Help solving r''(t) * r2(t) = k ?

[Some1]

I have the equation:

r''(t) * r²(t) = k

Where k is a constant and I want to find r(t) in terms of k and t.
Any help would be appreciated.

PS.
I was wondering whether I could square root it?:

d²/(dt)² * r(t) * r²(t) = k

d/(dt) * r^1/2(t) * r(t) = k^1/2

r^3/2(t) * d = k^1/2 * dt

∫r^3/2(t) * d = ∫k^1/2 * dt

r^3/2(t) = k^1/2t

r(t) = (kt²)^1/3

However I have a feeling I did something wrong here...

 

[HallsofIvy]

No, the second derivative is NOT a square so the square root of a second derivative is NOT a first derivative. $\displaystyle \sqrt{\frac{d^2f}{dx^2}}$ is NOT $\displaystyle \frac{df}{dx}$.

What you can do is use a method called "quadrature". Let v= dr/dt. Then $\displaystyle \frac{d^2r}{dt^2}= \frac{dv}{dt}$ and, by the chain rule, $\displaystyle \frac{dv}{dt}= \frac{dv}{dr}\frac{dr}{dt}= v\frac{dv}{dr}$.

The differential equation becomes $\displaystyle r^2v\frac{dv}{dr}= k$, a separable first order differential equation. We can write that as $\displaystyle v dv= \frac{k}{r^2}dr$ and integrate both sides: $\displaystyle \frac{1}{2}v^2= -\frac{k}{r}+ C$. Solve that for v: $\displaystyle \frac{dr}{dt}= v= \sqrt{\frac{2k}{r}+ 2C}$.

Now you just need to (try to) integrate $\displaystyle \frac{dr}{\sqrt{\frac{2k}{r}+ 2C}}= dt$.

It's doing that integration on the left that is challenging! It looks like it might well be an "elliptic" integral-
http://en.wikipedia.org/wiki/Elliptic_integral.

 

[Deleted member 4993]

Wolframalpha gives a solution w/o elliptic integrals.

 

[Ishuda]

A sightly different approach to get the same answer:
r''(t) * r²(t) = k
or, rearranging and multiplying through by r',
r' r'' = k r' r^-2
Integrate to get
(r')² = 2 (- k r^-1 + c)
where c is the (collected) constant of integration. Thus
r' = [2 (c - k r^-1 )]^1/2