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I have an ODE : $y' = f(x, y)$

I change for coordinates $(r, s) = (g(x, y), h(x, y))$. What is the equation like in terms of r and s ?

If it can help, in my case, $(r, s) = (y.x^{-k}, ln(x))$. I can express x(r, s) and y(r, s) but I don't know what $y'$ becomes and how to get $^{ds}_{dr}$ in. Thanks

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    $\begingroup$ Try doing $dy=\frac{\partial y}{\partial r} dr +\frac{\partial y}{\partial s} ds$ and $dx=\frac{\partial x}{\partial r} dr +\frac{\partial x}{\partial s} ds$? $\endgroup$
    – KittyL
    Jan 21, 2015 at 10:38
  • $\begingroup$ great technique ! it works well. you should write it as an answer $\endgroup$
    – Thomas
    Jan 21, 2015 at 11:09
  • $\begingroup$ Too short as an answer :) $\endgroup$
    – KittyL
    Jan 21, 2015 at 11:13

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