I am trying to find a displacement function for the following scenario: A particle starting at some $(0,r_0)$ for positive $r_0$ is given an initial velocity $v_0$ upward, but experiences a variable acceleration towards the x-axis. This acceleration is given by $a=\frac{Gm}{r^2}$, where G and m are constants and $r=r(t)$ is the straight-line distance between the particle and x-axis.
I have tried first solving $a=\frac{dv}{dt}=\frac{d^2r}{dt^2}=\frac{Gm}{r^2}$ for r(t) via Wolfram Alpha but this yields something that looks impossible to find an explicit expression for r(t) with. Any help would be appreciated.
Following the movement law
$$ \ddot y = -\frac{Gm}{y^2}\Rightarrow \ddot y \dot y= -\frac{Gm\dot y}{y^2}\Rightarrow \frac 12\dot y^2 = \frac{Gm}{y}+C_1 $$
we get at
$$ \dot y = \pm\sqrt{2\frac{Gm}{y}+C_2} $$
this ODE is separable
$$ \pm\frac{dy}{\sqrt{2\frac{Gm}{y}+C_2}}=dt $$
giving
$$ \pm\left(\frac{2 G m \tanh ^{-1}\left(\frac{\sqrt{C_2+\frac{2 G m}{y}}}{\sqrt{C_2}}\right)}{C_2^{3/2}}-\frac{y \sqrt{C_2+\frac{2 G m}{y}}}{C_2}\right) = t + C_3 $$
The problem is to find the inverse function to determine $y=f(t,C_2,C_3)$
NOTE
Here $C_2$ can be determined as follows
$$ v_0 = \pm\sqrt{2\frac{Gm}{r_0}+C_2}\Rightarrow C_2 = v_0^2-2\frac{Gm}{r_0} $$