Perpendicular Fields and Equations of Motion I have the following problem to solve:

A particle of mass $m$ and charge $e$ moves in the laboratory in crossed, static, uniform, electric and magnetic fields.  $\mathbf{E}$ is parallel to the $x$-axis; $\mathbf{B}$ is parallel to the $y$-axis.  Find the EOM for $|\mathbf{E}|<|\mathbf{B}|$ and $|\mathbf{B}|<|\mathbf{E}|$.

I was planning on using the following:
$$\vec{E}'=\gamma(\vec{E}+\vec{\beta}\times\vec{B})-\frac{\gamma^2}{\gamma +1}\vec{\beta}(\vec{\beta}\cdot \vec{E})$$
$$\vec{B}'=\gamma(\vec{B}-\vec{\beta}\times\vec{E})-\frac{\gamma^2}{\gamma +1}\vec{\beta}(\vec{\beta}\cdot \vec{B})$$
along with 
$$t'=\gamma (t-\vec{\beta}\cdot\vec{x})$$
$$ \vec{x}'=\vec{x}+\frac{(\gamma -1)}{\beta^2}(\vec{\beta}\cdot \vec{x})\vec{\beta}-\gamma \vec{\beta}t$$
with the addition constraints
$$\frac{d\vec{p}}{dt}=e\left[ \vec{E}+\frac{\vec{u}}{c}\times \vec{B}\right]$$
and
$$\frac{dU}{dt}=e\vec{u}\cdot \vec{E}$$
To solve this I am going to switch to a frame with
$$\vec{\beta}=\frac{E}{B}\hat{z}$$for the first case.  With this case $dU/dt=0$ and i can solve the equations of motion to find $\vec{x}(t)$ directly, and then boost back to get the trajectories in the original frame.  However for the the second case I was wondering if my procedure is correct.  I am going to switch to a frame with $\vec{\beta}=(B/E) \hat{z}$ to remove the magnetic field.  Now here it seems that $\vec{u}_0$ is perdepdicular to $\vec{E}$ to start in the new frame, but that it will be accelerated in the $x$ direction and hence $dU/dt\neq 0$ and I can't just straightforwardly solve the EOMs.  How would I proceed from here?
Thanks,
 A: After switching to the frame mentioned above, we are left with only a static electric field, perpendicular to the initial velocity of the particle.  Now we consider
$$\frac{d u^\alpha}{d\tau}=\frac{e}{mc}F^{\alpha\beta}u_\beta$$  This decomposes as
$$\frac{du^0}{d\tau}=\frac{e}{mc}F^{0\beta}u_\beta=\frac{e}{mc}F^{0i}u_i=\frac{e\gamma}{mc}\vec{E}\cdot\vec{v}$$
and
$$\frac{du^i}{d\tau}=\frac{e}{mc}F^{i\beta}u_\beta=\frac{e}{mc}(F^{i0}u_0 -F^{ij}u_j)=\frac{e}{mc}(\gamma c)\vec{E}$$
since $\vec{B}$ is zero in this frame.
Now I write all the velocities as parallel and perpendicular to the electric field and define $\omega_E=\frac{eE}{mc}$
$$\frac{d(\gamma c)}{d\tau}=\omega_E (\gamma v_{||})$$
$$\frac{d(\gamma v_{||})}{d\tau}=\omega_E (\gamma c)$$
$$\frac{d(\gamma v_{\perp})}{d\tau}=0$$
Then differentiating the second equation by $d/d\tau$ we get
$$\frac{d^2(\gamma v_{||})}{d\tau^2}=\omega_E \frac{d(\gamma c)}{d\tau}=\omega_{E}^2(\gamma v_{||})$$
solutions to this are $(\gamma v_{||})=A\sinh(\omega_E \tau)+B\cosh(\omega_E \tau)$ which implies $(\gamma c)=A\cosh(\omega_E \tau)+B\sinh(\omega_E \tau)$ and $\gamma v_{\perp}=\text{const.}$.  Now at $\tau=0$ we know that $v_{\perp}=v_0$ and $v_{||}=0$.  Let 
$$\gamma_0=\frac{1}{\sqrt{1-\frac{v_{0}^{2}}{c^2}}}$$
then the initial conditions demand that $B=0$, $A=\gamma_0 c$, and $\text{const.}=\gamma_0 v_0$.  Then we have
$$(\gamma c)=\gamma_0 c \cosh(\omega_E \tau)\implies \gamma=\gamma_0 \cosh(\omega_E \tau)$$
$$(\gamma v_{||})=\gamma_0 \cosh(\omega_E \tau)v_{||}=(\gamma_0 c)\sinh(\omega_E \tau)\implies v_{||}=c\tanh(\omega_E \tau)$$
$$\gamma v_{\perp}=\gamma_0 \cosh(\omega_E \tau)v_{\perp}=\gamma_0 v_0\implies v_{\perp}=\frac{v_0}{\cosh(\omega_E \tau)}$$
Now $dt/d\tau=\gamma$ so that 
$$t=\int_{0}^{\tau}\gamma d\tau'=\int_{0}^{\tau}\gamma_0 \cosh(\omega_E \tau')d\tau'$$
then
$$\frac{t \omega_E}{\gamma_0}=\sinh(\omega_E \tau)$$  The important one though is 
$$\cosh(x)=\sqrt{1+\sinh^2(x)}\implies \cosh(\omega_E \tau)=\sqrt{1+\frac{t^2 \omega_{E}^{2}}{\gamma_{0}^{2}}}$$
after plugging these into the above equations for $v_{\perp}$ and $v_{||}$ we can solve for $x_{||}$ and $x_{\perp}$ from
$$dx=v\, dt\implies x_{||}=\int_{0}^{t}v_{||}(t')dt'\quad \text{and}\quad x_{\perp}=\int_{0}^{t}v_{\perp}(t')dt'$$
These are what I was looking for.
