How to solve system of differential equations? I would like to solve a system of differential equations
\begin{align*}
&x''(t) = -a_0(a_1 - bz'(t))\cos(wt), &&x(t_o)= 0,  &&x'(t_o)=0\\
&z''(t)= -a_0 bx'(t)\cos(wt), &&z(t_o) =0,  &&z'(t_o)= 0
\end{align*}
It reduces to a third order equation
$z'''(t) = a(1-cz'(t))\cos^2(wt)-\tan(wt) dz''(t), z(t_o)=0,z'(t_o)= 0$
I tried mathematica and matlab but they do not want to return analytical solution. 
This is free electron in electromagnetic field.
If the field is complex mathematica find some solution. But the answer is also complex. Is it possible to get real Z for cos(wt) field out of it.
eqns = {z'''[t] == a*Exp[I*t*2*w]*(1 - z'[t]*c) + z''[t]*(I*w), 
   z[t0] == 0, z'[t0] == 0};

soln = Real[DSolve[eqns, z, t][[1]]]


$$x''(t) = -a_0(a_1 - bz'(t))\cos(wt),\ \ \ \  x(t_o)= 0,\ \ \ \ \  x'(t_o)=0$$
let $x'(t)=y(t) ,z'(t)=\beta(t)$
than we get 
$$y'(t)=-a_0(a_1 - b\beta(t))\cos(wt)$$
the second equation converts to $$\beta'(t)= -a_0b y(t)\cos(wt) $$
i think it can be solved now
on solving and referring results from work of @okrzysik Solution of a system of linear odes
$$\beta(t) = A \cos\left(\frac{c \sin(\omega t)}{\omega}\right) + B \sin\left(\frac{c \sin(\omega t)}{\omega}\right) + \frac{d}{c^2} $$
and
$$y'(t)=(-a_0a_1\cos(wt)+a_0b A \cos\left(\frac{c \sin(\omega t)}{\omega}\right) \cos(wt)+ a_0Bb \sin\left(\frac{c \sin(\omega t)}{\omega}\right)\cos(wt) +a_0 \frac{bd}{c^2}\cos(wt))$$
on integrating 

$$y(t)=\frac{1}{w}(a_0 \frac{bd}{c^2}-a_0a_1)\sin(wt)+a_0bA\frac{w^2}{c}\sin(\frac{c\sin(wt)}{w})-a_0Bb\frac{w^2}{c}\cos(\frac{c\sin(wt)}{w})+K$$

now $z(t),x(t)$ could be calculated from $\beta(t),y(t)$

$$z(t) = A \int \cos\left(\frac{c \sin(\omega t)}{\omega}\right) dt+ B \int\sin\left(\frac{c \sin(\omega t)}{\omega}\right)dt +\int \frac{d}{c^2} dt$$
