Finding particular solution to inhomogeneous system of differential equations I am asked to find the general solution set of the following system of differential equations:
$$\begin{cases}
x' = 3x -2y-2 \\ 
y' = 6x-4y-1
\end{cases}
$$
I found the general solution set of the corresponding homogeneous system to be:
$$\{ \left[
    \begin{array}{c}
      x\\                    
      y
    \end{array}
\right] = k_1\left[
    \begin{array}{c}
      2\\                    
      3
    \end{array}
\right] + k_2e^{-t}\left[
    \begin{array}{c}
      1\\                    
      2
    \end{array}
\right] : k_1, k_2 ∈ R\}$$
Now I have to find a particular solution. Since $\left[
    \begin{array}{c}
      2\\                    
      1
    \end{array}
\right]$ is a particular solution of the homogeneous equation I can't use the method of undetermined coefficients, correct? If so, how can I find a particular solution?
 A: A particular solution is not 
$\left[
    \begin{array}{c}
      2\\                    
      1
    \end{array}\right]$ 
but is 
$\left[
    \begin{array}{c}
      2\\                    
      3
    \end{array}\right]$
 . The method of undetermined coefficients continue to be usable.
A particulat solution of the non-homogeneous equation is to be searched on the form :
$$\left[
    \begin{array}{c}
      x\\                    
      y
    \end{array}\right]=\left[
    \begin{array}{c}
      2\\                    
      3
    \end{array}\right]a\:t+
\left[
    \begin{array}{c}
      b\\                    
      c
    \end{array}\right]
 $$
Bring it back into the ODE system and identify the coefficients.( I obtained  $a=-3$ , $b=0$ , $c=2$ , to be checked )
For more explanation, see http://en.wikipedia.org/wiki/Method_of_undetermined_coefficients and example [2].
Alternatively, one could use the method of "variation of coefficients", but it should be more arduous. In this method, one remplace $k_1$ and $k_2$ by $f(t)$ and $g(t)$ respectively and solve the system for $f(t)$ and $g(t)$.
A: if you multiply the first by $2$ and subtract from the second, you get 
$$2x' - y' = -3$$ we can satisfy this by picking $$x = at+b, y = (2a+3)t +b $$ where $a, b$  are to fixed later.  we will try to satisfy $$x' = 3x - 2y -2 \to a = 3(at+b) -2[(2a+3)t + b]-2=-t(a+6)+b \tag 1$$  we will choose $$a = b = -6  $$ to make $(1)$ an identity.
therefore, a particular solution is $$x = -6(t+1), \, y = -3(3t+2) $$
