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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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    $y=s+y_0$ gives you exactly the same curves, just shifted in $s$.2012-10-10

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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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    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...2012-10-07