Finding an Extremal for a function.

I need to find the extremals for the following function :

$$I(y) = \displaystyle \int_{x_0}^{x_1} \dfrac{1 + y^2}{(y')^3} dx$$

So, by Euler Lagrange Equations

$$I_{y}$$ -$$d/dx(I_{y'}) = 0$$

Now, using this I get :

$$\dfrac{2y}{y'} + \dfrac{2y}{3} = \dfrac{4(1+y^2)y''}{(y')^3}$$

At, this point I am stuck, Please tell me how should I proceed ?

Thank You.

• Oh! you forgot about your problem,! Dec 11 '20 at 12:22

$$I(y) = \displaystyle \int_{x_0}^{x_1} \dfrac{1 + y^2}{(y')^3} dx$$ reduced Euler-Lagrange equationd for $$I=\int F(x,y,y') dx$$ is given as $$F-y'\frac{\partial F}{\partial y'}=C.$$ So here we have $$\frac{1+y^2}{y'^3}+3y'\frac{1+y^2}{y'^4}=C \implies y'= D(1+y^2)^{1/3}.$$ $$\implies \int \frac{dy}{(1+y^2)^{1/3}}=Dx+E.$$ $$\implies y~_2F_1(1/2,1/3,3/2;y^2)=Dx+E.$$ Here $$~_2F_1(a,b;c,z)$$ is the Gauss hypergeometric function.