How would I solve this system of differential equations? Find a general solution of 
$$
x' =  \begin{pmatrix}
  11 & -25 \\
  4 & -9 
 \end{pmatrix} x + \begin{pmatrix}
  e^{-t} \\
  0 
 \end{pmatrix} $$
I found the general solution of the homogeneous system to be 
$$ c_1e^t \begin{pmatrix}
  5 \\
  2 
 \end{pmatrix} + c_2 \left[te^t\begin{pmatrix}
  5 \\
  2
 \end{pmatrix} +e^t \begin{pmatrix}
  3 \\
  1 
 \end{pmatrix} \right] 
$$
I am not sure how to go about solving the nonhomogeneous form. The only method I know is variation of parameters but that seems very impractical in this situation. Any help would be greatly appreciated. 
 A: I agree with the general solution you found.
See my post here for details of this process, resolve an non-homogeneous differential system.
We have:
$$\phi(t) = e^t \left(
\begin{array}{cc}
 5 & 5 t+3 \\
 2 & 2 t+1 \\
\end{array}
\right)$$
$$\phi^{-1}(t) = \left(
\begin{array}{cc}
 -e^{-t} (2 t+1) & e^{-t} (5 t+3) \\
 2 e^{-t} & -5 e^{-t} \\
\end{array}
\right)$$
$$\phi^{-1}(t).f(t) = \left(
\begin{array}{cc}
 -e^{-t} (2 t+1) & e^{-t} (5 t+3) \\
 2 e^{-t} & -5 e^{-t} \\
\end{array}
\right).\left(
\begin{array}{c}
 e^{-t} \\
 0 \\
\end{array}
\right) = \left(
\begin{array}{c}
 -e^{-2 t} (2 t+1) \\
 2 e^{-2 t} \\
\end{array}
\right)$$
Integrating this previous result yields:
$$\left(
\begin{array}{c}
 -e^{-2 t} (-t-1) \\
 -e^{-2 t} \\
\end{array}
\right)$$
We now multiply the previous result with $\phi(t)$, yielding a particular solution of:
$$\left(
\begin{array}{c}
 2 e^{-t} \\
 e^{-t} \\
\end{array}
\right)$$
Finally, our solution is:
$$x(t) = x_h(t) + x_p(t) =  c_1e^t \begin{pmatrix}
  5 \\
  2 
 \end{pmatrix} + c_2 \left[te^t\begin{pmatrix}
  5 \\
  2
 \end{pmatrix} +e^t \begin{pmatrix}
  3 \\
  1 
 \end{pmatrix} \right] 
 + \left(
\begin{array}{c}
 2 e^{-t} \\
 e^{-t} \\
\end{array}
\right)$$
