Coupled differential equation $\ddot{y}=\omega \dot{z}$
$\ddot{z}=\omega (\frac{E}{B}-\dot{y})$
These are the coupled differential equation i came across . They have already been asked here
How to Solve the Coupled Differential Equations?
but i am not satisfied with answer because i can not proceed further. Please give me its complete solution.
Answer: 

 A: Just as, in solving a system of linear equations, one reduces to a single equation in one unknown, so in a system of linear differential equations, one reduces to a single equation, of higher order.
The first of these equations says that $y''= \omega z'$.  Differentiating again, $y'''= \omega z''$ so that $z''= \frac{y'''}{\omega}$.  Replace z'' in the second equation by that: $\frac{y'''}{\omega}= \omega(\frac{B}{E}- y')$ so that $y'''= \omega^2(\frac{E}{B}- y')= \frac{\omega^2 E}{B}- \omega^2 y'$
That is the "linear, non-homogeneous, differential equation with constant coefficients" $y'''+ \omega^2 y'= \frac{\omega^2 E}{B}$.
The "associated homogeneous equation" is $y'''+ \omega^2 y'= 0$ which has characteristic equation $r^3+ \omega^2 r= r(r^2+ \omega^2)= r(r+ i\omega)(r- i\omega)= 0$.  That has roots 0, $i\omega$, and $-i\omega$ so then general solution to the associated homogeneous equation is $y(t)= A+ Bcos(\omega t)+ C sin(\omega t)$.
The "non-homogenous" part of the equation is then constant, $\omega\frac{E}{B}$.  Normally, because that is a constant, we would "try" a constant as a solution but, because the constant, A, is a solution to the associated homogeneous equation, new try $y= Dt$ for D a constant.  Then $y'= D$, $y''= 0$, and $y'''= 0$ so the equation becomes $\omega^2 D= \frac{\omega^2 E}{B}$ so $D= \frac{E}{B}$ and the general solution to the entire equation is $y(t)= A+ Bcos(\omega t)+ C sin(\omega t)+ \frac{E}{B}$. 
A: Continuing with the solution of How to Solve the Coupled Differential Equations?
The general solution of :
$$\ddot q + \omega^2q=\omega^2\frac{E}{B}$$
is :
$$q(t)=C_1\cos(\omega t)+C_2\sin(\omega t) +\frac{E}{B}$$
then
$$y(t)=\frac{C_1}{\omega}\sin(\omega t)-\frac{C_2}{\omega}\cos(\omega t)+\frac{E}{B}t+C_3$$
Now I guess that you can find $z$
