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Linear Partial Differential Eq.

Posted September 8th, 2007 by vic777
in

How can I find a solution for this eq. with Cauchy's conditions ?
$ xu_x+yu_y=u+1 $ with $ u(x,y)=x^2 $ on $ y=x^2 $

‹ Linear Partial Differential eqn Linear partial diff. eqn. ›

Comments

The same way you get

On September 8th, 2007 Structure says:

$ \frac{dx}{x}=\frac{dy}{y}=\frac{du}{u+1} $
$ lnx=lny-lnc\; so \frac{y}{x}=c $
The same way you get $ \frac{u+1}{x}=k $
Now solution can be written k=g(c) or $ u(x,y)+1=xg(\frac{y}{x}) $
$ u(x,x^2)=-1+xg(x)=x^2 $
so $ g(x)=\frac{x^2+1}{x} $and $ u(x,y)=-1+x(\frac{y}{x}+\frac{x}{y}) $
In few hours I shall leave for Wien for 10 days.See you soon.

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