We have in the form $x'' = Ax$:
$\left[ \begin{array}{cccc} x_1''\\x_2'' \end{array} \right] = \left[ \begin{array}{cccc} -10&6\\6&-10 \end{array} \right]\left[ \begin{array}{cccc} x_1\\x_2 \end{array} \right]$
The eigenvalues of the matrix $A$ are $\lambda = -4$ and $\lambda = -16$, with corresponding eigenvectors $v_1 = \left[ \begin{array}{cccc} 1\\-1 \end{array} \right]$ and $v_2 = \left[ \begin{array}{cccc} 1\\1 \end{array} \right]$. Write the general solution to the homogeneous problem.
Why wouldn't the answer just be $x = c_1e^{-4t}\left[ \begin{array}{cccc} 1\\-1 \end{array} \right] + c_2e^{-10t}\left[ \begin{array}{cccc} 1\\1 \end{array} \right]$?
What am I not understanding about the problem that actually makes the solution $x = c_1\left[ \begin{array}{cccc} 1\\-1 \end{array} \right]\cos2t+c_2\left[ \begin{array}{cccc} 1\\-1 \end{array} \right]\sin2t+c_3\left[ \begin{array}{cccc} 1\\1 \end{array} \right]\cos4t+c_4\left[ \begin{array}{cccc} 1\\1 \end{array} \right]\sin4t$?
Anyone know how to do this?