I've got some problem with solving system of second order differential equations which solves the move of 3 masses at 4 springs - horizontally.
There is no problem with arrranging the equation, it looks like that(masses of each object and stiffness factors are the same for every spring): $$ m \frac{d^{2}x_{1}}{dt^{2}} = -2kx_{1} + kx_{2} $$ $$ m \frac{d^{2}x_{2}}{dt^{2}} = -kx_{1} + 2kx_{2} -kx_{3} $$ $$ m \frac{d^{2}x_{3}}{dt^{2}} = -kx_{2} + 2kx_{3} $$
Then i define $\omega ^{2} = \frac{k}{m}$ and divide both sides of all of the equations by $m$:
$$ \frac{d^{2}x_{1}}{dt^{2}} = -2\omega ^{2}x_{1} + \omega ^{2}x_{2} $$ $$ \frac{d^{2}x_{2}}{dt^{2}} = -\omega ^{2}x_{1} + 2\omega ^{2}x_{2} -\omega ^{2}x_{3} $$ $$ \frac{d^{2}x_{3}}{dt^{2}} = -\omega ^{2}x_{2} + 2\omega ^{2}x_{3} $$
I am rewritting it using matrices: $$ \begin{bmatrix} \ddot{x_{1}}\\\ddot{x_{2}}\\ \ddot{x_{3}} \end{bmatrix} = \omega^{2} \begin{bmatrix} -2 & 1 & 0\\ -1 & 2 & -1\\ 0 & -1 & 2 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix}$$
And right there I am stuck.
Make the Ansatz
$x_i = A_i e^{\Omega t}$ (i=1,2 or 3) and Substitute it into this System. After that you can rewrite this System in a form where a Matrix (this Matrix depends on $\Omega$) Multiplied by the vector $\vec{A} = (A_1,A_2,A_3)^T$ is equal to Zero.
That means you obtain an Eigenvalue Problem; set the determinant of this Matrix to Zero. You will obtain a cubic equation that yields 3 different Solutions. Let these Solutions be denoted by $\Omega_1,\Omega_2,\Omega_3$. To every single solution you can determine the eigenvectors $\vec{A}_1,\vec{A}_2,\vec{A}_3$ (by substituting the corresponding solution into the Matrix and solving the linear algebraic system).
Finally, you can write the solution in the form:
$\vec{x}= (x_1,x_2,x_3)^T = \sum_{i=1}^3 \vec{A}_i e^{\Omega_i t}$.