Solving system of ODE $\frac{d^2x}{dt^2}+3\frac{dx}{dt}-2x+\frac{dy}{dt}-3y=2e^{-t}$
$2\frac{dx}{dt}-x+\frac{dy}{dt}-2y=0$ 
given that $x(0)=\frac{dx}{dt}_{|x=0}=0, y(0)=4$
I tried eliminating $\frac{dx}{dt}$ in the 1st eqn.
Eqn 1 becomes
$$\frac{d^2x}{dt^2}+\frac{dx}{dt}-x=y+2e^{-t}$$
I thought of decoupling this system to eliminate $y$.
$$\frac{d^3x}{dt^3}+\frac{d^2x}{dt^2}-\frac{dx}{dt}=\frac{dy}{dt}-2e^{-t}$$
But all  this seems to complicate the problem even more.
 A: Subtract the second from the first, solve that for $y$. Take the derivative to get an expression for $y'$. Substitute those two in the second equation and solve for $x$.
You will get:
$$y = x'' + x' - x - 2e^{-t}\\y' = x''' + x'' - x' + 2 e^{-t}$$
Substituting into the second equation, you get:
$$x'''-x''-x'+x +6e^{t} = 0$$
Also, you need to figure out one more IC, but you have:
$$x''(0) = 6$$
This leads to:
$$x(t) = \dfrac{3}{2}t(e^t-e^{-t})$$
I am sure you can take it from here.
A: $\mathbf{ x'' +3x'-2x+y'-3y=2e^{-t}}$    $            $    $\mathbf{...(1)}$
$\mathbf{ 2x' -x+y'-2y=0}$               $    $    $\mathbf{............(2)}$
Differential operators, 
$(D^2+3D-2)x+(D-3)y=2e^{-t}$    $        $ $.(1)$
$(2D-1)x+(D-2)y=0$          $          $    $..........(2)$
To eliminate $y$,
Multiply eqn $(1)$ by $(D-2)$ and
eqn $(2)$ by $(D-3)$.
After elimination
$(D^3+D^2-D+1)x=(D-2)(2e^{-t})$
$[(D-1)^{2}(D+1)]x=2(e^{-t}-2e^{-t})$
$[(D-1)^{2}(D+1)]x=-6e^{-t}$
Particular Integral
$x=-6[\frac{e^{-t}}{(D-1)^2(D+1)}]$
$x=-\frac{6}{D+1}[\frac{e^{-t}}{(D-1)^2+1}]$
$x=-\frac{e^{-t}}{2}[\frac{1}{D}]=-\frac{3}{2}te^{-t}$
Auxiliary equation
$x=(m-1)^2(m+1)=0 \rightarrow m=1,1,-1$
$x=Ae^{-t}+(Bt+C)e^{t}$
$\therefore x=Ae^{-t}+(Bt+C)e^{t}-\frac{3}{2}te^{-t}$
From $(1)$ and $(2)$, we can't make y subject of formula since $y'$ present. 
$\therefore$ Eliminate $y'$ by $(1)-(2)$
$\therefore y=x''+x'-x-2e^{-t}$
$x'=-Ae^{-t}+e^t(C)+(B+Ct)e^t+\frac{3}{2}te^{-t}-\frac{3}{2}e^{-t}$
$x''=Ae^{-t}+2Ce^t+(B+Ct)e^t+3e^{-t}-\frac{3}{2}te^{-t}$
$y=Ae^{-t}+2Ce^t+(B+Ct)e^t+3e^{-t}-\frac{3}{2}te^{-t}$
$  $ $+-Ae^{-t}+Ce^t+(B+Ct)e^t+\frac{3}{2}te^{-t}-\frac{3}{2}e^{-t}$
$  $ $-Ae^{-t}-(B+Ct)e^t+\frac{3}{2}te^{-t}-2e^{-t}$

$y=-Ae^{-t}+3Ce^t+(B+Ct)e^t-\frac{1}{2}e^{-t}+\frac{3}{2}te^{-t}$
