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 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 $g(t)$.