# How to solve $y'' = -\frac{k}{y^2}$, with $k > 0$?

I've been staring at this for a while and looking around the internet to see if I can find a solution, but no success. I think it probably has an exact solution, since I got it from a first year college physics test. In particular, this equation describes the motion of a body falling in a gravitational field.

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For gravity where $y$ is the distance to the point of attraction, you probably want $k$ negative. – Henry Feb 28 '12 at 23:22
@Henry: I know, but mathematically does it make any difference? – Javier Feb 28 '12 at 23:42
it makes a difference with simple harmonic motion of the form $y''=ky$ so I would not be surprised if it made a difference here – Henry Feb 28 '12 at 23:56

Mutiply by $2y'$ to get

$$(y'^2)' = \left(\frac{-2k}{y}\right)'$$

Thus giving

$$y'^2 = C - \frac{2k}{y}$$

which should be solveable by taking square roots etc (the sign depending on the physical assumptions)

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@Javier: Did this answer help? – Aryabhata Feb 29 '12 at 0:09
Well, I get $y' = \pm \sqrt{2k(\frac1{y}-\frac1{y(0)})}$ (I made it clearer in the title that the right hand side of the equation should be negative), but I don't know how to go from there. – Javier Feb 29 '12 at 0:47
@JavierBadia: Have you heard of the separation of variables method? en.wikipedia.org/wiki/Separation_of_variables – Aryabhata Feb 29 '12 at 1:06
I end up with $\int \frac{\mathrm{d}y}{\sqrt{\frac{2k}{y}+C}} = \int \mathrm{d}x$. According to WolframAlpha, the integral on the left hand side is quite complicated, and it seems impossible to solve for $y=y(x)$. Am I doing something wrong? – Javier Feb 29 '12 at 2:03
@JavierBadia: No, you aren't doing anything wrong. It might be difficult to solve for $y$ in terms of $x$. Does the physics test really need you do that? – Aryabhata Feb 29 '12 at 2:09