Find the solution to the given differential equation I am trying to find the general solution to the system $x' = Ax + at$ where,
$$ A = \begin{pmatrix} 
1 & -1\\
9 & 1
\end{pmatrix} $$
and $$ a = \begin{pmatrix} 
-1\\
1
\end{pmatrix} $$
I began by finding the eigenvectors and eigenvalues of $A$ and got the eigevalues $\lambda = 1 \pm 3i$ to correspond to the eigenvectors $  \begin{pmatrix} 
i\\
3
\end{pmatrix} $ and $  \begin{pmatrix} 
i\\
-3
\end{pmatrix} $
I am confused about how to proceed to finding the final solution here. Any guidance is greatly appreciated!

Typically, the general method of solving such problems is the Laplace transform which yields
$$
x'=Ax+at\implies sx-x(0)=Ax+L(at)\implies (sI-A)x=x(0)+L(at)\implies x=(sI-A)^{-1}(x(0)+L(at))
.
$$
In our case,
$$x{=
\begin{bmatrix}
s-1&1\\-9&s-1
\end{bmatrix}^{-1}
(x(0)+L(at))
\\
=\begin{bmatrix}
\frac{s-1}{(s-1)^2+9}&-\frac{1}{(s-1)^2+9}\\\frac{9}{(s-1)^2+9}&\frac{s-1}{(s-1)^2+9}
\end{bmatrix}
(x(0)+L(at))
}.
$$
If out experiment starts at $t=0$, we can substitute $t$ with $r(t)$ which is the ramp function with the Laplace transform of $\frac{1}{s^2}$. Hence
$$
x
=\begin{bmatrix}
\frac{s-1}{(s-1)^2+9}&-\frac{1}{(s-1)^2+9}\\\frac{9}{(s-1)^2+9}&\frac{s-1}{(s-1)^2+9}
\end{bmatrix}
\begin{bmatrix}
x_1(0)+\frac{a}{s^2}\\
x_2(0)+\frac{a}{s^2}
\end{bmatrix}
$$
