Suppose you are given:
$x'' = \left( \begin{array}{ccc} -50&10 \\ 30&-30 \\ \end{array} \right)\ x + \left( \begin{array}{ccc} 5 \\ 0 \\ \end{array} \right)\ \sin10t$
Find the particular solution of the system.
A separate part of the problem asked to find the natural frequencies of the masses, which I found to be $\omega_1^2 = 20$ and $\omega_2^2 = 60$. I tried solving the problem using $(A+\omega^2I)c = -F_0$ such as:
$\left( \begin{array}{ccc} \omega^2-50&10 \\ 30&\omega^2-30 \\ \end{array} \right)\ \left( \begin{array}{ccc} c_1 \\ c_2 \\ \end{array} \right)\ = \left( \begin{array}{ccc} -5 \\ 0 \\ \end{array} \right)\ $
But I don't know how to continue. I thought I should plug in $\omega^2 = 20$ (since the force is acting on the first mass), but that results in the system:
$\left( \begin{array}{ccc} -30&10 \\ 30&-10 \\ \end{array} \right)\ \left( \begin{array}{ccc} c_1 \\ c_2 \\ \end{array} \right)\ = \left( \begin{array}{ccc} -5 \\ 0 \\ \end{array} \right)\ $
Which doesn't have any solutions for $c_1$ and $c_2$ in which I would then be able to write the particular solution as $x(t) = c_1\sin20t + c_2\sin20t$, correct?
Any help would be great. Thanks.