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