Note that if [tex]{x_2}'=x_1[/tex], then [tex]{x_2}''={x_1}'[/tex], and so we can collapse the system of ODEs into a linear ODE:
[tex]{x_2}''=3{x_2}'-x_2+e^t[/tex]
[tex]{x_2}''-3{x_2}'+x_2=e^t[/tex]
which is a pretty standard linear ODE with constant coefficients. We have characteristic equation
[tex]r^2-3r+1=\left(r-\dfrac{3+\sqrt5}2\right)\left(r+\dfrac{3+\sqrt5}2\right)=0[/tex]
so that the characteristic solution is
[tex]{x_2}_C=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}[/tex]
Now let's suppose the particular solution is [tex]{x_2}_p=ae^t[/tex]. Then
[tex]{x_2}_p={{x_2}_p}'={{x_2}_p}''=ae^t[/tex]
and so
[tex]ae^t-3ae^t+ae^t=-ae^t=e^t\implies a=-1[/tex]
Thus the general solution for [tex]x_2[/tex] is
[tex]x_2=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}-e^t[/tex]
and you can find the solution [tex]x_1[/tex] by simply differentiating [tex]x_2[/tex].
