Find a particular solution for this linear system $\mathbf{x'}=\begin{bmatrix}3 &amp; 1 \\ 0 &amp; 3\end{bmatrix}\mathbf{x}+\begin{bmatrix} e^{3t}\\e^{3t}\end{bmatrix}$

My attempt:

We first find a solution to the associated homogeneous system $\mathbf{x'}=\begin{bmatrix}3 &amp; 1 \\ 0 &amp; 3\end{bmatrix}\mathbf{x}$. 

My answer is $\mathbf{x}_c(t)=a_0\begin{bmatrix}1 \\ 0\end{bmatrix}e^{3t}+b_0\left(\begin{bmatrix}1     \\ 0\end{bmatrix}t+\begin{bmatrix}0 \\ 1\end{bmatrix}\right)e^{3t}$

In order to eliminate duplication, let $\mathbf{x}_p(t)=\begin{bmatrix}a_1 \\ a_2\end{bmatrix}t^2e^{3t}+\begin{bmatrix}b_1 \\ b_2\end{bmatrix}te^{3t}+\begin{bmatrix}c_1 \\ c_2\end{bmatrix}e^{3t}$

When I substitute this into the given ODE, I get 6 equations in 6 variables but I do not manage to obtain values for all of the variables. How do 
I proceed?
$$X'=AX+g(t)$$
Your particular solution is correct but since the matrix A is already in Jordan form.
$$\pmatrix {x \\y}'=\pmatrix{3 &amp; 1 \\ 0 &amp; 3}\pmatrix  {x \\ y}+\begin{bmatrix} e^{3t}\\e^{3t}\end{bmatrix}$$
$$y'=3y+e^{3t} \implies (ye^{-3t})'=1$$
$$y(t)=e^{3t}(t+K)$$
$$x'=3x+y+e^{3t}$$
$$x'=3x+e^{3t}(t+K+1) \implies (xe^{-3t})'=t+K+1$$
$$x(t)=(\frac {t^2}2+(K+1)t+C)e^{3t}$$
It appears that the variables that don't have values can take on arbitrary values that are consistent with the 6 equations. I checked one case using MATLAB and it works.