Dyson equation from the equation of motion of the one-body Green's function Starting from the equation of motion of the one-body Green function is:
$$\left[ {i\hbar {\partial  \over {\partial {t_1}}} - {h_0}\left( 1 \right)} \right]G\left( {12} \right) - \int {\Sigma \left( {13} \right)G\left( {32} \right)d3}  = \delta \left( {12} \right)\tag{1}$$
and that for the noninteracting system:
$$\left[ {i\hbar {\partial  \over {\partial {t_1}}} - {h_0}\left( 1 \right)} \right]{G_0}\left( {12} \right) = \delta \left( {12} \right)\tag{2}$$
The authors in [1] got the Dyson equation:
$$G\left( {12} \right) = {G_0}\left( {12} \right) + \int\!\!\!\int {{G_0}\left( {13} \right)\Sigma \left( {34} \right)G\left( {42} \right)d3d4}\tag{3}$$
by multiplication of Eq. (1) with $G_0$ and Eq. (2) with $G$ from the left followed by integration.
How to get eq. (3) by following the same steps?
[1] Friedrich, Christoph, and Arno Schindlmayr. "Many-body perturbation theory: the GW approximation." NIC Series 31 (2006): 335.
 A: I actually found something strange as discussed in the following.
Let us first change the label of Equation 1,
$$\big[i\hbar{\partial \over \partial t_3}-h_0(3)\big]G(32)-\int{\Sigma(34)G(42)d4}=\delta(32)$$
And in a similar way we change the label of Equation 2,
$$\big[i\hbar{\partial \over \partial t_1}-h_0(1)\big]G_0(13)=\delta(13)$$
Now we multiply $G_0(13)$ to the first one and $G(32)$ to the second one,
we have
\begin{align}
& G_0(13)\bigg(\big[i\hbar{\partial \over \partial t_3}-h_0(3)\big]G(32)\bigg)-G_0(13)\int{\Sigma(34)G(42)d4}=G_0(13)\delta(32) \\
& G(32)\bigg(\big[i\hbar{\partial \over \partial t_1}-h_0(1)\big]G_0(13)\bigg)=G(32)\delta(13)
\end{align}
If we integrate both equations over $3$, since $\int{G_0(13)\delta(32)d3}=G_0(12)$ and $\int{G(32)\delta(13)d3}=G(12)$, we have
\begin{align}
& \int{G_0(13)\bigg(\big[i\hbar{\partial \over \partial t_3}-h_0(3)\big]G(32)\bigg)d3}-\iint{G_0(13)\Sigma(34)G(42)d3d4}=G_0(12) \\
& \int{G(32)\bigg(\big[i\hbar{\partial \over \partial t_1}-h_0(1)\big]G_0(13)\bigg)d3}=G(12)
\end{align}
The Equation 3 in the question is true if and only if
$$\int{G_0(13)\bigg(\big[i\hbar{\partial \over \partial t_3}-h_0(3)\big]G(32)\bigg)d3}=\int{G(32)\bigg(\big[i\hbar{\partial \over \partial t_1}-h_0(1)\big]G_0(13)\bigg)d3}$$
I am not sure why this should be true although the Dyson's equation must be correct.
A: You are missing the counterpart of $(1)$. We also have another form of the Dyson equation which is
$$
G\left( {12} \right) \left[
{-i\hbar {\partial  \over {\partial {t_2}}} - {h_0^*}\left( 2 \right)} \right] 
- \int {G\left( {13} \right) \Sigma \left( {32} \right)   d3}  = \delta \left( {12} \right)
\tag{A}
$$
here it is understood that that $\partial  \over {\partial {t_2}}$ and $h_0^*(2)$ are applied on $G(12)$. We get the following relation for the non-interacting Green's function
$$
G_0\left( {12} \right) G_0^{-1}(2) = \delta \left( {12} \right)
\tag{B}
$$
here $G_0^{-1}(1)=\left[ {i\hbar {\partial  \over {\partial {t_1}}} - {h_0}\left( 1 \right)} \right]$ and $G_0^{-1}(2)=\left[
{-i\hbar {\partial  \over {\partial {t_2}}} - {h_0^*}\left( 2 \right)} \right]$. Change label $1$ with $4$ in your equ $(1)$:
$$
 G_0^{-1}(4) G\left( {42} \right) - \int d3 {\Sigma \left( {43} \right)G\left( {32} \right)}  = \delta \left( {42} \right)
$$
now apply $G_0(14)$ from the left hand side and take integration over $d4$:
$$
 \int d4 G_0(14) G_0^{-1}(4) G\left( {42} \right)  - \int d4  \int d3{G_0(14)\Sigma \left( {43} \right)G\left( {32} \right)}  = \int d4 G_0(14)\delta \left( {42} \right)
$$
now use $(B)$ on the first term on the left-hand side
$$
 \int d4 \delta \left( {14} \right) G\left( {42} \right)  - \int d4  \int d3{G_0(14)\Sigma \left( {43} \right)G\left( {32} \right)}  = \int d4 G_0(14)\delta \left( {42} \right)
$$
solve integrations that involve delta functions
$$
 G\left( {12} \right)  - \int d4  \int d3{G_0(14)\Sigma \left( {43} \right)G\left( {32} \right)}  = G_0(12)
$$
interchange label $3$ and $4$
$$
 G\left( {12} \right)  = G_0(12) + \int d3  \int d4{G_0(13)\Sigma \left( {34} \right)G\left( {42} \right)}
$$
