Finding the general solution of a quasilinear PDE This is a homework that I'm having a bit of trouble with:
Find a general solution of:
$(x^2+3y^2+3u^2)u_x-2xyuu_y+2xu=0~.$
Of course this should be done using the method of characteristics but I'm having trouble solving the characteristic equations since none of the equations decouple:
$\dfrac{dx}{x^2+3y^2+3u^2}=-\dfrac{dy}{2xyu}=-\dfrac{du}{2xu}$
Any suggestions?
 A: Follow the method in http://en.wikipedia.org/wiki/Characteristic_equations#Example:
$(x^2+3y^2+3u^2)u_x-2xyuu_y+2xu=0$
$2xyuu_y-(x^2+3y^2+3u^2)u_x=2xu$
$2yu_y-\left(\dfrac{x}{u}+\dfrac{3y^2}{xu}+\dfrac{3u}{x}\right)u_x=2$
$\dfrac{du}{dt}=2$ , letting $u(0)=0$ , we have $u=2t$
$\dfrac{dy}{dt}=2y$ , letting $y(0)=y_0$ , we have $y=y_0e^{2t}=y_0e^u$
$\dfrac{dx}{dt}=-\left(\dfrac{x}{u}+\dfrac{3y^2}{xu}+\dfrac{3u}{x}\right)=-\dfrac{x}{2t}-\dfrac{3y_0^2e^{4t}}{2xt}-\dfrac{6t}{x}$
$\dfrac{dx}{dt}+\dfrac{x}{2t}=-\left(\dfrac{3y_0^2e^{4t}}{2t}+6t\right)\dfrac{1}{x}$
Let $w=x^2$ ,
Then $\dfrac{dw}{dt}=2x\dfrac{dx}{dt}$
$\therefore\dfrac{1}{2x}\dfrac{dw}{dt}+\dfrac{x}{2t}=-\left(\dfrac{3y_0^2e^{4t}}{2t}+6t\right)\dfrac{1}{x}$
$\dfrac{dw}{dt}+\dfrac{x^2}{t}=-\dfrac{3y_0^2e^{4t}}{t}-12t$
$\dfrac{dw}{dt}+\dfrac{w}{t}=-\dfrac{3y_0^2e^{4t}}{t}-12t$
I.F. $=e^{\int\frac{1}{t}dt}=e^{\ln t}=t$
$\therefore\dfrac{d}{dt}(tw)=-3y_0^2e^{4t}-12t^2$
$tw=\int(-3y_0^2e^{4t}-12t^2)~dt$
$tx^2=-\dfrac{3y_0^2e^{4t}}{4}-4t^3+f(y_0)$
$x^2=-\dfrac{3y_0^2e^{4t}}{4t}-4t^2+\dfrac{f(y_0)}{t}$
$x=\pm\sqrt{-\dfrac{3y_0^2e^{4t}}{4t}-4t^2+\dfrac{f(y_0)}{t}}$
$x=\pm\sqrt{-\dfrac{3y^2}{2u}-u^2+\dfrac{2f(ye^{-u})}{u}}$
