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Math Help - how to find p and q (Charpit method) for this equation

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    how to find p and q (Charpit method) for this equation

     2uq^2+y^2(q-p)=0

    p= \partial u/\partial x ;   q= \partial u/\partial y

    We get the charpit equations:
    \frac{dx}{-y^2}=\frac{dy}{4uq+y^2}=\frac{du}{-py^2+q(4uq+y^2)}=\frac{dp}{-2pq^2}=\frac{dq}{-2y(q-p)-2q^3}

    I don't know how to find the value of p and q ??

    (i have tried the following ways but i cannot:

    from the original equation we have:  (q-p) =-\frac{2uq^2}{y^2}

    substitute into the last pair equation \frac{dq}{-2y(-\frac{2uq^2}{y^2})-2q^3}=\frac{dp}{-2pq^2}

    or \frac{dq}{dp}=\frac{1}{p}(\frac{-2u}{y}+q) => q = \frac{2u}{y}+p.c1

    but i don't know how to find p? because any pairs will lead to a complicated equation. please experts help!! thanks )
    Last edited by lotusquantum; April 27th 2013 at 03:35 PM.
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