Consider the following functional which is as follows:

Question

I am stuck on the following problem:

I tried using Euler's formula which is as follows:

But my calculation gets complicated and I could not get the results. Can someone help me in this regard? Thanks in advance for your time.

Answer 1

Let $f(x,y(x),y'(x)):=\frac{1+y^2}{(y')^2}$. Then

$$\frac{\partial f}{\partial y}=\frac{2y}{(y')^2},$$

$$\frac{d}{dx}(\frac{\partial f}{\partial y'})=\frac{d}{dx}(-2\frac{1+y^2}{(y')^3})= 6\frac{(1+y^2)y^{''}}{(y')^4}-4\frac{yy'}{(y')^3};$$

The Euler Lagrange equations are then

$$y(y')^2-(1+y^2)y^{''}=0$$