Impulse Response Function-Green's Function

Question

Assuming a input-output system
$$u(t) = c \cdot \big(h(t)\circledast m(t) \big)$$

where its impulse response function is:

$$ h(t) = \begin{cases} \frac{A}{\tau}te^{-\frac{t}{\tau}}, &\quad t\geq0 \\ 0, &\quad t<0 \\ \end{cases} $$

Given only these, how can one derive the differential equation below, whose Green's function is the above?:

$$ \ddot{u}(t)+\frac{2}{\tau}\dot{u}(t)+\frac{1}{\tau^2}u(t) = \frac{A}{\tau}m(t) $$

Would, considering an inverse n-order L differential operator and performing the consequent integrals, be a possible way?

Answer 1

A simple way to derive the differential equation from the impulse response is to transform the latter to the frequency domain, rewrite the input/output relation, and then transform the resulting equation back to the time domain.

The Laplace transform of the given impulse response $h(t)$ is

$$\begin{align} H(s) &=\frac{A}{\tau}\frac{1}{\left(s+\frac{1}{\tau}\right)^2} \\ &=\frac{A}{\tau}\frac{1}{s^2+\frac{2}{\tau}s+\frac{1}{\tau^2}} \\ &=\frac{U(s)}{M(s)}\tag{1} \end{align}$$

where $U(s)$ and $M(s)$ are the Laplace transforms of the output $u(t)$ and input $m(t)$, respectively.

From $(1)$ we get

$$U(s)\left[s^2+\frac{2}{\tau}s+\frac{1}{\tau^2}\right]=\frac{A}{\tau}M(s)\tag{2}$$

Transforming $(2)$ back to the time domain finally gives the desired differential equation

$$\ddot{u}(t)+\frac{2}{\tau}\dot{u}(t)+\frac{1}{\tau^2}u(t)=\frac{A}{\tau}m(t)\tag{3}$$