# A Newtonian differential equation I'm having trouble with

As an illustration to an answer over at Physics, I have been trying to work out what, for $r(0)=1$, $r'(0)=0$ and $r''(t)={{-1}\over{r(t)^2}}$, the (real) solution $r(t)$ is. Actually, I only need to figure out $r'(t)$, with $t$ such that $r(t)=0$.

It has been fun to try for a while, but now I need help. Well, I don't need it, but I would certainly be pleased if somebody could open my eyes.

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The initial conditions make this one tough. You'd think that $r(t) = (a+b t)^{2/3}$ would do the trick, but I can't get the condition on $r'(t)$ to work. – Ron Gordon Jan 15 '13 at 19:45
@rigordonma Would you say that that is conclusive? There is no analytic solution, not even to the $r"(t^*)$ question alone? – Keep these mind Jan 15 '13 at 21:03
No, not at all. This is a nonlinear equation, so unfortunately there aren't too many conclusions you can draw from my observation. (Or at least any that I know about.) – Ron Gordon Jan 15 '13 at 21:12
No, I do not, because the posted "solution" violates the condition $r'(0) = 0$. I like Marek's solution below and you should at least upvote it. – Ron Gordon Jan 16 '13 at 11:54
Done that, thanks. You can tell that the derivative of $y(t)=\sum_{n=1}^\infty[...]$ on Wiki is not $0$ for $t=0$? Or is it not applicable to the ODE? – Keep these mind Jan 16 '13 at 12:07

We have $$r'' = -1 / r^2,$$ $$((r')^2)' = -2 r' / r^2,$$ $$(r')^2 = 2/r + A.$$ $$r' = \sqrt{2/r + A}.$$ When $A = 0$ we have $r = (9/2)^{1/3} (t + B)^{2/3}$.
For $A \neq 0$ we need to evaluate $$R(r) =\int {dr \over \sqrt {2/r + A}} = \int {rdr \over \sqrt {2r + Ar^2}},$$ which is a standard integral that I leave to you. When you find the antiderivative put $R(r) = t + C$ and invert to obtain $r$ as a function of $t$.
@Gugg antiderivative to $f$ is a function $F$ such that $F' = f$ and is obtained by the process of integration. Does that help? – Marek Jan 15 '13 at 20:15
I think $A=-2$ from the initial conditions stated. This might help simplify things, stress on "might". – Ron Gordon Jan 15 '13 at 21:15