A simple ordinary equation?

Question

How to solve the following ordinary equation? $$-\frac{d^2f(x)}{dx^2}+a f(x)^2=0,$$ where $a>0\in\mathbb{R}$.

Does anybody know of this equation?

Answer 1

I will expand on LutzL's comment, and I will rewrite: $y=f(x)$: $$-y''+ay^2=0.$$

Use the integrating factor $2y'$, multiplying it by every term to get: $$-2y'y''+2ay'y^2=0.$$

Recognize that this is equivalent to: $$\frac{d}{dx}\left((y')^2 + \frac{2a}{3}y^3\right)=0$$ by the chain rule. If a derivative is zero, then the function itself is constant, i.e.: $$(y')^2+\frac{2a}{3}y^3=c.$$

As LutzL mentioned, this is indeed separable; rearranging gives: $$ \frac{dy}{\pm\sqrt{c-\frac{2a}{3}y^3}}=dx $$ which you must now integrate. Mathematica readily does so, but as mrtaurho noted elliptic functions are involved.