= Solution
Put $\Delta=\theta_1-\theta_2$ and $\omega_0^2=g/l$. The Euler--Lagrange equations, after division by the appropriate powers of $l$, are
$$
\boxed{(M+m)\ddot\theta_1+m\cos\Delta\,\ddot\theta_2
+m\sin\Delta\,\dot\theta_2^2+(M+m)\omega_0^2\sin\theta_1=0,}
$$
$$
\boxed{\ddot\theta_2+\cos\Delta\,\ddot\theta_1
-\sin\Delta\,\dot\theta_1^2+\omega_0^2\sin\theta_2=0.}
$$
Time-translation invariance conserves the total energy
$$
E=\frac12(M+m)l^2\dot\theta_1^2+\frac12ml^2\dot\theta_2^2
+ml^2\cos\Delta\,\dot\theta_1\dot\theta_2
-(M+m)gl\cos\theta_1-mgl\cos\theta_2.
$$
Setting both angles and velocities to zero satisfies the equations, so the downward configuration is an equilibrium. Writing $\theta_i=z_i$ and retaining linear terms gives
$$
\boxed{(M+m)\ddot z_1+m\ddot z_2+(M+m)\omega_0^2z_1=0,}
$$
$$
\boxed{\ddot z_1+\ddot z_2+\omega_0^2z_2=0.}
$$
Equivalently, the mass and stiffness matrices are
$$
\mathsf M=l^2\begin{pmatrix}M+m&m\\m&m\end{pmatrix},
\qquad
\mathsf K=gl\begin{pmatrix}M+m&0\\0&m\end{pmatrix}.
$$
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