Solving the differential equation $q \vec{v} \times \vec{B} + q \vec{E} = m \frac{d\vec{v}}{dt}$

$$q \pmb{v} \times \pmb{B} + q \pmb{E} = m \frac{\mathrm d\pmb{v}}{\mathrm dt}\tag 1$$ If $\pmb{B}=B\hat z$ an $\pmb E=E\hat y$, we have $$\left\{ \begin{align} \dot v_x&=\omega v_y\\ \dot v_y&=-\omega v_x+\gamma \end{align}\right.\tag 2 $$ with $\omega=\frac{qB}{m}$ (cyclotron frequency) and $\gamma=\frac{qE}{m}$. The system (2) can be written as

$$\left\{ \begin{align} \ddot v_x+\omega^2 v_x&=\omega\gamma\\ \ddot v_y+\omega^2 v_y&=0 \end{align}\right.\tag 3 $$ and the solution might be written as $$\left\{ \begin{align} v_x(t)&=a\cos(\omega t+\phi)+\tfrac{E}{B}\\ v_y(t)&=b\cos(\omega t+\theta) \end{align}\right.\tag 4 $$ and finally $$\left\{ \begin{align} x(t)&=\frac{a}{\omega}\sin(\omega t+\phi)+\tfrac{E}{B}t\\ y(t)&=\frac{b}{\omega}\sin(\omega t+\theta) \end{align}\right.\tag 5 $$ Assuming for simplicity that initially $(t = 0)$ the particle is at rest at the origin of a Cartesian coordinate system, we have $$\left\{ \begin{align} v_x(0)=0&=a\cos(\phi)+\tfrac{E}{B}\\ x(0)=0&=\frac{a}{\omega}\sin(\phi) \end{align}\right. $$ $$ \phi=k\pi, \, k\in\Bbb Z\quad\text{and}\quad a=-\tfrac{E}{B} $$ that is $$ x(t)=\frac{E}{B}\left[t-\frac{1}{\omega}\sin(\omega t)\right]\tag 6 $$ Analogously we find $$ y(t)=\frac{E}{B}\frac{1}{\omega}\left[1-\cos(\omega t)\right]\tag 7 $$ Putting $\omega t=\psi$ and $\frac{E}{B\omega}=\rho$ we have $$\left\{ \begin{align} x&=\rho\left[\psi-\sin(\psi)\right]\\ y&=\rho\left[1-\cos(\psi)\right] \end{align}\right.\tag 8 $$ which gives us the parametric equations of a cycloid.

Observe that from (6) and (7) we have $$ (x-\rho \omega t)^2+(y-\rho)^2=\rho^2\tag 9 $$ that is a circle of radius $\rho$ whose center $C:=(\rho \omega t,\rho)$ travels in the $x$ direction at a constant speed $$v_c=\omega \rho=\frac{E}{B}$$ The particle moves as though it were a spot on the reem of a wheel, rolling down the $x$ axis at speed $v_c$. And the curve generated this way is a cycloid.

Notice that the overall motion is not in the direction of $\pmb E$, as one might suppose, but perpendicular to it.

You just missed one more step. You need to find the $x$ and $y$ coordinates. Just integrate $\vec{v}$ with respect to time. The cycloid contains the sine and cosine terms, plus in the $\hat{i}$ direction you have a term proportional to time: $$\frac{E}{B}t$$