Impulse Response Function-Green's Function Assuming a input-output system
$$u(t) = c \cdot \big(h(t)\circledast m(t) \big)$$

where its impulse response function is:

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

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?
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

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

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

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