Differential Equations Lecture Work Solutions 256

# Differential Equations Lecture Work Solutions 256 - 6. The...

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Unformatted text preview: 6. The set of ODEs is du dx = t2 u =5 dt dt The solution of the ﬁrst ODE requires the yet unknown u thus we tackle the second ODE du = 5 dt Now integrate this u(x, t) = 5t + K At t = 0 the solution is u(x0 , 0) = f (x0 ) = x0 = K Thus u(x, t) = x0 + 5t Now substitute this solution in the characteristic equation (ﬁrst ODE) dx = 5t3 + x0 t2 dt Integrate x= 1 54 t + t3 x0 + C 4 3 For t = 0 we get x0 = 0 + 0 + C Thus C = x0 and the characteristics are x= 13 t +1 3 54 t+ 4 Solve this for x0 x0 = x− 1+ The solution is then given by u(x, t) = 5t + 54 t 4 13 t 3 x− 1+ 256 54 t 4 13 t 3 x0 ...
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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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