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
$$H(s)=\frac{H}{\tau}\frac{1}{\left(s+\frac{1}{\tau}\right)^2}=\frac{H}{\tau}\frac{1}{s^2+\frac{2}{\tau}s+\frac{1}{\tau^2}}=\frac{U(s)}{M(s)}\tag{1}$$$$\begin{align} H(s) &=\frac{A}{\tau}\frac{1}{\left(s+\frac{1}{\tau}\right)^2} \\ &=\frac{A}{\tau}\frac{1}{s^2+\frac{2}{\tau}s+\frac{1}{\tau^2}} \\ &=\frac{U(s)}{M(s)}\tag{1} \end{align}$$
where $U(s)$ and $M(s)$ are the Laplace transforms of the output $u(t)$ and input $m(t)$, respectively.
From $(1)$ we get
$$U(s)\left[s^2+\frac{2}{\tau}s+\frac{1}{\tau^2}\right]=\frac{H}{\tau}M(s)\tag{2}$$$$U(s)\left[s^2+\frac{2}{\tau}s+\frac{1}{\tau^2}\right]=\frac{A}{\tau}M(s)\tag{2}$$
Transforming $(2)$ back to the time domain finally gives the desired differential equation
$$\ddot{u}(t)+\frac{2}{\tau}\dot{u}(t)+\frac{1}{\tau^2}u(t)=\frac{H}{\tau}m(t)\tag{3}$$$$\ddot{u}(t)+\frac{2}{\tau}\dot{u}(t)+\frac{1}{\tau^2}u(t)=\frac{A}{\tau}m(t)\tag{3}$$