Solution to Differential Equation

Question

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.

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 01:18
I don't have any reason to believe there is a solution in elementary functions, and I should have been more specific about its periodicity. I know that there are situations to this equation where the result is periodic. However, I also know there are results where the y is just a constant (y describes the orbit of certain planetary orbits). — palindromicPrimes, Jan 21 '15 at 01:37
@palindromicPrimes I calculated a few solutions numerically with various parameters, and ended up with oscillating solutions where the amplitude appears to grow unbounded. Is that what you're expecting? — hasnohat, Jan 28 '15 at 05:19

score 1 · Answer 1 · answered Jan 20 '15 at 21:53

$$y' y'' = \frac12 \frac{d}{dx} (y'^2) = c_1 \frac{y'}{y} - c_2 \frac{y'}{y^2} $$

Integrate both sides to get

$$\frac12 y'^2 = c_1 \log{y} + \frac{c_2}{y} + K_1$$

where $K_1$ is a constant of integration. Now take the square root of both sides and integrate to get

$$\pm x+K_2 = \frac1{\sqrt{2}}\int \frac{dy}{\sqrt{K_1+c_1 \log{y}+c_2 y^{-1}}}$$

At this point, nothing further comes to mind.

1 Answers1