Atwood's machine: $(m_1+m_2)\ddot x=(m_1-m_2)g$

If the Lagrangian $L(x,\dot x)$ corresponding to Atwood's machine is given as $L(x,\dot x)=\frac12(m_1+m_2)\dot x^2+m_1gx+m_2g(l-x)$, where $m_1,m_2,l$ and $g$ are constants. Then the equation of motion is: