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For a PDE $$(x-y^{2}) u_{x} + u_{y} = 0$$ I've tried to use method of characteristics. But I've failed to do so. It was because of the term $x-y^{2}$; I don't know how to integrate this on the characteristic line. Should I try another method than method of chracteristic? Or is there a clever trick for this?

Related equatons: $dx/ds = x-y^{2}$, $dy/ds = 1$, $du/ds = 0$.

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If you integrate the $dy/ds=1$ and obtain $y=s$ then $dx/ds = x-s^2$ which is integrable by partial fractions. –  James S. Cook Oct 7 '12 at 22:09
    
ah I think it becomes $dx/ds = x-s^2$... –  julypraise Oct 7 '12 at 22:10
    
strike the partial fractions. –  James S. Cook Oct 7 '12 at 22:12
    
hm.. I'm not sure how that really works.. since $s$ is not a constant –  julypraise Oct 7 '12 at 22:14
    
Integrating factor method. Look at it as $dx/ds-x=-s^2$. –  James S. Cook Oct 7 '12 at 22:16

1 Answer 1

up vote 3 down vote accepted

Solving your "related equations" should give you the solutions: $u$ is constant on the curves $y = s$, $x = 2+2 s+s^2+ c \exp(s)$, i.e. $u = F((x - 2 - 2 y - y^2) \exp(-y))$.

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Thanks for your answer. But how did you get the $x$ part? –  julypraise Oct 7 '12 at 21:53
    
Ah.. just one short question. Is it okay to use $y=s$ rather than $y=y_{0} +s$ because if I use the latter, the problem gets much harder... –  julypraise Oct 7 '12 at 22:41

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