Jtdylktuy

$mathbf{x'}=begin{bmatrix}3 & 1 \ 0 & 3end{bmatrix}mathbf{x}+begin{bmatrix} e^{3t}\e^{3t}end{bmatrix}$

My attempt:

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

My answer is $mathbf{x}_c(t)=a_0begin{bmatrix}1 \ 0end{bmatrix}e^{3t}+b_0left(begin{bmatrix}1 \ 0end{bmatrix}t+begin{bmatrix}0 \ 1end{bmatrix}right)e^{3t}$

In order to eliminate duplication, let $mathbf{x}_p(t)=begin{bmatrix}a_1 \ a_2end{bmatrix}t^2e^{3t}+begin{bmatrix}b_1 \ b_2end{bmatrix}te^{3t}+begin{bmatrix}c_1 \ c_2end{bmatrix}e^{3t}$

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.
$$pmatrix {x \y}'=pmatrix{3 & 1 \ 0 & 3}pmatrix {x \ y}+begin{bmatrix} e^{3t}\e^{3t}end{bmatrix}$$
$$y'=3y+e^{3t} implies (ye^{-3t})'=1$$
$$y(t)=e^{3t}(t+K)$$
$$x'=3x+y+e^{3t}$$
$$x'=3x+e^{3t}(t+K+1) implies (xe^{-3t})'=t+K+1$$
$$x(t)=(frac {t^2}2+(K+1)t+C)e^{3t}$$

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.