Let $s(t)=s(r(t),\theta(t))$ be a solution curve of $\ddot{x}(t)$=$\frac{1}{m}F(x(t))$, where $F(x)$ is the Newtonian gravitational field on $\mathbb{R}^2$, and h is non-zero along $s(t)$, where $h=mr^2\dot\theta$.
$u(t)=\frac{1}{r(t)}$, i.e. $u(t)=-V(s(t))$. r is a function of $\theta$ along the curve $s(t)$, i.e. $r=r(\theta)$.
I want to prove the following ode but I don't know how to start: Along $s(t)$ the functions u and $\theta$ satisfy the following ode: $\frac{d^2u}{d\theta^2}+u=\frac{m}{h^2}$