Lagrange Equation

${\displaystyle {\frac {d}{dt}}\left({\frac {\partial L}{\partial {\dot {x}}}}\right)-{\frac {\partial L}{\partial x}}=0}$

where ${\displaystyle {\dot {x}}={\frac {dx}{dt}}}$
The equation above is called Lagrange Equation.

Let the kinetic energy of the point mass be ${\displaystyle T}$ and the potential energy be ${\displaystyle U}$.
${\displaystyle T-U}$ is called Lagrangian. Then the kinetic energy is expressed by

${\displaystyle T={\frac {1}{2}}m{\dot {x}}^{2}+{\frac {1}{2}}m{\dot {y}}^{2}}$

${\displaystyle ={\frac {m}{2}}({\dot {x}}^{2}+{\dot {y}}^{2})}$

Thus

${\displaystyle T=T({\dot {x}},{\dot {y}})}$

${\displaystyle U=U(x,y)}$

Hence the Lagrangian ${\displaystyle L}$ is

${\displaystyle L=T-U}$

${\displaystyle =T({\dot {x}},{\dot {y}})-U(x,y)}$

${\displaystyle ={\frac {m}{2}}({\dot {x}}^{2}+{\dot {y}}^{2})-U(x,y)}$

Therefore ${\displaystyle T}$ relies on only ${\displaystyle {\dot {x}}}$ and ${\displaystyle {\dot {y}}}$. ${\displaystyle U}$ relies on only ${\displaystyle x}$ and ${\displaystyle y}$. Thus

${\displaystyle {\frac {\partial L}{\partial {\dot {x}}}}={\frac {\partial T}{\partial {\dot {x}}}}=m{\dot {x}}}$

${\displaystyle {\frac {\partial L}{\partial {\dot {y}}}}={\frac {\partial T}{\partial {\dot {y}}}}=m{\dot {y}}}$

In the same way, we have

${\displaystyle {\frac {\partial L}{\partial x}}=-{\frac {\partial U}{\partial x}}}$

${\displaystyle {\frac {\partial L}{\partial y}}=-{\frac {\partial U}{\partial y}}}$