\[ \begin{eqnarray} \theta_1'' & = & \frac{-g\left(2m_1+m_2\right)\sin \theta_1 - m_2 g \sin\left(\theta_1 - 2 \theta_2\right) - 2\sin\left(\theta_1 - \theta_2\right)\,m_2 \big[\theta_2'^2 L_2 + \theta_1'^2L_1 \cos\left(\theta_1 - \theta_2\right)\big]}{L_1\big[2m_1+m_3-m_2\cos \left(2\theta_1 - 2 \theta_2\right) \big]}\\ \theta_2'' & = & \frac{ 2\sin\left(\theta_1 - \theta_2\right) \big[ \theta_1'^2L_1(m_1+m_2) + g (m_1+m_2) \cos \theta_1 + \theta_2'^2L_2m_2\cos\left(\theta_1 - \theta_2\right) \big]}{L_2\big[2m_1+m_3-m_2\cos \left(2\theta_1 - 2 \theta_2\right) \big]}\\ \end{eqnarray} \]
Source: www.myphysicslab.com