Can we solve an equation of the form $$(\ddot{r}-\frac{A}{r^3})=B$$ where
- $A$, $B$ are constants subjected to the initials conditions $r=R$,
- and $\dot{r}=v$ at $t=0$?
Overhead dots represent derivatives w.r.t time $t$. This is the equation I arrived at while solving a physics problem, and it remains to solve this equation.
$$\ddot{r}-\frac{A}{r^3}=B$$ $$2\ddot{r}\dot{r}-\frac{2A\dot{r}}{r^3}=2B\dot{r}$$ $$\dot{r}^2+\frac{A}{r^2}=2Br+c_1$$ At $t=0\quad\to\quad v^2+\frac{A}{R^2}=2BR+c_1\quad$ leads to : $$\dot{r}^2+\frac{A}{r^2}=2Br+v^2+\frac{A}{R^2}-2BR$$ $$\dot{r}=\pm\sqrt{-\frac{A}{r^2}+2Br+v^2+\frac{A}{R^2}-2BR}$$ $$\frac{dr}{\sqrt{-\frac{A}{r^2}+2Br+v^2+\frac{A}{R^2}-2BR}}=dt$$ $$\int\frac{dr}{\sqrt{-\frac{A}{r^2}+2Br+v^2+\frac{A}{R^2}-2BR}}=t+c_2$$ At $r(0)=R$ leads to : $$t(r)=\int_{\rho=R}^{\rho=r}\frac{d\rho}{\sqrt{-\frac{A}{\rho^2}+2B\rho+v^2+\frac{A}{R^2}-2BR}}$$ The closed form $t(r)$ of this elliptic integral is very complicated. Moreover, the inverse function $r(t)$ should be even more nightmarish, of the kind of generalized Jacobi elliptic functions. It is doubtful than a closed form could be derived.