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Differential Equations Lecture Work Solutions 235

# Differential Equations Lecture Work Solutions 235 - u t = 3...

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2. e. The ODEs in this case are dx dt = t 2 du dt = u Solve the characteristic equation x = t 3 3 + x 0 Now solve the second ODE. To do that we rewrite it as du u = dt Therefore the solution as in 2c ln u = t + ln K and the constant is 3 e x 0 Plug this K in the solution and substitute for x 0 from the characteristic curve ln u ( x, t ) = ln 3 e x + 1 3 t 3 t u ( x, t ) = 3 e x + 1 3 t 3 e t To check the answer, we differentiate
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Unformatted text preview: u t = 3 e x t 2 − 1 e 1 3 t 3 − t u x = 3 e x e 1 3 t 3 − t Substitute in the PDE u t − t 2 u x = 3 e x e 1 3 t 3 − t − t 2 n 3 e x t 2 − 1 e 1 3 t 3 − t o = 3 e x e 1 3 t 3 − t h t 2 − 1 − t 2 i = − 3 e x + 1 3 t 3 − t = − u 235...
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