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-\frac{3+\sqrt5}2\right)\left(r+\frac{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`.