How to proceed further on this electromagnetism problem? I am taking intro classes to electromagnetism and our textbook gives a problem like this:
$\mathbf Q.$ A positively charged particle of charge $q$ and mass $m$ is released at origin. There are uniform electric field and magnetic field in the space given by $\vec{E}=E_0\hat{j}$ and $\vec{B}=B_0\hat{k}$, where $E_0$ and $B_0$ are constants. Find the y-coordinate of the particle as a function of time.
My attempt at the solution:
As $\vec{E}=E_0\hat{j}$, the force due to the electric field on the particle will be given by:
$$\vec{F_E}=q.\vec{E}=q.E_0\hat{j}$$ As the particle is initially at rest it will not have any force acting on it due to the magnetic field at $t=0$. At some time $t$, the particle will have gained some velocity due to the action of electric field. Let the velocity at a time $t$ be: $$\vec{v}=v_x\hat{i}+v_y\hat{j}+v_z\hat{k}$$ Since there is no force along the z-axis, hence $v_z=0$ at any time t. So velocity becomes: $$\vec{v}=v_x\hat{i}+v_y\hat{j}$$ Now force due to magnetic field will be given by: $$\vec{F_M}=q.( {\vec{v}} \times {\vec{B}})=qv_yB_0\hat{i}-qv_xB_0\hat{j}$$ So net force on the particle in the y-direction will be: $$F_y=qE_0-qv_xB_0$$ Thus, net acceleration in the y-direction will be given by: $$a_y=\frac{q}{m}(E_0-v_xB_0)$$ which can also be written as: $$\frac{dv_y}{dt}=\frac{q}{m}(E_0-v_xB_0).......(1)$$ Also net acceleration along x-direction will be given by:$$a_x=\frac {dv_x}{dt}=\frac{qv_yB_0}{m}.......(2)$$ Differentiating equation $(1)$ with respect to time we get: $$\frac{d^2v_y}{dt^2}=-\frac{qB_0}{m}.\frac{dv_x}{dt}=-\frac{q^2B_0^2}{m^2}.v_y$$ This second order differential equation seems similar to the equation of Simple harmonic motion: $$\frac{d^2x}{dt^2}=-\omega^2x$$ So we can write the equation as: $$v_y=v_{0y}.\sin(\frac{qB_0}{m}.t+\phi)$$ $$\implies\frac{dy}{dt}=v_{0y}.\sin(\frac{qB_0}{m}.t+\phi)$$ $$\implies dy=v_{0y}.\sin(\frac{qB_0}{m}.t+\phi).dt ........(3)$$ Integrating equation $(3)$ we get: $$\int_{0}^{y}dy=\int_{0}^{t}v_{0y}.\sin(\frac{qB_0}{m}.t+\phi).dt$$ $$\implies y=v_{0y}.\frac{m}{qB_0}.[\cos(\phi)-\cos(\frac{qB_0}{m}.t+\phi)]$$ I am not able to solve this further and I am left with two variables to solve for, $\phi$ and $v_{0y}$. I think there is some boundary condition that I am not able to find here to solve for $\phi$ and I have no idea how to solve for $v_{0y}$.
 A: Perhaps you may want to start with a different method.
We start with 
\begin{equation}
\textbf{F} = m\textbf{a} = q(\textbf{E} + \textbf{v} \times \textbf{B}) \tag{i}
\end{equation}
In terms of components, we have
\begin{equation}
m(a_1, a_2, a_3) = q(0, E_0 , 0) + q((v_1, v_2, v_3) \times (0, 0, B_0)) \tag{ii}
\end{equation}
Hence,
\begin{equation}
m(a_1, a_2, a_3) = q(0, E_0 , 0) + q(B_0 v_2, -B_0 v_1, 0) \tag{iii}
\end{equation}
Simplifying equation (iii) yields
\begin{equation}
m(a_1, a_2, a_3) = q(B_0 v_2, E_0 -B_0 v_1 ,0 ) \tag{iv}
\end{equation}
We can now extract two ODE's:
\begin{equation}
ma_1= q(B_0 v_2) \tag{v}
\end{equation}
\begin{equation}
ma_2 = q(E_0 -B_0 v_1) \tag{vi}
\end{equation}
We can rewrite equation (vi) as
\begin{equation}
\frac{d^2 y}{dt^2}= \frac{qE_0}{m}- \frac{qB_0}{m}\frac{dx}{dt} \tag{vii}
\end{equation}
We can also recast equation (v) as
\begin{equation}
\frac{d^2 x}{dt^2}= \frac{qB_0}{m}\frac{dy}{dt} \tag{viii}
\end{equation}
Integrating equation (viii) once with respect to time yields
\begin{equation}
\frac{d x}{dt}= \frac{qB_0}{m}y + C \tag{ix}
\end{equation}
Here $C$ is some constant. Now, at $t = 0$, the particle is released from rest at the origin so $y = 0$ and $\frac{dx}{dt} = 0$. Hence, $C = 0$. 
Therefore, we have that $\frac{d x}{dt}= \frac{qB_0}{m}y$. Substituting for $\frac{d x}{dt}$ in equation (vii) gives
\begin{equation}
\frac{d^2 y}{dt^2} +  \frac{q^2B_0^2}{m^2}y - \frac{qE_0}{m} = 0 \tag{x}
\end{equation}
Using the initial conditions $y = 0$ and $\frac{dy}{dt} = 0$ at $t=0$, you can check that
\begin{equation}
y = \frac{E_0 m}{B_0 ^2 q} \left(1- \cos\left(\frac{B_0 q}{m} t\right)\right) \tag{xi}
\end{equation}
