Namely, is there a function $f(x)$ that satisfies
$$\frac{d^2f(x)}{dx^2} = \frac{1}{(f(x))^2}$$
I've been messing with an overly simplified physical system in which a body has a gravitational attraction towards another body much more massive (allowing me to assume that the second body doesn't move), and I got the following differential equation:
$$\frac{d^2x(t)}{dt^2} = \frac{Gm}{(x(t))^2}$$
That got me wondering if such differential equation has a known solution.
Best Answer
hint
$$y'' = \frac{1}{y^2} \implies$$
$$y'y''=\frac{y'}{y^2} \implies$$
$$\frac{y'^2}{2}= -\frac 1y +C$$
In the particular case $ C=0 $, we have
$$y'=\frac{\sqrt{2}}{\sqrt{-y}}$$ and
$$\frac 23(-y)^\frac 32=-x\sqrt{2}$$
From here, we can see that the function $$Y:x\mapsto x^{\frac 23}$$ satisfies $$Y'=\frac 23x^{-\frac 13} \; \text { and }\;Y''=-\frac 29x^{-\frac 43}$$ $$=\frac{K}{Y^2}$$