 1 e ct
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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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