Originally Posted by

**bugatti79** I was hoping that by solving the pde the normal way which is part of the question that it might give a clue as to how the change of variables method should should turn out.

Solving the original PDE $\displaystyle \displaystyle y \frac{\partial u}{\partial x}-x\frac{\partial u}{\partial y}=1$ I get

$\displaystyle \displaystyle \frac{dy}{dx}=\frac{b(x,y)}{a(x,y)}=\frac{-x}{y} \implies \frac{y^2}{2}=\frac{-x^2}{2}+k$. Taking the positive solution of y

$\displaystyle \displaystyle \frac{du}{dx}=\frac{h(x,y)}{a(x,y)}=\frac{1}{y} \implies u=\int \frac{1dx}{\sqrt{2k-x^2}}$

which gives

$\displaystyle \displaystyle u=Sin^{-1} [\frac{x}{\sqrt {2k}}]+f(\frac{y^2}{2} +\frac{x^2}{2})$

The IC's are u(x,0)=0, therefore

$\displaystyle \displaystyle 0=Sin^{-1} [\frac{x}{\sqrt {2(\frac{x^2}{2})}}]+f(\frac{x^2}{2})$. Let $\displaystyle \displaystyle t=\frac{x^2}{2}$ therefore

$\displaystyle \displaystyle f(t)=\sqrt {2t}=-\frac {\pi}{2}$

$\displaystyle \displaystyle u(x,y)=Sin^{-1} [\frac{x}{\sqrt {2(\frac{x^2}{2}+\frac{y^2}{2})}}}]-\frac {\pi}{2}$

I am not sure of this is right either...