Suppose the double Atwood machine is composed of three masses, $m_1\, m_2\, m_3$ connected by two chords of length $L$ and $L\'$ respectively through two ideal pulleys. The system has two degrees of freedom, since the height of $m_3$ can be found from the height of $m_2$ and the second pulley. The most convenient set of generalized coordinates to describe the motion are $y_1$: the distance from $m_{1 }$to the the top pulley, and $y_2$: the distance from $m_2$ to the second pulley. Given these coordinates, the distance from $m_2$ to the top pulley is $L-y_1 \+y_2$, and the distance from $m_3$ to the top pulley is $L\+L\' - y_1- y_2$.

The kinetic energy of the system is then:

$$$T\=\frac{1}{2}{m}_1{\stackrel{\.}{y}}_1^2\+\frac{1}{2}{m}_2{\left(-\stackrel{\.}{y_1}\+\stackrel{\.}{y_2}\right)}^2\+\frac{1}{2}{m}_3{\left(-\stackrel{\.}{y_1}-\stackrel{\.}{y_2}\right)}^2$,

where the dot denotes a time derivative. By setting the potential energy to zero at the height of the top pulley, the total potential energy of the system is:

$U \= -m_{1}g y_1-m_{2}g\left(L-y_1\+y_2\right)-m_{3}g\left(L-y_1\+L\' - y_2\right)$.

 The Lagrangian is then $L \= T - U$ or:

$L \= \frac{1}{2}{m}_1{\stackrel{\.}{y}}_1^2\+\frac{1}{2}{m}_2{\left(-\stackrel{\.}{y_1}\+\stackrel{\.}{y_2}\right)}^2\+\frac{1}{2}{m}_3{\left(-\stackrel{\.}{y_1}-\stackrel{\.}{y_2}\right)}^2\+m_{1}g y_1\+m_{2}g\left(L-y_1\+y_2\right)\+m_{3}g\left(L-y_1\+L\'-y_2\right)$.

The equations of motion then follow from evaluating the Euler-Lagrange equations:

$\frac{\partial }{\partial t}\left(\frac{\partial L}{\partial {\stackrel{\.}{y}}_1}\right)-$$\frac{\partial L}{\partial y_1} \= 0$,

$\frac{\partial }{\partial t}\left(\frac{\partial L}{\partial {\stackrel{\.}{y}}_2}\right)-$$\frac{\partial L}{\partial y_2} \= 0$.

Differentiating and simplifying these equations give the following equations of motion for the double Atwood machine:

$m_2\stackrel{..}{y_1}\+m_2\left(\stackrel{..}{y_1}- \stackrel{..}{y_2}\right)\+m_3-g\left(m_1-m_2-m_3\right)\=0$,

$m_2\left(-\stackrel{..}{y_1}\+\stackrel{..}{y_2}\right)\+m_3\left(\stackrel{..}{y_1}\+\stackrel{..}{y_2}\right)-g\left(m_2-m_3\right)\=0$.

Rearranging terms gives the accelerations of each block as:

$\stackrel{..}{y_1}\=-g\frac{m_1\left(m_2 \+ m_3\right)- 4 m_2{m}_3}{m_1\left(m_2\+m_3\right) \+ 4 m_2{m}_3}$,

$\stackrel{..}{y_2}\=-g\frac{m_1\left(m_2 - 3 m_3\right)\+ 4 m_2{m}_3}{m_1\left(m_2\+m_3\right) \+ 4 m_2{m}_3}$,

$\stackrel{..}{y_3}\=-g\frac{m_1\left(m_3 - 3 m_2\right)\+ 4 m_2{m}_3}{m_1\left(m_2\+m_3\right) \+ 4 m_2{m}_3}$.$$

where we used that $\stackrel{..}{y_1}\=-\frac{\left(\stackrel{..}{y_2}\+\stackrel{..}{y_3}\right)}{2}$ and solved for $\stackrel{..}{y_3}$, the position of the third mass.