I am trying to solve the period of the solution to this ordinary differential equation: $$k \cdot r \cdot arctan(r'/r) \cdot (r^2+r'^2)^2= r'(r \cdot r''-2r'^2-r^2)$$
The numerical solution when $k=-20/9$ and initial condition $r(0)=r_0=0.9, r'(0)=0$ is
So I tried this perturbation: $$r(t)=A \cdot [1-\epsilon \cdot cos(\nu t)]$$ where $A=r_0/(1-\epsilon)$.
After plugging this into the ODE, I got $$A^3 (1+A^2k) \cdot \nu sin(\nu t) \cdot \epsilon -\frac{1}{2} A^3 (2+4A^2k+\nu^2) \cdot \nu sin(2\nu t) \cdot \epsilon^2 + O(\epsilon^3) = 0$$
It seems there is no way for me to eliminate the $\epsilon$ term.
Could you tell me what I should do? I'd really appreciate your help!