# Finding the general solution of this differential equation

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.

• What is $x$? Is it a constant? Is it a function of $t$? Is $y=y(t)$ only or $y=y(x,t)$? – bartgol Mar 18 '16 at 20:56
• What homogeneous solution? This is not a linear equation. – Robert Israel Mar 18 '16 at 21:18
• 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. – alfdc80 Mar 18 '16 at 21:55
• 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 – alfdc80 Mar 19 '16 at 0:57

$$y(x,t) = \dfrac{G(t)}{\sqrt{A - c \ln(x)}}$$
$$\dfrac{dG}{dt} = \sqrt{B - c \ln G(t)}$$
$A, B, c$ arbitrary constants.