<s>[INST] How do I determine the general solution of $L=\frac{\partial ^2}{\partial t^2}$? How do I determine the general solution of $L=\frac{\partial ^2}{\partial t^2}$? $L\begin{pmatrix}x\\ y\\ z\end{pmatrix}=A\begin{pmatrix}x\\ y\\ z\end{pmatrix}$. $A=\begin{pmatrix} 4 &0 &2 \\ 0&4 &6 \\ 2& 6 &1 \end{pmatrix}$. I calculated eigenvalues and those are -4,4,9, but what is next that I should do?
 [/INST] Let $X(t) = \pmatrix{x(t) \\y(t)\\z(t)} = \pmatrix{x_1(t) \\ x_2(t) \\ x_3(t)} $. The equation is equivalent to $X''(t) = AX(t)$.
As you calculated,  the eigenvalues of $A$ are $-4, 4$ and $9$. Three associated eigenvectors are
$$ \begin{pmatrix}-\frac{1}{4}\\ -\frac{3}{4}\\ 1\end{pmatrix},\:\begin{pmatrix}-3\\ 1\\ 0\end{pmatrix},\:\begin{pmatrix}\frac{2}{5}\\ \frac{6}{5}\\ 1\end{pmatrix}. $$
Thus, one has 
$$A = \underbrace{\begin{pmatrix}-\dfrac{1}{4}&-3&\dfrac{2}{5}\\ -\dfrac{3}{4}&1&\dfrac{6}{5}\\ 1&0&1\end{pmatrix}}_{P} \underbrace{\begin{pmatrix}-4&0&0\\ 0&4&0\\ 0&0&9\end{pmatrix}}_D P^{-1}.$$
Let $Y(t) = \pmatrix{y_1(t)\\y_2(t)\\y_3(t)} = P^{-1}X(t) \Leftrightarrow X = PY$, hence $Y'' = P^{-1}X''$ and $X'' = PY''$.
$$ \begin{align} X'' & = AX \\  PY'' & = APY \\  Y'' & = P^{-1}APY \\ Y'' &= DY \tag{*}\end{align}$$
The new system (*) is equivalent to
$$ \cases{y_1''(t) = -4y(t) \\ y_2''(t) = 4y_2(t) \\ y_3''(t) = 9y_3(t)} $$
Using for instance each equation's characteristic polynomial, one can easily find its solution:
$$ \cases{y_1(t) = c_1 \cos(2 t) + c_2 \sin(2 t) \\ y_2(t) = c_3 e^{2t}+c_4 e^{-2t} \\ y_3(t) = c_5 e^{3t}+c_6 e^{-3t} } \qquad \text{with} \: c_1, c_2, c_3, c_4, c_5, c_6 \in \mathbb{R}.$$
Using $X = PY$, one can eventually get to the solution of the initial system:
$$\pmatrix{x_1(t) \\ x_2(t) \\ x_3(t)} = \begin{pmatrix}-\dfrac{1}{4}&-3&\dfrac{2}{5}\\ -\dfrac{3}{4}&1&\dfrac{6}{5}\\ 1&0&1\end{pmatrix} \pmatrix{y_1(t)\\y_2(t)\\y_3(t)},$$
which yields, using your initial notation,
$$\bbox[5px,border:2px solid green,lightgreen]{\cases{x(t) = -\frac{1}{4}\left(c_1 \cos(2t) + c_2 \sin(2t)\right) - 3 \left(c_3 e^{2t} + c_4 e^{-2t}\right) + \frac{2}{5} \left(c_5e^{3t} + c_6 e^{-3t}\right) \\ y(t) = -\frac{3}{4}\left(c_1 \cos(2t) + c_2 \sin(2t)\right) + c_3 e^{2t} + c_4 e^{-2t} + \frac{6}{5} \left(c_5e^{3t} + c_6 e^{-3t}\right) \\ z(t) = c_1 \cos(2t) + c_2 \sin(2t) + c_5e^{3t} + c_6 e^{-3t}}}.$$
Of course, to determine the constants $c_1, c_2, c_3, c_4, c_5$ and $c_6$, you'll need initial conditions.
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