I have this system of diferential equations
$$ \begin{cases} q'+q+i=50e^{-t}u_1(t) \\ i'+i-q=0 \end{cases}$$
$q(0)=i(0)=0$
** $ u_1(t) $ is the Heaviside step function
Solution
Rewriting: $$ \begin{cases} q'=50e^{-t}u_1(t)-q-i \\ i'=q-i \end{cases} $$ Laplace transform:
$$ \begin{cases} sL[q]-q(0)=\frac{50e^{-(s+1)}}{s+1}-L[q]-L[i] \\ sL[i]-s(0)=L[q]-L[i] \end{cases} $$
Rewriting:
$$ \begin{cases} (s+1)L[q]+L[i]=\frac{50e^{-(s+1)}}{s+1} \\ -L[q]+(s+1)L[i]=0 \end{cases} $$
By Cramer:
$$L[q]=\frac{\left|\begin{array}{cc}\frac{50e^{-(s+1)}}{s+1} & 1 \\ 0 & (s+1) \end{array}\right|}{\left|\begin{array}{cc}(s+1) & 1 \\ -1 & (s+1) \end{array}\right|}=\frac{50e^{-(s+1)}}{(s+1)^2+1}$$ so $\qquad \qquad q=50L^{-1}[\frac{e^{-(s+1)}}{(s+1)^2+1}]$
$$ $$
$$L[i]=\frac{\left|\begin{array}{cc}(s+1) & \frac{50e^{-(s+1)}}{s+1}\\ -1 & 0 \end{array}\right|}{\left|\begin{array}{cc}(s+1) & 1 \\ -1 & (s+1) \end{array}\right|}=\frac{\frac{50e^{-(s+1)}}{s+1}}{(s+1)^2+1}$$ so $\qquad \qquad i=50L^{-1}[\frac{e^{-(s+1)}}{(s+1)((s+1)^2+1}]$
$$ $$
I don't have any idea how I can solve this inverse Laplace transform.
It seems that it is somewhat easier to first obtain an equation for $i$ : \begin{equation*} i^{\prime \prime }+2i^{\prime }+i=50e^{-t},\;t\geqslant 0 \end{equation*} Then, with \begin{eqnarray*} f(z) &=&\int_{0}^{\infty }dt\exp [izt]i(t),\;Imz>0, \\ i(t) &=&\frac{1}{2\pi }\int_{\Gamma }dz\exp [-izt]f(z) \end{eqnarray*} where $\Gamma $ is a straight line above and parallel to the real axis, You obtain, using your initial conditions, which imply that also $i^{\prime }(0)=0$, \begin{eqnarray*} f(z) &=&-50i(z+i)^{-3} \\ i(t) &=&-50i\frac{1}{2\pi }\int_{\Gamma }dz\exp [-izt](z+i)^{-3} \end{eqnarray*} from which $i(t)$ follows in the usual way. By the way, you can also proceed by setting $i(t)=g(t)\exp [-t]$ which leaves you with \begin{equation*} g^{\prime \prime }(t)+g^{\prime }(t)-g(t)=50 \end{equation*} that can also be handled by standard methods.