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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.

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  • $\begingroup$ Oh! you forgot about your problem,! $\endgroup$
    – Z Ahmed
    Dec 11 '20 at 12:22
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$I(y) = \displaystyle \int_{x_0}^{x_1} \dfrac{1 + y^2}{(y')^3} dx$ 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.

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