Differential Equations Lecture Work Solutions 234

# Differential Equations Lecture Work Solutions 234 - 1 2 e 2...

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2. d. The ODEs in this case are dx dt = 2 du dt = e 2 x Solve the characteristic equation x = 2 t + x 0 Now solve the second ODE. To do that we have to plug in for x du dt = e 2( x 0 2 t ) = e 2 x 0 e 4 t u ( x, t )= e 2 x 0 1 4 e 4 t ± + K The constant of integration can be found from the initial condition cos( x 0 )= u ( x 0 , 0) = 1 4 e 2 x 0 + K Therefore K =co s ( x 0 )+ 1 4 e 2 x 0 Plug this K in the solution and substitute for x 0 from the characteristic curve u ( x, t )= 1 4 e 2( x +2 t ) e 4 t +cos( x +2 t )+ 1 4 e 2( x +2 t ) u ( x, t )= 1 4 e 2 x e 4 t 1 +co s ( x +2 t )
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Unformatted text preview: 1 2 e 2 x e 4 t − 1 − sin( x + 2 t ) u t = 1 4 e 2 x 4 e 4 t − 2 sin( x + 2 t ) Substitute in the PDE u t − 2 u x = e 2 x e 4 t − 2 sin( x + 2 t ) − 2 ² 1 2 e 2 x e 4 t − 1 − sin( x + 2 t ) ³ = e 2 x e 4 t − 2 sin( x + 2 t ) − e 2 x e 4 t + e 2 x + 2 sin( x + 2 t ) = e 2 x which is the right hand side of the PDE 234...
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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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