A particle moves with a central acceleration $\mu\left(r+\frac{2a^3}{r^2}\right)$, being projected from an apse at a distance $a$ with twice the velocity for a circle at that distance; find the other apsidal distance and show that the equation to the path is, $$\frac{\theta}{2}=\tan^{-1}(t\sqrt{3})-\frac{1}{\sqrt5}\tan^{-1}\left(t\sqrt{\frac53}\right)$$
Here,
$$ \begin{align} v\frac{dv}{dx}&=-\mu\left(x+\frac{2a^3}{x^2}\right)\\ vdv&=-\mu\left(x+\frac{2a^3}{x^2}\right)dx\\ \frac{V^2}{2}&=-\mu\left(\frac{x^2}{2}-\frac{2a^3}{x}\right)+C \end{align} $$
Question: I didn't understand the line, "...being projected from an apse at a distance $a$ with twice the velocity for a circle at that distance". That's why couldn't get the initial values for this DE.
I knew that then need to solve the pedal equation, $$\frac{d^2u}{du^2}+u=\frac{f}{h^2u^2}$$ where $u=\frac1r$.
Any help will be appreciated. TIA.