Think of $t$ and $r$ as two independent variables.
Suppose $E$ be a function of $r$ and $V~$ be a function of $(t,r)$ such that both go to $0$ at $r=0$.
There exists a positive function $M(r)$ such that $M(0)=0$ and $V(t,r) = -\dfrac{M(r)}{R(t,r)}$ where $R$ is another positive function such that $R(0,r)=r$.
Let $p(r) = \dfrac{E(r)}{V(0,r)}$ be a function regular at $r=0$ such that $p(0) \in (-\infty,1)$.
Also define a function $a$ of $r$ such that, $a(r) = \dfrac{M(r)}{\dfrac{4}{3}\pi r^3}$. Then $a$ is also a positive definite function with a well-defined value at $r=0$.
Define $\alpha = a(0)$
Now look at this differential equation,
$$\frac{\dot{R}^2}{2} + V(t,r) = E(r)$$
Apparently this differential equation has a solution of the form,
$$\frac{t}{t_0} = \sqrt{\frac{\alpha}{a(r)}}\frac{F(p(r))}{F(p(0))} \left [1 - \left ( \dfrac{R(t,r)}{r} \right)^{\dfrac{3}{2}}~\cdot~\dfrac{F\left(~~ \dfrac{p(r)R(t,r)}{r} \right) }{F(p(r))} \right ] $$
where $t_0 = \sqrt {\dfrac{3}{8\pi \alpha}} F(p(0))$
and the function $F$ is defined over the interval $(-\infty,1)$ as,
$$F(x) = \left\{ \begin{array}{c c} -\frac{\sqrt{1-x}}{x} - \frac{1}{(-x)^{\frac{3}{2}}} \tanh^{-1} \left [ \sqrt{\frac{x}{x-1}} \right ] & x<0 \\ \frac{2}{3} & x =0 \\ \frac{1}{x^{\frac{3}{2}}}tan^{-1} \left [ \sqrt{\frac{x}{1-x}} \right ] - \frac{\sqrt{1-x}}{x} & 0<x<1 \end {array} \right. $$
How does one get the above solution?