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Linear partial diff. eqn.

Posted September 8th, 2007 by vic777
in

How can I find a solution for this eq. with Cauchy conditions ?
$ u_x+xu_y=0 $ with $ u(0,y)=\sin(y) $

‹ Linear Partial Differential Eq. linear partial differential eq. ›

Comments

So

On September 8th, 2007 Structure says:

$ \frac{dx}{1}=\frac{dy}{x} $
$ xdx=dy $; $ \frac{1}{2}x^2-y=c $
$ u(x,y)=f(y-\frac{1}{2}x^2) $; $ u(0,y)=f(y)=\sin(y) $
So $ u(x,y)=sin(y-\frac{1}{2}x^2) $

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