System of differential equations Can you help me solve this system of differential equations
$ x'= 4x - 2y $
$ y'= 3x - y - 2e^{3t} $
Initial conditions are  $ x(0) = y(0) = 0 $
 A: $\mathbf x' + \begin{bmatrix} -4&2\\-3&1\end{bmatrix} \mathbf x = \begin{bmatrix}0\\-2e^{3t}\end{bmatrix}$
$e^{At} \mathbf x' + A e^{At} \mathbf x = e^{At}g(t)$
$A = P^{-1}DP$
$A = \begin{bmatrix}1&2\\1&3\end{bmatrix}\begin{bmatrix}-2&0\\0&-1\end{bmatrix}\begin{bmatrix}3&-2\\-1&1\end{bmatrix}\\
e^{At} = P^{-1} e^{Dt}P$
$\mathbf x =$$ e^{-At}C + e^{-At} \int e^{At}g(t) dt\\
e^{-At}C + P^{-1}e^{-Dt} \int e^{Dt}Pg(t) dt$
$e^{Dt}Pg(t) =\begin{bmatrix}4e^t\\-2e^{2t}\end{bmatrix}$ 
$\mathbf x =e^{-At}C + P^{-1}\begin{bmatrix}4\\-1\end{bmatrix} e^{3t}\\
P\mathbf x(0) =C + \begin{bmatrix}4\\-1\end{bmatrix}\\
C = \begin{bmatrix}-4\\1\end{bmatrix}$
$\mathbf x = \begin{bmatrix}-4e^{2t}+2e^{t}+2e^{3t}\\-4e^{2t}+3e^t + e^{3t}\end{bmatrix}$
A: Subtracting both equations, you get
$$x'-y'=x-y+2e^{-3t}$$ which you can solve for $x-y$, giving
$$x-y=Ce^t-\frac{e^{-3t}}2,$$ where $C=\frac12$.
Then solve
$$y'=3\left(y+\frac{e^t-e^{-3t}}2\right)-y+2e^{-3t}.$$
A: With the operator $D:=d/dt$, $$\begin{cases}(D-4)x+2y=0\\3x-(D+1)y=2e^{-3t}.\end{cases}$$
Then by elimination of $y$,
$$6x+(D+1)(D-4)x=(D-1)(D-2)x=2e^{-3t},$$ which is an ordinary second order equation.
Its solution is readily obtained as
$$x(t)=C_1e^t+C_2e^{2t}+\frac{2e^{-3t}}{(-3-1)(-3-2)}.$$
The rest is straightforward.
A: By solution is more ad-hoc than formal. I used a general solution of the form
$$\begin{aligned}
x(t) &= \left( X_0 + X_1 {\rm e}^{\alpha t} \right) {\rm e}^{3 t} + X_2 {\rm e}^{\beta t} \\
y(t) &= \left( Y_0 + Y_1 {\rm e}^{\alpha t} \right) {\rm e}^{3 t} + Y_2 {\rm e}^{\beta t}
\end{aligned}$$
which I pluged into the IC first to solve for $X_0$ and $Y_0$ in terms of the rest of the coefficients, and the used this in the D.E.
The solution I got was
$$\begin{aligned}
x(t) &= \left( 2 -4 {\rm e}^{-t} \right) {\rm e}^{3 t} + 2 {\rm e}^{ t} \\
y(t) &= \left( 1 -4 {\rm e}^{- t} \right) {\rm e}^{3 t} + 3 {\rm e}^{ t}
\end{aligned}$$
NOTE: I solved for $X_2$ and $Y_2$ first, and then used the intermediate result $X_1 = \frac{2 Y_1}{1-\alpha}$ to find the coefficients $\alpha$ and $\beta$ that solve the D.E.
