# Solution to Differential Equation

I'm looking for a solution to the following differential equation:

$$y'' = \frac{c_1}{y} - \frac{c_2}{y^2}$$

where $c_1$ and $c_2$ are non-zero constants, and y is always positive.

The resulting function should be periodic. Any help is appreciated.

• Well technically $y = c_{2}/c_{1}$ is periodic . . . – user14717 Jan 20 '15 at 21:15
• Do you have some boundary conditions? – Alex R. Jan 20 '15 at 21:16
• I think the typical trick here is to multiply by $y'$. The first integration will be easy. The second would not be something I'd want to attempt... – Mike Jan 20 '15 at 21:45
• Do you have some more conditions, as otherwise the solution won't be periodic. For instance, if $c_1 = c_2 = 1$ and $y(0) = 1, y'(0) = 2$, then $y''(0) = 1/4 \Rightarrow y'(t) > 0 \Rightarrow y(t) > 2 \Rightarrow y''(t) > 0$ for all $t > 0$. – Simon S Jan 20 '15 at 21:57
• Do you have any reason to believe that the solution can be expressed in terms of elementary functions? – hasnohat Jan 21 '15 at 1:18

$$y' y'' = \frac12 \frac{d}{dx} (y'^2) = c_1 \frac{y'}{y} - c_2 \frac{y'}{y^2}$$
$$\frac12 y'^2 = c_1 \log{y} + \frac{c_2}{y} + K_1$$
where $K_1$ is a constant of integration. Now take the square root of both sides and integrate to get
$$\pm x+K_2 = \frac1{\sqrt{2}}\int \frac{dy}{\sqrt{K_1+c_1 \log{y}+c_2 y^{-1}}}$$