= Solution
Use the first-order <action>
$$
S[q,p]=\int_{t_1}^{t_2}\bigl(p\dot q-H(q,p,t)\bigr)\,dt.
$$
Varying $q$ with fixed endpoints and varying $p$ freely gives
$$
\delta S=\int_{t_1}^{t_2}
\left[(\dot q-H_p)\delta p+(-\dot p-H_q)\delta q\right]dt.
$$
The <principle of stationary action> therefore yields <Hamilton's equations>
$$
\dot q=H_p,\qquad\dot p=-H_q.
$$
For $H=p^2+q^{-2}$ these become $\dot q=2p$ and $\dot p=2q^{-3}$. Along their <Hamiltonian flow>, $H=E$ is conserved and
$$
\frac d{dt}(pq)=\dot p\,q+p\dot q=2q^{-2}+2p^2=2H.
$$
Thus $F=pq-ctH$ is an <integral of motion> exactly when \b[$c=2$].
Write its constant value as $F_0$. The surfaces $pq-2tE=F_0$ in extended $(q,p,t)$-space give
$$
pq=F_0+2Et.
$$
Using $Eq^2=p^2q^2+1$,
$$
\boxed{\ q(t)=\pm\sqrt{\frac{(F_0+2Et)^2+1}{E}},
\qquad
p(t)=\frac{F_0+2Et}{q(t)}\ }.
$$
The sign of $q$ is fixed. In <phase space>, each energy orbit is one of the two branches
$$
p^2+q^{-2}=E,\qquad
q=\pm(E-p^2)^{-1/2},\qquad |p|<\sqrt E.
$$
Each branch approaches $p=-\sqrt E$ and $p=+\sqrt E$ as $|q|\to\infty$ and turns at $(p,q)=(0,\pm E^{-1/2})$.
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