I am training to solve the following physical system via lagrangian equations for equilibrium Points: 
I have got the following boundary condition: $\frac{1}{2}(y_1+y_2) = l_2-y_3$ $\Rightarrow$ $y_3 = l_2 - \frac{1}{2} (y_1 - y_2)$. $y_1, y_2, y_3$ are the vertical positions of $m_1, m_2, m_3$, and $l_1$ and $l_2$ are the lengths of the ropes on the small and the big wheel.
I get the following equations for the kinetic energy
$$
T = \frac{1}{2} \left( m_1\dot{y_1} + m_2 \dot{y_2} + m_3 \dot{y_3} \right) = \frac{1}{2} \left( \left( m_1 + \frac{m_3}{4}\right)\dot{y_1} + \left(m_2 + \frac{m_3}{4} \right) \dot{y_2} - \frac{m_3}{2} \dot{y_1}\dot{y_2}\right)
$$
and the potential Energy
$$
V = mg(h-y_1) + m_2g(h-y_2) + m_3g(h-y_3) = \left( \frac{m_3}{2} - m_1\right)y_1g + \left( \frac{m_3}{2} - m_2\right) y_2g + const.
$$
Which leads to the following 2 lagrangian equations: $$ (i) \quad \frac{d}{dt} \frac{\partial L}{\partial \dot{y_1}} - \frac{\partial L}{\partial y_1} = \left( m_1 + \frac{m_3}{4}\right)\ddot{y_1} - \frac{m_3}{4}\ddot{y_2} + \left(\frac{m_3}{2} - m_1\right) g = 0 $$ and $$ (ii) \quad \frac{d}{dt}\frac{\partial L}{\partial \dot{y_2}} - \frac{\partial L}{\partial y_2} = \left( m_2 + \frac{m_3}{4} \right) \ddot{y_2} - \frac{m_3}{4}\ddot{y_1} + \left( \frac{m_3}{2} - m_2 \right)g = 0. $$
Now for one of the equibilirium points i guessed the following condition: $m_1 = m_2 \Rightarrow$
$$ (i) - (ii) = m_1\ddot{y_1} - m_2\ddot{y_2} + \frac{m_3}{4} \left( \ddot{y_1} - \ddot{y_2}\right) + (m_2 - m_1) g = 0 $$ with $m_1 = m_2 = m$ $$ \Rightarrow m \ddot{y_1} - m \ddot{y_2} + \frac{m_3}{4} ( \ddot{y_1} - \ddot{y_2}) = 0\\ \Rightarrow m(\ddot{y_1} - \ddot{y_2}) = -\frac{m_3}{4} ( \ddot{y_1} - \ddot{y_2}) $$ $$ \Rightarrow m = -\frac{m_3}{4} $$ Where is my error? Please help me. D:
