Function where second derivative is equal to reciprocal squared I was recently interested in whether there exists a closed-form solution for the position of an object (say a spacecraft) in freefall as a function of time. To complicate things, I wanted to take into account the increase in acceleration as you get nearer to the surface.
The differential equation is straightforward:
$ x''(t) = \frac{\mu}{x(t)^2} $ where $\mu = G \cdot m_{planet}$
How, if at all, would I go about solving this? I've tried to use the common functions that return themselves when differentiated (trig, hyperbolic trig, exponential, etc.) but none of them obviously equal the reciprocal of themselves when differentiated twice.
The closest I've gotten was $B\cosh(D \cdot t)$, but it doesn't work in the differential equation and doesn't quite match the analytical curve.

If you switch variables
$$x''(t) = \frac{\mu}{x(t)^2} \quad \implies \qquad -\frac {t''(x)}{[t'(x)]^3}=\frac{\mu}{x^2}$$ Reduction of order $p(x)=t'(x)$ leads to
$$p(x)=t'(x)=\pm\frac{\sqrt{x}}{\sqrt{2} \sqrt{c_1 x-\mu }}$$
$$t(x)+c_2=\pm  \frac{1}{\sqrt{2} c_1^{3/2}}\Big[\sqrt{c_1} \sqrt{x} \sqrt{c_1 x-\mu }+\mu  \log \left(\sqrt{c_1} \sqrt{c_1 x-\mu
   }+c_1 \sqrt{x}\right) \Big]$$
