I have the differential equations:
$$m\ddot x_1=(8a^2b+k)x_1-kx_2+(8a^3b+2ka)$$ $$m\ddot x_2=(8a^2b+k)x_2-kx_1-(8a^3b+2ka)$$
I need to put into the form $M\ddot X = KX$, but those constants first need to be dealt with. How do I do this?
I tried saying that $\ddot x_1(0)=\ddot x_2(0)=0$ and this gave me a system of 2 equations with 2 unknowns. When I solved for $x_1(0)$ and $x_1(0)$, I treated the resulting values as equilibrium points. Then I created new coordinates $x'_1(t)=x_1(t)-x_1(0)$ and $x'_2(t)=x_2(t)-x_2(0)$. The problem is that when I substituted in the new prime coordinates, the constants did not cancel. Was my approach wrong?
Let $\;X=\begin{bmatrix}x_1\\x_2\end{bmatrix}\:$ and $\ddot X=\begin{bmatrix}\ddot x_1\\\ddot x_2\end{bmatrix}.$ Then the system $\;\begin{cases} m\ddot x_1 = \left(8a^2b+k\right)x_1-kx_2+\left(8a^3b+2ka\right)\\m\ddot x_2=\left(8a^2b+k\right)x_2-kx_1-\left(8a^3b+2ka\right)\end{cases}$ can be rewritten as $\:\;%m\,\ddot{X} = m\begin{bmatrix} \ddot{x}_1 \\ \ddot{x}_2 \end{bmatrix} =\begin{bmatrix} 8a^2b+k & -k \\ -k & 8a^2b+k \end{bmatrix} \begin{bmatrix} {x}_1 \\ {x}_2 \end{bmatrix} + \begin{bmatrix} \phantom{-}8a^3b+2ka \\ - 8a^3b-2ka \end{bmatrix} .$
Adding together two original equations will result in $\;m\left(\ddot{x}_1+\ddot{x}_2\right) = 8a^2b\left(x_1+x_2\right).\,$ Subtracting the second equation from the first one yeilds $\;m\left(\ddot{x}_1-\ddot{x}_2\right) = \left(8a^2b + 2k\right)\left(x_1-x_2+2a\right).\,$
Let $\ y_1:=x_1+x_2\,$ and $\,y_2 := x_1 - x_2 + 2a.\,$ Then $\,\ddot y_1 = \ddot{x}_1+\ddot{x}_2\,$ and $\,\ddot y_2 = \ddot{x}_1-\ddot{x}_2,\,$ and so the system of equations can be rewritten as
$$\left\lbrace\begin{aligned} m\,\ddot y_1 =& 8a^2b\,y_1\\ m\,\ddot y_2 =& \left(8a^2b + 2k\right)y_2 \end{aligned}\right. \iff m\begin{bmatrix} \ddot{y}_1 \\ \ddot{y}_2 \end{bmatrix} =\begin{bmatrix} 8a^2b & 0 \\ 0 & 8a^2b+2k \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} \iff \bbox[1ex, border:solid 2pt #e10000]{m\,\ddot{Y} = KY} $$
where $\;\displaystyle Y = \begin{bmatrix} y_1 \\ y_2 \end{bmatrix}\,$ and $\,K=\begin{bmatrix} 8a^2b & 0 \\ 0 & 8a^2b+2k \end{bmatrix}.\,$