Solving the following PDE? The PDE is given as 
$$y^2 u_x-xy u_y=x(u-2y)$$
What is the shortest method solving the PDE?
 A: $y^2u_x-xyu_y=x(u-2y)$
$\dfrac{u_y}{y}-\dfrac{u_x}{x}=\dfrac{2}{y}-\dfrac{u}{y^2}$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dy}{dt}=\dfrac{1}{y}$ , letting $y(0)=0$ , we have $\dfrac{y^2}{2}=t$
$\dfrac{dx}{dt}=-\dfrac{1}{x}$ , we have $-\dfrac{x^2}{2}=t+x_0=\dfrac{y^2}{2}+x_0$
$\dfrac{du}{dt}=\dfrac{2}{y}-\dfrac{u}{y^2}=\pm\dfrac{\sqrt2}{\sqrt t}-\dfrac{u}{2t}$ , we have $u(x,y)=\dfrac{F(x_0)}{\sqrt t}\pm\sqrt{2t}=\dfrac{f\left(x^2+y^2\right)}{y}\pm y$
A: $$y^2 u_x-xy u_y=x(u-2y)$$
$$\frac{dx}{y^2}=\frac{dy}{-xy}=\frac{du}{x(u-2y)}$$
$\frac{dx}{y^2}=\frac{dy}{-xy}\quad\to\quad xdx+ydy=0\quad\to\quad x^2+y^2=c_1$
$\frac{dy}{-xy}=\frac{du}{x(u-2y)}\quad\to\quad \frac{du}{dy}+\frac{u}{y}-2=0\quad\to\quad yu-y^2=c_2$
General solution on the form of implicite equation, any differentiable function $\Phi$ of two variables :
$$\Phi\left((yu-y^2),(x^2+y^2)\right)=0$$
Or, equivalent explicite form :
$$uy-y^2=f(x^2+y^2)$$
$$u(x,y)=\frac{y^2+f(x^2+y^2)}{y}$$
Any differentiable function $f$.
