Differential Equations Lecture Work Solutions 243

Differential Equations Lecture Work Solutions 243 - v x t t...

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8. Solve the initial value problem v t + e x v x = 0 v ( x, 0) = x The ODEs are dx dt = e x , dv dt = 0 and the initial condition for each x (0) = ξ, v ( ξ, 0) = ξ Solve the characteristic equation e x dx = dt e x = t + C Now use the initial condition e ξ = 0 + C or e x = t + e ξ The solution of the other ODE
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Unformatted text preview: v ( x ( t ) , t ) = K = v ( x (0) , 0) = ξ Solve the characteristic equation for ξ ξ = − ln( t + e − x ) and substitute in v to get v ( x, t ) = − ln( t + e − x ) 243...
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