Question

Find the general solution of x'1 = 3x1 - x2 + et, x'2 = x1.

Answer 1

Note that if ${x_2}'=x_1$, then ${x_2}''={x_1}'$, and so we can collapse the system of ODEs into a linear ODE:

${x_2}''=3{x_2}'-x_2+e^t$
${x_2}''-3{x_2}'+x_2=e^t$

which is a pretty standard linear ODE with constant coefficients. We have characteristic equation

$r^2-3r+1=\left(r-\dfrac{3+\sqrt5}2\right)\left(r+\dfrac{3+\sqrt5}2\right)=0$

so that the characteristic solution is

${x_2}_C=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}$

Now let's suppose the particular solution is ${x_2}_p=ae^t$. Then

${x_2}_p={{x_2}_p}'={{x_2}_p}''=ae^t$

and so

$ae^t-3ae^t+ae^t=-ae^t=e^t\implies a=-1$

Thus the general solution for $x_2$ is

$x_2=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}-e^t$

and you can find the solution $x_1$ by simply differentiating $x_2$.