Help with non-equilibrium Green functions with a time-dependent Hamiltonian 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.
Edit : 
 of math I'm trying to reproduce and understand
 A: The expression below is close to, but not exactly the answer you have. But perhaps it can help.
Start with the Green's function eq. in the form 
$$
i \frac{\partial G(t, t')}{\partial t} = [ H_0 + V(t) ] G(t, t') + \delta(t - t')
$$
and look for a solution of the form
$$
G(t, t') = G_0(t - t') W(t, t')
$$
where $G_0(t-t')$ is the Green's function for $H_0$, satisfying
$$
i \frac{\partial G_0(t-t')}{\partial t} = H_0 G_0(t-t') +  \delta(t - t') 
$$
and $W$ satisfies the equal-time condition $W(t, t) = I$. This gives
$$
i \frac{\partial G_0}{\partial t}W + i G_0\frac{\partial W}{\partial t}  = H_0 G_0 W + V(t) G_0 W +  \delta(t - t')
$$
$$
H_0 G_0 W + \delta(t - t') I + i G_0\frac{\partial W}{\partial t}  = H_0 G_0 W + V(t) G_0 W + \delta(t - t')
$$
$$
i \frac{\partial W}{\partial t} = G_0^{-1} V(t) G_0 W
$$
and eventually
$$
W = I - i \int_{t'}^t{dt_1 G_0^{-1}(t_1-t')V(t_1) G(t_1, t') }
$$
Take this back into the expression for $G$, 
$$
G(t, t') = G_0(t-t') - i \int_{t'}^t{dt_1 G_0(t-t')G_0^{-1}(t_1-t')V(t_1) G(t_1, t') } 
$$
use the fact that $G_0(t-t')G_0^{-1}(t_1-t') = G_0(t-t_1)$ and obtain
$$
G(t, t') = G_0(t-t') - i \int_{t'}^t{dt_1 G_0(t-t_1)V(t_1) G(t_1, t') } 
$$
As I already mentioned, this is formally similar to your answer, but not exactly identical. Maybe you can use a hermitian conjugate to rearrange, but there still remains a factor of (-i). Also, if you want to check that $G(t, t')$ is indeed a solution of the Green's function eq. keep the factors $G_0(t-t')G_0^{-1}(t_1-t')$ under the integral separate when you take the derivative on t.
