Nonhomogeneous Linear ODE $$
x'
=\left(\begin{array}{rr}4 & 8 \\ -2 & -4\end{array}\right)x + \left(\begin{array}{rr}t^{-3} \\ -t^{-2}\end{array}\right), t>0
$$
To find the general solution of the given system above, my solution is:
$det(\left(\begin{array}{rr}4-r & 8 \\ -2 & -4-r\end{array}\right)) = r^2$
, eigenvalues are $r_{1,2}=0$
$\left(\begin{array}{rr}4 & 8 \\ -2 & -4\end{array}\right)\left(\begin{array}{rr}v_{11} \\ v_{12}\end{array}\right) = \left(\begin{array}{rr}0 \\ 0\end{array}\right)$ Therefore $v_{1} =\left(\begin{array}{rr}-2 \\ 1\end{array}\right)$
$\left(\begin{array}{rr}4 & 8 \\ -2 & -4\end{array}\right)\left(\begin{array}{rr}v_{21} \\ v_{22}\end{array}\right) = \left(\begin{array}{rr}-2 \\ 1\end{array}\right)$ From there, I have and equation $2v_{21}+4v_{22}=-1$
What should I do now? Thanks.

Since $\begin{pmatrix}
4 & 8\\ 
-2 & -4
\end{pmatrix}\cdot \binom{-1/2}{0} = \binom{-2}{1}$ works, we get a homogeneous solution to the ODE of:
$c_{1}\binom{-2}{1}+c_{2}(t\binom{-2}{1}+\binom{-1/2}{0})$. Now it suffices to find a particular solution.
