 Solution

Use the first-order actionVarying $q$ with fixed endpoints and varying $p$ freely givesThe principle of stationary action therefore yields Hamilton's equations

For $H=p^2+q^{-2}$ these become $\dot{q}=2p$ and $\dot{p}=2q^{-3}$. Along their Hamiltonian flow, $H=E$ is conserved andThus $F=pq-ctH$ is an integral of motion exactly when $c=2$.

Write its constant value as $F_0$. The surfaces $pq-2tE=F_0$ in extended $(q,p,t)$-space giveUsing $E{q}^2=p^2{q}^2+1$,The sign of $q$ is fixed. In phase space, each energy orbit is one of the two branchesEach branch approaches $p=-\sqrt{E}$ and $p=+\sqrt{E}$ as $|q|\to \infty$ and turns at $(p,q)=(0,\pm E^{-1/2})$.