Second Order ODE De-coupling I have two coupled second order DE that can be written as
$$\begin{pmatrix}
\ddot y_1  \\
\ddot y_2  \\
\end{pmatrix}  =
\begin{pmatrix}
-10 & 4  \\
4 & -4  \\
\end{pmatrix}
 \begin{pmatrix}
 y_1  \\
 y_2  \\
\end{pmatrix}$$
I need to find the solutions. I have tried to find the eigenvalues and I find that they are -12 and -2. But then I am not sure how to write the solution.
The eigenvectors are for -12
$$\begin{pmatrix}
 2  \\
 -1  \\
\end{pmatrix}$$
and for 
-2
$$ \begin{pmatrix}
 1  \\
 2  \\
\end{pmatrix}$$
I cannot form a solution after this. Simple I dont know how to proceed. Can anyone help me please..
 A: Per my comment, we get a linear 1st order system:
$$\begin{bmatrix} \dot{y_1} \\  \dot{y_2} \\  \dot{y_3} \\  \dot{y_4} \end{bmatrix}=\begin{bmatrix} 0 & 0 & 1 & 0 \\  0 & 0 & 0 & 1 \\  -10 & 4 & 0 & 0 \\ 4 & -4 & 0 & 0 \end{bmatrix} \begin{bmatrix} y_1 \\  y_2 \\  y_3 \\  y_4 \end{bmatrix} $$
With eigenvalues:
$$\lambda_{1,2}=\pm i\sqrt{2} \\ \lambda_{3,4}=\pm 2i\sqrt{3}$$
For eigenvectors you can look at Wolfram Alpha output.
A: Your system can be written in factored matrix/vector form,
$$\ddot Y=T\Lambda T^{-1}Y,$$
or
$$T^{-1}\ddot Y=\Lambda T^{-1}Y,$$ which is of the form
$$\ddot Z=\Lambda Z.$$
Hence the solutions in $Z$ will be linear combinations of $e^{\pm i\sqrt2x}$ and $e^{\pm i\sqrt{12}x}$, and so will the solutions in $Y$.
A: With the (left) eigenvectors you now know which linear combinations to choose that decouple the system, $u_1=2y_1-y_2$ and $u_2=y_1+2y_2$. Then 
$$
\ddot u_1=2\ddot y_1-\ddot y_2=-12(2y_1-y_2)=-12u_1
$$
and
$$
\ddot u_2=\ddot y_1+2\ddot y_2=-2(y_1+2y_2)=-2u_2
$$
These are both harmonic oscillators and the solutions thereof recombine to solutions of the original system as $y_1=\frac15(2u_1+u_2)$, $y_2=\frac15(2u_2-u_1)$.
