Let $$x_1' = 3x_1 - x_2 + e^t$$ And $$x_2' = x_1$$
To prevent confusion for myself, I will let $x = x_1$ and $y = x_2$. So,
$$x' = 3x - y + e^t$$ and $$y' = x$$
Differentiate: $y'' = x'$ So that means $y'' = 3x -y + e^t$. Let's re-write this as a non-homogeneous second order differential equation:
$$y'' -3y' + y = e^t$$ The characteristic equation is: $r^2 -3r + 1 = 0$. The complementary solution $y_c$ is $$y_c = C_1 e^{\bigg(\frac{3+\sqrt{5}}{2}\bigg)\bigg(t\bigg)}+C_2 e^{\bigg(\frac{3-\sqrt{5}}{2}\bigg)\bigg(t\bigg)}$$
Okay, so now let our particular solution $y_p$ be: $$y_p = Ae^t$$ Since the particular solution is not in the complementary solution, we don't need to multiple by t.
$$y_p' = Ae^t$$ $$y_p'' = Ae^t$$
Plugging back into the DE: $$Ae^t - 3Ae^t + Ae^t = e^t$$
So, A = -1.
Thus, $$y(t) = C_1 e^{\bigg(\frac{3+\sqrt{5}}{2}\bigg)\bigg(t\bigg)}+C_2 e^{\bigg(\frac{3-\sqrt{5}}{2}\bigg)\bigg(t\bigg)} - e^t$$
Now that we found $y(t)$, we need to find $x(t)$, but $x(t) = y'(t)$, so we can just take the first derivative of $y$.
Doing so yields:
$$x(t) = \bigg(\frac{3+\sqrt{5}}{2}\bigg)C_1 e^{\bigg(\frac{3+\sqrt{5}}{2}\bigg)\bigg(t\bigg)}+ \bigg(\frac{3-\sqrt{5}}{2}\bigg)C_2 e^{\bigg(\frac{3-\sqrt{5}}{2}\bigg)\bigg(t\bigg)} - e^t$$
Could someone check if this is correct? Thanks