Find the solution to the given differential equation

Question

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!

Answer 1

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 $$ x =\begin{bmatrix} \frac{s-1}{(s-1)^2+9}&-\frac{1}{(s-1)^2+9}\\\frac{9}{(s-1)^2+9}&\frac{s-1}{(s-1)^2+9} \end{bmatrix} \begin{bmatrix} x_1(0)+\frac{a}{s^2}\\ x_2(0)+\frac{a}{s^2} \end{bmatrix} $$