Solution
= Solution
For $H=p\cosh(2q)$,
$$
\boxed{\dot q=\cosh(2q),\qquad \dot p=-2p\sinh(2q).}
$$
The derivatives of $O=e^{qp}$ are $O_q=pe^{qp}$ and $O_p=qe^{qp}$. Its <Poisson bracket> with $H$ is therefore
$$
\boxed{\dot O=\{O,H\}=p e^{qp}\left[\cosh(2q)-2q\sinh(2q)\right].}
$$
Direct differentiation, $\dot O=e^{qp}(p\dot q+q\dot p)$, gives the same expression and checks the bracket sign.