= Solution
The <canonical momenta> are $p_x=m\dot x$, $p_y=m\dot y+qBx$, $p_z=m\dot z$. The <Legendre transform in mechanics> gives
$$
\boxed{H=\frac{p_x^2+(p_y-qBx)^2+p_z^2}{2m}+qEy.}
$$
<Hamilton's equations> are
$$
\dot x=p_x/m,\quad\dot y=(p_y-qBx)/m,\quad\dot z=p_z/m,\qquad\dot p_x=qB\dot y,\quad\dot p_y=-qE,\quad\dot p_z=0.
$$
Thus $\ddot x=\Omega\dot y$, $\ddot y=-\Omega\dot x-qE/m$, and $\ddot z=0$, where $\Omega=qB/m$. The signs also follow from $\mathbf E=-E\mathbf e_y$ and $\mathbf B=B\mathbf e_z$.
For $Bq\ne0$, put $v_d=-E/B$, $a=v_{x0}-v_d$, $b=v_{y0}$. The motion is
$$
\boxed{\begin{aligned}x(t)&=x_0+v_dt+\frac{a\sin\Omega t+b(1-\cos\Omega t)}\Omega,\\y(t)&=y_0+\frac{a(\cos\Omega t-1)+b\sin\Omega t}\Omega,\\z(t)&=z_0+v_{z0}t.\end{aligned}}
$$
Differentiation verifies both equations and all initial conditions. The centre of the circular motion has the <electric drift> <velocity> $-E\mathbf e_x/B$. For $B=0$, take $x=x_0+v_{x0}t$, $y=y_0+v_{y0}t-qEt^2/(2m)$ and the same $z$. For $q=0$, the particle moves freely.
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