Solving a nonhomogeneous system of equations without matrices

I have the given system, posted here earlier today, which I here try to solve without using matrix.

I start with the system at hand:

$\left[\begin{array}{@{}c@{}} x' \\ y' \end{array} \right]= \left[\begin{array}{@{}c@{}} 3&2 \\ -2 & -1 \end{array} \right]\left[\begin{array}{@{}c@{}} x' \\ y' \end{array} \right]+\left[\begin{array}{@{}c@{}} 2e^{-t} \\ e^{-t} \end{array} \right]$

The homogeneous form is solved as given in the original post, but when it comes to the specific solution, I do the following.

\begin{equation} \begin{array} f x'=3x+2y+2e^{-t} \\ y'=-2x-y+e^{-t} \end{array} \end{equation}

I move the x's and y's to each side:

\begin{equation} \begin{array} fx'-3x=2y+2e^{-t} \\ 2x=y'-y+e^{-t} \end{array} \end{equation}

I then call the first order derivative for x, x', for D, and get:

\begin{equation} \begin{array} fD-3x=2y+2e^{-t} \\ 2x=y'-y+e^{-t} \end{array} \end{equation}

Multiplying above with 2, and below with 3, I eliminate the x term, and obtain

\begin{equation} \begin{array} f2D=4y+4e^{-t} \\ 0=-3y'-3y+3e^{-t} \end{array} \end{equation}

Rearranging, and renaming D to dx/dt, and y' to dy/dt gives:

\begin{equation} \begin{array} f2\frac{dx}{dt}=4y+4e^{-t} \\ 3\frac{dy}{dt}+3y=3e^{-t} \end{array} \end{equation}

Simplifying,

\begin{equation} \begin{array} f\frac{dx}{dt}=2y+2e^{-t} \\ \frac{dy}{dt}+y=e^{-t} \end{array} \end{equation}

Summing up to one eqn:

\begin{equation} \frac{dx}{dt}+\frac{dy}{dt}=y+3e^{-t} \end{equation}

Splitting it up to a system again, we obtain a simpler system than above:

\begin{equation} \begin{array} f\frac{dx}{dt}=3e^{-t} \\ \frac{dy}{dt}-y=0 \end{array} \end{equation}

If we solve each separately, we have that the first has the solution $x(t)=-3e^{-t}$ and the second has the solution $y(t)=ce^{t}$. If we merge these with the homogeneous solution from the top, we get:

\begin{equation} y_g=y_h+y_p=e^{t}\left[\begin{array}{@{}c@{}} 1 \\ -1 \end{array} \right]+e^{t}\left[\begin{array}{@{}c@{}} \frac{1}{2} \\ 0 \end{array} \right]-3e^{t}+ce^{t} \end{equation}

but I fear that solving the two last eqns separately was not a good idea. Can this approach benefit to find the solution to the nonhomogeneous system and how can we find the specific solution to the last pair of eqns?

Thanks

Hint.

Adding the two ODEs we get

$$ (x+y)'=x+y+3e^{-t} $$

then

$$ x+y = c_0 e^{-t}-\frac 32 e^{-t} $$

etc.