Discussion
The extremal of the functional
\[ I(y(x)) = \int_{0}^{1} \frac{\dot{y}^{2}}{x^{3}}\, dx, \]
when \(x(0) = 0\), \(x(1) = 1\) is :
Answer: C
The extremal is y = x4.
The boundary conditions are printed as x(0) = 0 and x(1) = 1; they are the conditions \(y(0) = 0\), \(y(1) = 1\) on the unknown function. The integrand \(F = y'^{2}/x^{3}\) does not contain y, so the Euler-Lagrange equation gives
\[ \begin{aligned} \frac{\partial F}{\partial y'} = \frac{2y'}{x^{3}} &= c \\ y' &= \frac{c}{2}x^{3} \\ y &= Ax^{4} + B \end{aligned} \]
\(y(0) = 0\) gives \(B = 0\) and \(y(1) = 1\) gives \(A = 1\), so \(y = x^{4}\). The line \(y = x\) and the constants c and 0 do not satisfy the Euler-Lagrange equation with these end values.
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