Discussion
The extremal of the functional
\[ I(y(x)) = \int_{0}^{4} \left(xy' - (y')^{2}\right) dx \]
determined by the boundary conditions \(y(0) = 0\) and \(y(4) = 3\) is :
Answer: D
The extremal is \(4y = x(x - 1)\).
The integrand \(F = xy' - y'^{2}\) does not contain \(y\), so the Euler-Lagrange equation reduces to \(F_{y'} = \text{const}\):
\[ \begin{aligned} x - 2y' &= c \\ y' &= \frac{x - c}{2} \\ y &= \frac{x^{2}}{4} - \frac{cx}{2} + d \end{aligned} \]
The condition \(y(0) = 0\) gives \(d = 0\), and \(y(4) = 3\) gives \(4 - 2c = 3\), so \(c = 1/2\). Then \(y = x^{2}/4 - x/4\), that is \(4y = x(x - 1)\).
Checking the others: \(y = x^{2}/4\) gives \(y(4) = 4\), \(x^{2}/2 + 3x\) gives 20, and \(x^{2}/4 - x + 1\) gives \(y(0) = 1\). Each fails a boundary condition.
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