I have to prove that if the central acceleration is inversely proportional to the square of distance then the central orbit is a conic section. I tried proving it using the polar form of differential equation of a central orbit i.e. $\frac{d^2u}{d\theta^2}+u=\frac{p}{h^2u^2}$
where $u=\frac{1}{r}$ and $p$ is central acceleration given by $p=\frac{\mu}{r^2}$,$\mu$ being a constant of proportionality. Also $h$ is a constant given by $h=r^2\frac{d\theta}{dt}$.
Upon putting $p=\mu u^2$ in the differential equation, I get
$\frac{d^2u}{d\theta^2}+u=\frac{\mu}{h^2}$
This is a second order constant coefficient linear differential equation in $u$ and $\theta$ whose solution is given by:
$u=c_1\cos(\theta)+c_2\sin(\theta)+\frac{\mu}{h^2}$
where $c_1$ and $c_2$ are arbitrary constants. Now the conics in polar form are given by:
$u=\frac{1+e\cos(\theta)}{l}$
where $e$ and $l$ are the eccentricity and length of latus rectum of conic respectively. But I can't think of any way in which I can remove the $\sin(\theta)$ term to get to the required polar form. Please suggest as to how to go about it.
This is just the transformation to the amplitude-phase form of the trigonometric terms, $$ c_1\cos(θ)+c_2\sin(θ)=a\cos(θ-θ_p) $$ where $a^2=c_1^2+c_2^2$ and $θ_p=\arg(c_1+ic_2)$. The latter gives the angle of the pericenter/periapsis which is set to zero in the referenced normal form of the ellipse.