Differential Equations Lecture Work Solutions 235

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

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2 .e . TheODEsinth iscaseare 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 h 3 e x + 1 3 t 3 i t u ( x, t )=3 e x + 1 3 t 3 e
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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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