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How can I find the general solution of this differential equation?

$$\frac{d^2y}{dt^2}=\frac{1}{y}-\frac{x\frac{dy}{dx}}{y^2}$$

Finding the homogeneous solution is very simple, so I'm asking for the particular solution.

I know the right hand side can be expressed as $\frac{d(\frac{x}{y})}{dx}$, giving,

$$\frac{d^2y}{dt^2}=\frac{d(\frac{x}{y})}{dx}$$

but I don't know what to do with this, if it is of help. Any hints would be appreciated.

Thanks in advance.

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    $\begingroup$ What is $x$? Is it a constant? Is it a function of $t$? Is $y=y(t)$ only or $y=y(x,t)$? $\endgroup$ – bartgol Mar 18 '16 at 20:56
  • $\begingroup$ What homogeneous solution? This is not a linear equation. $\endgroup$ – Robert Israel Mar 18 '16 at 21:18
  • $\begingroup$ I think I might have the wrong idea, but I thought that $y$ depended only on $t$, since I didn't see any partial derivatives. I also thought that $x$ was also a function of $t$, and since both $x$ and $y$ depend on $t$, $y$ could be an implicit function of $x$. Is this a plausible approach? The problem's statement doesn't say anything about $y$ depending on $x$ and $t$. In any case, by the hints given in the problem's statement you must be right: it is meant to be a PDE. $\endgroup$ – alfdc80 Mar 18 '16 at 21:55
  • $\begingroup$ There was a typo in the problem statement. It's not $\frac{d^2y}{dt^2},$ $\frac{dy}{dx},$ but $\frac{\partial y^2}{\partial t^2}$ , $\frac{\partial y}{\partial x}$ respectively $\endgroup$ – alfdc80 Mar 19 '16 at 0:57
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Assuming this is meant to be a PDE, separation of variables gives solutions of the form

$$ y(x,t) = \dfrac{G(t)}{\sqrt{A - c \ln(x)}} $$

where

$$ \dfrac{dG}{dt} = \sqrt{B - c \ln G(t)}$$

$A, B, c$ arbitrary constants.

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  • $\begingroup$ I've solved the PDE by using the method of separation of variables. There was a typo in the problem statement. A critical one, I should say. Thanks. $\endgroup$ – alfdc80 Mar 19 '16 at 0:46

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