I'm trying to solve a problem with a time dependent Hamiltonian $$H = H_0 + V(t)$$ where $H_0$ is a Hamiltonian with a non explicit time-dependence and $V(t)$ has an explicit time dependence. My system consists of three parts. Each is an harmonic oscillator and there's interaction with them, that is $$H_0 = H_L+ H_R + H_C +H_{coup}$$
where $L$ stands for left, $R$ for right, $C$ for center and $coup$ for the coupling and
$$H_\alpha = \sum_n \frac{p_{\alpha,n}^2}{2m_\alpha}+ \frac{k_\alpha}{2}(x_{\alpha,n}-x_{\alpha,n+1})^2$$
for $\alpha = L,R,C$ and
$$H_{coup}= \frac{k_L}{2}(x_{L,1}-x_{C,1})^2+\frac{k_R}{2}(x_{R,1}-x_{C,N})^2$$
that is, I have a chain of spring and masses where I can recognise three parts and a coupling.
so I want to solve $(\imath \partial_t -H)G=\delta(t-t')\delta_{l,l'}$ or in matricial form $$\begin{pmatrix} \omega-H_L &-H_{LC} &0 \\ -H_{LC} &\omega-H_C &-H_{RC} \\ 0& -H_{LC} & \omega-H_R \end{pmatrix}. \begin{pmatrix} G_L &G_{L,C} &0 \\ G_{L,C} & G_C &G_{R,C} \\ 0&G_{R,C} &G_R \end{pmatrix}= \begin{pmatrix} 1 & & \\ &1 & \\ & &1 \end{pmatrix}$$
The answer is $$G(t,t')=G_0+\int_{t'}^{t}dt_1G(t,t_1)V(t_1)G_0(t_1-t)$$.
where $G_0$ is the green function of $H_0$.
So to get $G$ I saw that is needed to make a gradient expansion and only can be solve to an order (for example at first order)
$$G(t,\omega) = G_0 + GVG_0+ \imath \partial_\omega G\partial_t V G_0$$ and so on.
My question are why I need to do a gradient expansion? Why isn't $$ G = [G_0^{-1}-V(t)]^{-1}$$ the solution? How can I compute the solution (fix point for example)? I saw that the solution is get by replacing $\tilde{G}=[G_0^{-1}-V(t)]^{-1}$ in the formula above so $$G = \tilde{G}+ \imath \partial_\omega \tilde{G} \partial_t V G_0$$ but I don't realise how to get it.