Find extremum of functional $$\int\limits_{x_0}^{x_1}{1+y^2\over \dot{y}^2}$$ Euler-Lagrange equation is: $${y \over \dot{y}^2} - {3\ddot{y}\over \dot{y}^4} - {3y^2\ddot{y}\over\dot{y}^4} + {2y\dot{y}\over\dot{y}^3} = 0$$ After simplifying I got $${y\dot{y}^2 - \ddot{y} - y^2\ddot{y} \over \dot{y}^4} = 0$$ I mentioned that $$y{d\over dx}({y\over \dot{y}^3}) = {y\dot{y}^2 - 3\ddot{y}y^2\over \dot{y}^4}$$ But I couldn't use it anyway. So I'm stuck and need help with this problem

  • $\begingroup$ As is typical for equations not containing $t$, look for $\dot y$ as a function of $y$. $\endgroup$ – Ivan Neretin Apr 22 '18 at 14:13

So you have the following equation:

$${y\dot{y}^2 - \ddot{y} - y^2\ddot{y} \over \dot{y}^4} = 0$$


$${y\dot{y}^2 - \ddot{y} - y^2\ddot{y}} = 0$$

$$y\dot{y}^2=(1+y^2)\ddot y$$


$$\ddot{y}=\frac{d\dot{y}}{dy} \dot{y}$$

...so the equation becomes:

$$ y\dot{y}^2=(1+y^2) \frac{d\dot{y}}{dy} \dot{y}$$

$$ \frac{y dy}{1+y^2} = \frac{d\dot{y}}{\dot y}$$

By integrating the last equation you get:

$$ \frac1 2 \ln({1+y^2}) = \ln{\dot{y}} - \ln{C_1}$$

$$ \sqrt{1+y^2} = \frac{1}{C_1}\dot{y}$$

$$ \sqrt{1+y^2} = \frac{1}{C_1}\frac{dy}{dt}$$


$$C_1 dt =\frac{dy}{\sqrt{1+y^2}}$$

By integrating the last equation you get:

$$ C_1 t=\sinh^{-1}{y}-C_2 $$

or, finally:

$$ y = \sinh (C_1 t + C_2)$$


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