Differential Equations Lecture Work Solutions 255

# Differential Equations Lecture Work Solutions 255 - b. The...

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b. The set of ODEs is dx dt = t 2 u du dt = 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 )= t + K or u ( x, t Ce t At t =0thesolutionis u ( x 0 , 0) = f ( x 0 C Thus u ( x, t f ( x 0 ) e t Now substitute this solution in the characteristic equation (Frst ODE) dx dt = t 2 f ( x 0 ) e t or Z dx = f ( x 0 ) Z t 2 e t dt Integrate and continue as in part a of this problem x = f ( x 0 ) h t 2 e t 2 te t 2 e t + C i ±or t =0weget x 0 = f ( x 0 )[ 2+ C ] Thus Cf ( x 0 x 0 +2 f ( x 0 ) and the characteristics are x = f ( x 0 ) h
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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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