Solving $\rm\dot v = \frac q m v \times B + \frac q m E$ I need to solve the following ode: $\rm\dot v = \frac q m v \times B + \frac q m E$, where $\mathrm{ v_0}=(0,0,v_0), \mathrm{B}= (0,B,0), \mathrm{E} = (0,E,0) $.
If it helps anyhow, this comes from a particle entering a magnetic/electric field with speed perpendicular to both fields.
I believe that $v_y=\frac q m E \,t $ and $v_x=v_0$. But I couldn't get $v_z$ because the equation yields $\dot v_z=\frac q m \sqrt{v_x^2+v_y^2+v_z^2}B \sin(\alpha)$ where $\alpha$ is the angle between $\rm v, B$, and I don't know how to solve that nonlinear ode...

After working out the cross product, we see that the system of ODEs is $$\dot v_x=-\frac{qB}{m} v_z,$$ $$\dot v_y=\frac{q E}{m},$$  $$\dot v_z=\frac{qB}{m}v_x.$$ The solution to $v_y$ can be found by direct integration: $$v_y=\frac{qE}{m}t.$$ Let $qB/m=w$. For the other two, take another derivative:
$$\ddot v_x=-w\dot v_z=-w^2 v_x$$ $$\ddot v_z=w\dot v_x=-w^2 v_z.$$ These are standard ODEs whose solutions are well known (simple harmonic oscillator) $$v_x=A\sin(wt)+B\cos(wt)$$ $$v_z=C\sin(wt)+D\cos(wt).$$ Since we know that $v_x(0)=0$, we know that $B=0$. Also, since $v_z(0)=v_0$, $D=-v_0$. Next, $\dot v_x(0)=-wv_z(0)=-wv_0$, so $A=-v_0$. Finally, $\dot v_z(0)=wv_x(0)=0$, so $C=0$. Thus, $$v_x=-v_0\sin(wt)$$ and $$v_z=-v_0\cos(wt).$$
