What are Hamilton's equations with respect to a nonstandard symplectic form? Hamilton's equations for a Hamiltonian $H(q,p)$  w.r.t. to a standard symplectic from $\omega = dq \wedge dp$ are $$\dot{q} = \partial H_{p}, \quad \dot{p} = - \partial H_{q}$$
How do Hamilton's equations write w.r.t. a nonstandard symplectic form $F(q,p) dq \wedge dp$, where $F(q,p)$ is some smooth function?
 A: *

*More generally, let there be given a Poisson manifold $(M,\pi)$, where
$$\pi ~=~ \frac{1}{2} \pi^{IJ} \frac{\partial}{\partial z^I} \wedge  \frac{\partial}{\partial z^J} \tag{1}$$
is a Poisson bi-vector, and
$$\{ f, g\}_{PB}~=~\frac{\partial f}{\partial z^I}\pi^{IJ}\frac{\partial g}{\partial z^J} \tag{2}$$
is the corresponding Poisson bracket. Let the Hamiltonian $H$ be a globally defined function on $M$. Then Hamilton's equations read
$$ \dot{z}^{I}~=~\{ z^I, H\}_{PB},\tag{3} $$
i.e. time-evolution is given by (minus) the Hamiltonian vector field
$$ X_H~=~\{H,\cdot\}_{PB}.\tag{4} $$


*If the Poisson structure is invertible, then $M$ is a symplectic manifold
with symplectic 2-form
$$\omega ~=~\frac{1}{2}  \omega_{IJ}~ \mathrm{d}z^I \wedge  \mathrm{d}z^J,\tag{5}$$
where $\omega_{IJ}$ is the inverse matrix:
$$ \pi^{IJ}\omega_{JK}~=~\delta^I_K. \tag{6}$$


*In canonical/Darboux coordinates
$$ (z^1, \ldots, z^{2n})~=~(q^1, \ldots, q^n,p_1,\ldots, p_n) ,\tag{7}$$
the above construction reduces to the standard Poisson bi-vector
$$\pi~=~\frac{\partial}{\partial q^i} \wedge  \frac{\partial}{\partial p_i},\tag{8}$$
and the standard symplectic 2-form
$$\omega ~=~ \mathrm{d}p_i \wedge  \mathrm{d}q^i.\tag{9}$$
A: A Hamiltonian $H:M\rightarrow \mathbb{R}$ defines a vector field $X_H$ through the equation
\begin{equation}
  \omega(X_H,\cdot)=dH.
\end{equation}
For $\omega=F(q,p)dq\wedge dp$ and substituting the components $X_H=X_{Hq}\partial_q+X_{Hp}\partial_p$ we get 
\begin{equation}
  F(q,p)(X_{Hq}dp-X_{Hp}dq)=(\partial_qH)dq+(\partial_pH)dp.
\end{equation}
The integral curves $t\mapsto(q(t),p(t))$ of  the vector field $X_H$ represent the Hamiltonian flow of the system. Therefore, we have
\begin{align}
  \dot{q}=\frac{\partial_qH}{F(q,p)};\;\;\;
  \dot{p}=-\frac{\partial_pH}{F(q,p)};
\end{align}
