Second Order ODE De-coupling

Question

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..

Answer 1

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.

Answer 2

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 $Y$.

Answer 3

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 $y_1=\frac15(2u_1+u_2)$, $y_2=\frac15(2u_2-u_1)$.