Solve These Irregular Coupled Second Order Differential Equations

Question

I have a question on how to solve this pair of coupled differential equations.
\begin{align*}
\omega_1^2 Q_1 &= -\ddot{Q}_1 – \frac{M}{L_1}\ddot{Q}_2 \\
\omega_2^2 Q_2 &= -\ddot{Q}_2 – \frac{M}{L_2}\ddot{Q}_1
\end{align*}
for $Q_1 (t)$ and $Q_2 (t)$. Generally I would solve these types of equations by using a matrix $M^{-1}K – \omega^2 I = 0$. However, in this case I don't really see how I can create matrices to solve this problem and I can't find any neat trick to solve these equations. If anyone could help that would be great!

Answer 1

Meant to be a comment

I have never solved coupled differential equations, but here goes my try:

Let $l_1=\frac{M}{L_1},l_2=\frac{M}{L_2}$. \begin{align*} \ \ \ \omega_1^2 Q_1 &= -\ddot{Q}_1 - l_1\ddot{Q}_2 \\ +bc\times(\omega_2^2 Q_2 &= -\ddot{Q}_2 - l_2\ddot{Q}_1)\\ \hline \omega_1^2 Q_1 +bc\omega_2^2 Q_2&=-\left(1+bcl_2\right)\ddot{Q}_1-\left(l_1+bc\right)\ddot{Q}_2\\ \omega_1^2 Q_1 +bc\omega_2^2 Q_2&=-\left(1+bcl_2\right)\ddot{Q}_1-\left(\frac{l_1}c+b\right)c\ddot{Q}_2\\ \star(Q_1 +cQ_2)&=-\star(\ddot{Q}_1+c\ddot{Q}_2)\\ \end{align*} is only possible when \begin{align*} b&=\frac{\omega_1^2}{\omega_2^2}\\ 1+bcl_2=\frac{l_1}c+b\Rightarrow bl_2c^2-(b-1)c-l_1=0\Rightarrow c&=\frac{(b-1)\pm\sqrt{(b-1^2)+4bl_1l_2}}{2bl_2}\\ \end{align*} So, the equations become \begin{align*} \omega_1^2(Q_1 +cQ_2)&=-(1+bl_2c)(\ddot{Q}_1+c\ddot{Q}_2)\\ -\omega^2Q&=\ddot{Q}\\ \end{align*} where $\omega=\sqrt{\frac{\omega_1^2}{1+bl_2c}}, Q=Q_1+cQ_2$.