Differential Equations Lecture Work Solutions 248

# Differential Equations Lecture Work Solutions 248 - t 2 x =...

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2. The set of ODEs are dx dt =0 and du dt = u The characteristics are x = constant and the ODE for u can be written du u = dt Thus u ( x, t )= k ( x ) e t On x = 2 t or x +2 t =0w ehav e 1+co s x = k ( x ) e t | x = 2 t = k ( x ) e x 2 Thus the constant of integration is k ( x )= e x 2 (1 + cos x ) Plug this in the solution u we get u ( x, t )=(1+co s x ) e x 2 + t Another way of getting the solution is by a rotation so that the line x +2 t
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Unformatted text preview: t 2 x = 0, which we call . So here is the transformation = x + 2 t = t 2 x. The PDE becomes: u + 1 2 u = 1 2 u and the intial condition is: u ( , = 0) = 1 + cos 2 5 | =0 = 1 + cos 2 5 Rewrite this as a system of two Frst order ODEs, d d = 1 2 (0) = du ( ( ) , ) d = 1 2 u 248...
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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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