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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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  • $\begingroup$ 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. $\endgroup$
    – doraemonpaul
    Jan 24, 2013 at 1:53

1 Answer 1

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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. $=e^{\int\frac{1}{t}dt}=e^{\ln t}=t$

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

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

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

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

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

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

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  • $\begingroup$ How to handle $y_0$ ? Treat it as a constant? Or what? Can somebody help me? $\endgroup$
    – doraemonpaul
    Jan 20, 2013 at 21:26
  • $\begingroup$ Actually there was a typo in the original post. It has been updated $\endgroup$
    – rmh52
    Jan 23, 2013 at 19:33
  • $\begingroup$ Thank you about the hint in math.stackexchange.com/questions/285384 given by @Fabian, in fact it just treat $y_0$ as a constant. $\endgroup$
    – doraemonpaul
    Jan 26, 2013 at 0:58

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