I've got the following ODE, and I'm just having trouble coming up with the form of the general solution. I'm really trying to find the particular solution, but in order to do that, I need to know the homogeneous solution.
Here's the problem: $$ y' = \begin{bmatrix} 3 & 1 & -1 \\ 3 & 5 & 1 \\ -6 & 2 & 4 \end{bmatrix} y + \begin{bmatrix} 3\\ 6\\ 8 \end{bmatrix} $$
I know that my eigenvalues are r_1 = 0 and r_2=6 and r_3 = 6
My eigenvectors should be: $$ v_1 = \begin{bmatrix} 1\\ -1\\ 2 \end{bmatrix} v_2 =\begin{bmatrix} 1\\ 3\\ 0 \end{bmatrix} v_3 = \begin{bmatrix} -1\\ 0\\ 3 \end{bmatrix} $$
This means my homogeneous solution should take the form: $$ y(t) = c_1\begin{bmatrix} 1\\ -1\\ 2 \end{bmatrix} + c_2\begin{bmatrix} 1\\ 3\\ 0 \end{bmatrix}e^{6t} + c_3(\begin{bmatrix} -1\\ 0\\ 3 \end{bmatrix}e^{6t} +\begin{bmatrix} 1\\ 3\\ 0 \end{bmatrix}te^{6t}) $$ Can anyone confirm this for me? I know that I'm right about the eigenvalues, but wasn't sure if my homogeneous component was in the correct form?
Thanks!