Green's function of an ODE (first order)

Question

What is the Green's Function of the following ODE (in general form)?

$$ \frac{dy}{dt} = f(t) y(t) + g(t) $$

Answer 1

By definition Green function is the solution of equation with specific RHS, namely $$\Bigl(\frac{d}{dt}- f(t)\Bigr)G(t) =\delta(t)$$ Where $\delta(t)$ is Dirac delta-function. If we find $G(t)$, the solution of the ODE with $g(t)$ in the RHS $$\Bigl(\frac{d}{dt}- f(t)\Bigr) y(t)=g(t)$$ can be presented in the form $$y(t)=\int_{-\infty}^{\infty}G(t-p)g(p)dp$$ Using the delta function property $\delta(t)=\frac{d}{dt}h(t)$, where $h(t)$ is the Heaviside step-function ($h(t)=1$ at $t>0$ and $h(t)=0$ at $t<0$). Let $y(t)$ be a solution of the free equation (equation with a zero RHS): $\Bigl(\frac{d}{dt}- f(t)\Bigr)y(t)=0$

In this case $$\Bigl(\frac{d}{dt}- f(t)\Bigr)h(t)y(t)=h(t)y'(t)+\delta(t)y(t)-f(t)h(t)y(t)$$ $$=h(t)\bigl(y'(t)-f(t)\bigr)+\delta(t)y(0)=\delta(t)y(0)$$ So, if we correctly norm the solution of the free equation (to get $y(0)=1$), we get $G(t)$.

The solution of the free equation is $y(t)=Ce^{\int^t f(s)ds}$, where C is an arbitrary constant. Let's choose it $C=\frac{1}{e^{\int^0 f(s)ds}}$, and out Green function looks $$G(t)=h(t)\frac{e^{\int^t f(s)ds}}{e^{\int^0 f(s)ds}}$$ Any solution of the equation $$\Bigl(\frac{d}{dt}- f(t)\Bigr) y(t)=g(t)$$ is $$y(t)=C_1e^{\int^t f(s)ds}+\frac{1}{e^{\int^0 f(s)ds}}\int_{-\infty}^te^{\int^{t-p} f(s)ds}g(p)dp$$ where $C_1$ is the constant defined by initial data of the problem (for example, function value in some point).