# 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?

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Because people have already answered the question you asked, you should simply start a new question with the correction. Otherwise, the answers that were already given for the question you asked will no longer appear to match the question. I edited this question to restore the original wording. It's perfectly fine to ask a second question, there is no limitation of resources or anything like that to worry about. –  doraemonpaul Jan 24 '13 at 1:53

$(x^2+3y^2+3u^2)u_x-2xyuu_y+2xu=0$

$2xyuu_y-(x^2+3y^2+3u^2)u_x=2xu$

$2yu_y-\left(\dfrac{x}{u}+\dfrac{3y^2}{xu}+\dfrac{3u}{x}\right)u_x=2$

$\dfrac{du}{dt}=2$ , 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. $=e^{\int\frac{1}{t}dt}=e^{\ln t}=t$

$\therefore\dfrac{d}{dt}(tw)=-3y_0^2e^{4t}-12t^2$

$tw=\int(-3y_0^2e^{4t}-12t^2)~dt$

$tx^2=-\dfrac{3y_0^2e^{4t}}{4}-4t^3+f(y_0)$

$x^2=-\dfrac{3y_0^2e^{4t}}{4t}-4t^2+\dfrac{f(y_0)}{t}$

$x=\pm\sqrt{-\dfrac{3y_0^2e^{4t}}{4t}-4t^2+\dfrac{f(y_0)}{t}}$

$x=\pm\sqrt{-\dfrac{3y^2}{2u}-u^2+\dfrac{2f(ye^{-u})}{u}}$

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How to handle $y_0$ ? Treat it as a constant? Or what? Can somebody help me? –  doraemonpaul Jan 20 '13 at 21:26
Actually there was a typo in the original post. It has been updated –  rmh52 Jan 23 '13 at 19:33
Thank you about the hint in math.stackexchange.com/questions/285384 given by @Fabian, in fact it just treat $y_0$ as a constant. –  doraemonpaul Jan 26 '13 at 0:58