Discussion
Let \(J(y) = \displaystyle\int_{-1}^{1} (y'e^{y} + xy^{2})\, dx\) be a functional defined on \(C^{1}[-1, 1]\). Then the variation of the integral \(J[y]\) is :
Answer: C
The variation is \(\int_{-1}^{1} [[y'e^{y} + 2xy]\delta y + e^{y}\delta y']\,dx\).
For \(J = \int F(x, y, y')\,dx\), the first variation is \(\delta J = \int (F_{y}\,\delta y + F_{y'}\,\delta y')\,dx\). Here \(F = y'e^{y} + xy^{2}\), so
\[ \begin{aligned} F_{y} &= y'e^{y} + 2xy \\ F_{y'} &= e^{y} \end{aligned} \]
which is option C. Option D adds a spurious \(y^{2}\) to \(F_{y}\), and options A and B are not variations at all: they contain no \(\delta y\) and look like attempts at differentiating the integrand.
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