How to derive the analytical expression for the retarded Green's function with quadratic Hamiltonian?

For two operators, $$A(t)$$ and $$B(t)$$ the retarded Green’s function is defined as $$$$G^R(t,t') \equiv \langle \langle A(t)|B(t) \rangle \rangle^R = -i\theta(t-t')\langle \{A(t),B(t')\} \rangle$$$$

then one can show that \begin{align*} i\dfrac{\partial}{\partial t} G^R(t,t') & = \delta(t-t')\langle \{A(t),B(t')\} \rangle - i \theta(t-t')\langle \{i\dfrac{\partial A(t)}{\partial t},B(t')\} \rangle \\ & = \delta(t-t')\langle \{A(t),B(t')\} \rangle - i \theta(t-t')\langle \{[A(t),H(t)],B(t')\} \rangle \\ & = \delta(t-t')\langle \{A(t),B(t')\} \rangle + \langle \langle [A(t),H(t)]|B(t') \rangle \rangle^R \end{align*} If the Hamiltonian $$H$$ in Schrodinger picture is independent of time, then the correlation functions depend on $$(t-t')$$, not on $$t$$ and $$t'$$ separately. We can go to the Fourier space,the EOM becomes $$$$\boxed{\omega \langle \langle A|B \rangle\rangle^R = \langle \{A,B\} \rangle + \langle \langle [A,H]|B \rangle\rangle^R}.$$$$

Starting with this formula, I want to derive the analytical expression for retarded Green's function with the following Hamiltonian $$H$$ (fermionic system): $$$$H = \sum_k x a_k^\dagger a_k + \sum_{m \neq n} y (a^\dagger_m a_n + a^\dagger_n a_m)$$$$

This following is my solution: $$$$\boxed{A=a_s,B=a^\dagger_t} \Rightarrow \omega \langle \langle a_s|a^\dagger_t \rangle \rangle^R = \langle \{a_s,a^\dagger_t\} \rangle + \langle \langle [a_s,H]|a^\dagger_t \rangle \rangle^R$$$$

\begin{align*} [a_s,H] & =\left[a_s,\sum_k x a_k^\dagger a_k + \sum_{m \neq n} y (a^\dagger_m a_n + a^\dagger_n a_m) \right ]\\ & = \sum_k x \{a_s,a_k^\dagger\}a_k + \sum_{m \neq n} y \{a_s,a_m^\dagger\}a_n +\sum_{m \neq n} y \{a_s,a_n^\dagger\}a_m \\ & = x a_s+ \sum_{n} y a_n \qquad (s=m \neq n) \end{align*}

$$$$\omega G_{st}^R = \delta_{st} + x G_{st}^R + \sum_n y \langle \langle a_n|a^\dagger_t \rangle \rangle^R \Rightarrow G_{st}^R = \dfrac{\delta_{st}+y\sum_n G^R_{nt}}{\omega-x}$$$$

But this result is the final solution? Or how can I further simplify my results?

$$\textbf{Quick hint:}$$ Using this answer. You just have to identify a specific form of $$\mathbb{H}=\left(x-y\right)\mathbb{I}_{N \times N}^{}+ y \mathbb{J}_{N \times N}^{}$$, with $$\mathbb{I}_{N \times N}^{}$$ and $$\mathbb{J}_{N \times N}^{}$$ being identity matrix and matrix of ones respectively. $$\mathbb{G}_{}^{R}(E)=\left[E\mathbb{I}-\mathbb{H}+i 0_{}^{+}\mathbb{I}\right]_{}^{-1}$$ can be found analytically using Sherman-Morrison formula.