How do I solve for r(theta) in central force motion with r''=(k^2)*r?

[DarkEternal]

I have a central force motion with vector form r''=(k^2)*r, where k^2>0. It's trivial to solve for the vector r(t), but I'm having a little trouble solving for r(theta). I get a second order differential equation of the form r''=A/r^3+Br. Any tips on solving this?

 

[Gokul43201]

$$r'' = \frac {A}{r^3} + Br = - \frac {d}{dr}(\frac{A}{2r^2} - \frac{Br^2}{2})$$

Multiply both sides with r' = dr/dt

$$LHS = r'r" = \frac{1}{2} \frac {d}{dt} r'^2$$
$$RHS = -\frac{dX}{dr}.\frac{dr}{dt} = -\frac{dX}{dt}$$
This gives :
$$r'^2 = -2(X) + const.$$

 

[DarkEternal]

think i got it, thanks gokul