Differential Equations Lecture Work Solutions 236

Differential Equations Lecture Work Solutions 236 - 3. The...

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Unformatted text preview: 3. The ODEs in this case are dx = 2u dt du =0 dt Since the first ODE contains x, t and u, we solve the second ODE first u(x, t) = u(x(0), 0) = f (x(0)) Plug this u in the first ODE, we get dx = 2f (x(0)) dt The solution is x = x0 + 2tf (x0 ) These are characteristics lines all with slope 1 2f (x0 ) Note that the characteristic through x1 (0) will have a different slope than the one through x2 (0). That is the straight line are NOT parallel. 236 ...
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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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