System of linear first order DEs Question:
$$3\dot{x} + \dot{y} + 5x - y = 2e^{-t}+4e^{-3t}$$
$$\dot{x} + 4\dot{y} - 2x + 7y = -3e^{-t}+5e^{-3t}$$
Subject to:
$$x(0)=y(0)=0$$
Attempt at a solution:
I have gotten to:
$$\left(\begin{array}{cc}1&-1&3&-3\\ 1&1&1&1\end{array}\right)\left(\begin{array}{cc}\dot{x}\\ \dot{y}\\ x \\ y \end{array}\right) = 2 \left( \begin{array}{cc}e^{-t}\\ e^{-3t}\end{array}\right) $$
But I am not sure what to do next.
 A: Note that your system can be rewritten in a more convenient form as follows:
$$A \,Y'(t) = B\, Y(t) + C(t),$$ where $Y(t) = (x(t),y(t))$, $C$ is the vector of independent terms and $A$ and $B$ are given by:
$$ 
A = \left( \begin{array}{cc}
3 & 1 \\
1 & 4
\end{array}
\right), \quad B = 
\left( \begin{array}{cc}
-5 & 1 \\
2 & -7
\end{array}
\right).
$$ Since $\det{A} \neq 0$ you may multiply both sides by $A^{-1}$ to get:
$$Y'(t) = M \, Y(t) + D(t), $$ with $M = A^{-1} B$ and $D = A^{-1} C$. I'm sure you can take it from here.
Spoiler:

If I remember well, the solution is then given by: 
  $$ Y(t) = Y_0 \Phi(t) + \Phi(t) \int^t_0 \Phi^{-1}(s) D(s) \, \mathrm{d}s, $$
  where $\Phi(t) = \exp{At}$ is the fundamental matrix of the system and $Y_0$ is the vector of initial conditions.

Cheers!
A: One way to rewrite:
$$
         \left[\begin{array}{cc} 3 & 1 \\ 1 & 4\end{array}\right]
             \frac{d}{dt}\left[\begin{array}{c} x \\ y\end{array}\right]+
         \left[\begin{array}{cc} 5 & -1 \\ -2 & 7\end{array}\right]
                   \left[\begin{array}{c} x \\ y\end{array}\right]=
         \left[\begin{array}{cc} 2 & 4 \\ -3 & 5\end{array}\right]
                   \left[\begin{array}{c} e^{-t} \\ e^{-3t}\end{array}\right].
$$
Multiplying by the inverse
$$
       \left[\begin{array}{cc} 3 & 1 \\ 1 & 4\end{array}\right]^{-1}=
       \frac{1}{11}\left[\begin{array}{cc} 4 & -1 \\ -1 & 3\end{array}\right]^{-1}
$$
gives
$$
       \frac{d}{dt}\left[\begin{array}{c}x \\ y\end{array}\right]+
       \left[\begin{array}{cc} 2 & -1 \\ -1 & 2\end{array}\right]
         \left[\begin{array}{c} x \\ y\end{array}\right]=
       \left[\begin{array}{cc} 1 & 1\\ -1 & 1\end{array}\right]
         \left[\begin{array}{c} e^{-t} \\ e^{-3t}\end{array}\right].
$$
The coefficient matrix $C$ on the left has characteristic polynomial $(\lambda-2)^{2}-1=(\lambda-3)(\lambda-1)$. So the coefficient matrix has diagonal representation:
$$
C=\left[\begin{array}{cc} 2 & -1 \\ -1 & 2\end{array}\right] =
     U^{T}\left[\begin{array}{cc}3 & 0 \\ 0 & 1 \end{array}\right]U\;\;\;
      \mbox{ where } U = \frac{1}{\sqrt{2}}\left[\begin{array}{cc}1 & -1 \\ 1 & 1\end{array}\right].
$$
Therefore,
$$
       e^{tC} = U^{T}\left[\begin{array}{c}e^{3t} & 0\\0 & e^{t}\end{array}\right]U
        = \frac{1}{2}\left[\begin{array}{cc}1 & 1 \\ -1 & 1\end{array}\right]
          \left[\begin{array}{c}e^{3t} & 0\\0 & e^{t}\end{array}\right]
          \left[\begin{array}{cc}1 & -1 \\ 1 & 1\end{array}\right]
$$
Multiplying both sides of the differential equation by the matrix $e^{tC}$ gives
$$
       \frac{d}{dt}\left(e^{tC}\left[\begin{array}{c}x \\ y\end{array}\right]\right)
    = e^{tC}\left[\begin{array}{cc} 1 & 1\\ -1 & 1\end{array}\right]
         \left[\begin{array}{c} e^{-t} \\ e^{-3t}\end{array}\right]
$$
Integrating,
$$
        e^{tC}\left[\begin{array}{c}x \\ y\end{array}\right]
        -\left[\begin{array}{c}x(0) \\ y(0)\end{array}\right]
      = \int_{0}^{t} e^{sC}\left[\begin{array}{cc} 1 & 1\\ -1 & 1\end{array}\right]
\left[\begin{array}{c} e^{-s} \\ e^{-3s}\end{array}\right]\,ds
$$
Applying $x(0)=y(0)=0$ and the specific form of $e^{sC}$ yields the final answer
$$
\begin{align}
    \left[\begin{array}{c} x \\ y\end{array}\right]  & = e^{-tC}\int_{0}^{t}
    \left[\begin{array}{cc} 1 & 1\\ -1 & 1\end{array}\right]
         \left[\begin{array}{cc} e^{2s}\\e^{-2s}\end{array}\right]\,ds \\
%%      & = \left[\begin{array}{cc} 1 & 1\\ -1 & 1\end{array}\right]
%%          \left[\begin{array}{cc} e^{-3t} & 0 \\ 0 & e^{-t}\end{array}\right]
%%          \int_{0}^{t}\left[\begin{array}{cc} e^{2s}\\e^{-2s}\end{array}\right]\,ds \\
%%      & = \left[\begin{array}{cc}e^{-3t} & e^{-t} \\ -e^{-3t} & e^{-t}\end{array}\right]
%%     \frac{1}{2}\left[\begin{array}{c}e^{2t}-1 \\ 1-e^{-2t}\end{array}\right] \\
%%      & = \frac{1}{2}\left[\begin{array}{cc}e^{-t}-e^{-3t}+e^{-t}-e^{-3t} \\
%%             -e^{-t}+e^{-3t}+e^{-t}-e^{-3t}\end{array}\right] \\
      & = \left[\begin{array}{c}e^{-t}-e^{-3t} \\ 0\end{array}\right].
\end{align}
$$
