In the context of signal processing, consider the following system: $$ x\left( t \right) \to System\, A \to z\left( t \right) \to System \, B \to y\left( t \right) .$$
Let the global system be the "union" of system A and system B. The input and output of System A are represented by the equations:
$$ \frac{d^2 \, z(t)}{d \, t^2} + 6 \frac{d\, z(t)}{d\, t} + 8 z\left( t \right) = \frac{d\, x(t)}{d\, t} + 2x\left( t \right) $$
and the unit impulse response of System B is defined as follows: $$ h_B \left( t \right) = t e^{-3t} u\left( t \right) .$$
Q1: What is the unit impulse response of the global system?
Q2: For a periodic input of the form
$$
x\left( t \right) =e^{-t} \quad ; \quad -1<t<0
.$$
with a period of $T=1$, what will the output signal $y\left( t \right) $ be?
Q3: What is the frequency spectrum of that last signal?
Note: In this problem $u(t)$ denotes the unit step function. And I found these three articles that I think are related.
Edit: With the help of @KwinvanderVeen , I tried to solve it by finding the transfer function for both systems. $$ H_A(s)= \frac{Z\left( s \right) }{X\left( s \right) } = \frac{\mathcal{L}\{z\} \left( s \right) }{\mathcal{L}\{ x\}\left( s \right) }= \frac{\int^{\infty}_{0} z\left( t \right) e^{-st} dt}{\int^{\infty}_{0} x\left( t \right) e^{-st} dt }$$
And for the system B: $$ H_B(s)= \frac{Y\left( s \right) }{Z\left( s \right) } = \frac{\mathcal{L}\{y\} \left( s \right) }{\mathcal{L}\{ z\}\left( s \right) }= \frac{\int^{\infty}_{0} y\left( t \right) e^{-st} dt}{\int^{\infty}_{0} z\left( t \right) e^{-st} dt }$$
Applying the properties of Laplace transforms:
$$ \mathcal{L}\{ \frac{d^2 \, z(t)}{d \, t^2} + 6 \frac{d\, z(t)}{d\, t} + 8 z\left( t \right) \} = \mathcal{L} \{ \frac{d\, x(t)}{d\, t} + 2x\left( t \right) \} $$ $$ \Rightarrow s^2 \mathcal{L}\{ z\} +6s \mathcal{L}\{ z\} +8 \mathcal{L}\{ z\} = s\mathcal{L} \{ x\} +2\mathcal{L} \{ x\} $$ $$ \Rightarrow H_A(s)= \frac{\mathcal{L} \{ z(t) \}}{\mathcal{L} \{ x(t) \}} = \frac{s+2}{(s+2)(s+4)}=\frac{1}{s+4} $$ $$ \Rightarrow h_A=e^{-4t} $$
(in this last line, the original expression was wrong. It was fixed thanks to @KwinvanderVeen) for $s\neq -2$. We also have that:
$$ H_G(s)=\frac{Y(s)}{X(s)} = \frac{Y(s)}{Z(s)} \frac{Z(s)}{X(s)} = H_B H_A = \mathcal{L} \{ (h_B*h_A )(t) \} $$ $$ \Rightarrow \mathcal{L}\{ h_G\}=H_G(s)= \mathcal{L}\{ (h_B*h_A)(t) \} $$ $$ \Rightarrow h_G=h_B*h_A = \int_{0}^{t} h_B(\tau ) h_A(t-\tau )d\tau = e^{-4t}(t e^t-e^t+1) $$
(this last line was also fixed thanks to @KwinvanderVeen). And I think that last line would the the unit impulse response, and the answer to question 1.
Now, I tried to solve question 2 with the suggestions of the same commenter. If the input is periodic, then I think it follows that:
$$ x(t)= e^{1-t} \quad ; \quad 0<t<1 \quad ; \quad T=1 $$ $$ x(t)=e^{2-t} \quad ; \quad 1<t<2 \quad ; \quad T=1 $$
And in general, for $n \in \mathbb{N} $: $$ x(t)=e^{n-t} \quad ; \quad n-1<t<n \quad ; \quad T=1 $$
Then using the formula for the convolution integral: $$ y(t)= \left( h_G*x \right) (t) =e^{n-t} \int_{n-1}^{n} e^{-3\tau } (\tau e^{\tau} -e^{\tau} +1 ) d\tau $$
for $n-1<t<n $. Is that the answer? Can i leave the expression like that?
The transfer function and impulse response for system A seem to be correct; good that you also noticed the pole zero cancellation. You could manually calculate the impulse response from the transfer function using the inverse Laplace transform, but in practice it is often easier/quicker to use a Laplace transform table such as this one. From such a table you should also be able to derive what the (minimal) transfer function should be for system B.
In order to answer question 1 you could indeed do the convolution of the two impulse responses, however I believe that one (me) can easily make mistakes when solving integrals. Instead I would recommend to decompose the total transfer function into a sum of simple terms that appear in a Laplace transform table. For example if the total transfer function is given by
$$ G(s) = \frac{a_2\,s^2 + a_1\,s + a_0}{(s + b_1)^2\,(s + b_2)}, $$
then you would want to solve for $A$, $B$ and $C$ such that
$$ G(s) = \frac{A}{s+b_1} + \frac{B}{(s+b_1)^2} + \frac{C}{s+b_2}. $$
The total impulse response can then be found by summing the impulse responses of each of the three terms, where impulse response of each term should be able to be found on a a Laplace transform table.
Question 2 does not have a completely well defined answer, because no initial conditions are given for system A and B. One way you could interpret this is with all initial conditions set to zero, which can be obtained by evaluating the convolution integral between the input and the impulse response of the total system. Another way is to assume that the output has settled into a period signal as well, which can be obtained by first evaluating the Fourier series of the input
$$ x(t) = \sum_{k=0}^\infty a_k \cos(2\,\pi\,k\,t) + b_k \sin(2\,\pi\,k\,t) = \sum_{k=0}^\infty c_k \cos(2\,\pi\,k\,t + \phi_k), $$
the output can then be obtained using
$$ y(t) = \sum_{k=0}^\infty c_k\,|G(j\,2\,\pi\,k)| \cos(2\,\pi\,k\,t + \phi_k + \angle G(j\,2\,\pi\,k)), $$
so the total transfer function is evaluated at $s=j\,2\,\pi\,k$, with $j$ the imaginary unit.
This second approach would basically already give the answer to question 3, so I guess the first approach, with the convolution integral, is probably the intended answer.