Discussion
The extremal of the functional :
\[ I(y(x)) = \int_{1}^{2} \frac{x^{3}}{(y')^{2}}\, dx \]
subject to the conditions \(y(1) = 0\), \(y(2) = 3\) is :
Answer: B
The extremal is \(y = x^{2} - 1\).
\(F = x^{3}/(y')^{2}\) does not contain \(y\), so the Euler equation reduces to \(F_{y'} = \text{constant}\):
\[ \frac{-2x^{3}}{(y')^{3}} = c \;\Rightarrow\; y' = kx \]
so \(y = \tfrac{k}{2}x^{2} + d\). From \(y(1) = 0\): \(k/2 + d = 0\). From \(y(2) = 3\): \(2k + d = 3\). Subtracting, \(\tfrac{3k}{2} = 3\), so \(k = 2\), \(d = -1\) and \(y = x^{2} - 1\).
Check the others against the boundary values: \((x^{2} - 1)x\) gives 6 at \(x = 2\), \((x - 1)x\) gives 2, and \(x - 1\) gives 1, none of them 3.
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