= Solution
The quadratic total <mechanical energy> is
$$
E=\frac12\dot{\boldsymbol\theta}^{\mathsf T}\mathbf T\dot{\boldsymbol\theta}+\frac12\boldsymbol\theta^{\mathsf T}\mathbf V\boldsymbol\theta,\qquad \frac{dE}{dt}=\dot{\boldsymbol\theta}^{\mathsf T}(\mathbf T\ddot{\boldsymbol\theta}+\mathbf V\boldsymbol\theta).
$$
For this frictionless conservative <double pendulum>, the <Euler-Lagrange equations> for $L=K-U$ give
$$
\boxed{\mathbf T\ddot{\boldsymbol\theta}+\mathbf V\boldsymbol\theta=0,}
$$
or $8\ddot\theta_1+2\ddot\theta_2+(4g/l)\theta_1=0$ and $2\ddot\theta_1+\ddot\theta_2+(g/l)\theta_2=0$. Substitution immediately verifies <energy conservation>.
\b[Logical qualification.] <Energy conservation does not uniquely determine equations of motion>: the scalar equation $dE/dt=0$ alone does not determine both equations. For example, replacing their right side by $\mathbf G\dot{\boldsymbol\theta}$ with any nonzero antisymmetric matrix $\mathbf G$ preserves the same energy because $\dot{\boldsymbol\theta}^{\mathsf T}\mathbf G\dot{\boldsymbol\theta}=0$, while changing the dynamics. Such a gyroscopic force is absent from the specified pendulum. The conservative-force structure, supplied here by the <Lagrangian>, is the additional information needed to justify the displayed equations rather than infer two independent equations from one scalar identity.
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