Given the definition of Green's function $G(x,s)$ by Wikipedia as the solution of $L ~G(x,s) = \delta(x-s)$. Consider the following equation
$$\Big( \frac{d}{dt} + i A + B \Big) f(t) = C$$
where $A$, $B$ and $C$ are constants.
How can one find the Green's function $G(t)$ for $f(t)$ in this case?
I am expected to obtain $G(t) = \Big( \frac{d}{dt} + i A + B \Big)^{-1}$. It seems that Wikipedia page doesn't mention the procedure to calculated the Green's function. Any help would be greatly appreciated.
It is pretty trivial. Consider $$\Bigg(\frac{d}{dt}+iA+B\Bigg)G(t-t')=\delta(t-t').$$ We use the Fourier transform representation $$G(t-t')=\int_{-\infty}^{\infty}\frac{d\omega}{2\pi}g(\omega)e^{i\omega t}, \quad \delta(t-t')=\int_{-\infty}^{\infty}\frac{d\omega}{2\pi}e^{i\omega t}.$$ It follows $$g(\omega)=\frac{1}{i\omega+iA+B}\implies G(t-t')=\int_{-\infty}^{\infty}\frac{d\omega}{2\pi i}\frac{e^{i\omega (t-t')}}{\omega+A-iB}$$ $$=\Theta(t-t')e^{-iA(t-t')}e^{-B(t-t')}.$$ It follows that the solution of $$\Bigg(\frac{d}{dt}+iA+B\Bigg)f(t)=h(t), \quad f(0)=f_{0},$$ has the form $$f(t)=f_{0}e^{-iAt}e^{-Bt}+\int_{0}^{t}dt' e^{-iA(t-t')}e^{-B(t-t')}h(t'). $$