Everything else is correct except equations (2.1) to (3.2). Equation (4) cannot be used to since the Hamiltonian is time dependent. What you have to do is take the derivative of (6.1) and substitute equation (6.2). By doing this you will get $$\frac{d^2X(t)}{dt^2}+\omega^2X(t)=\frac{qE_0}{m}\cos{\omega't}$$ I have solved this equation by assuming $\omega=\omega'=q=E_0=m=1$ and $X(0)=1,\dot{X}(0)=0$ for simplicity and got the result as $$X(t)=\frac{1}{4}(2\cos^3t+2\cos t+2t\sin t+\sin{2t}\sin{t})$$
Hope this helps.