Discussion
The curve which extremizes the functional
\[ I[y(x)] = \int_{0}^{\pi/4} (y''^{2} - y^{2} + x^{2})\,dx \]
under the conditions \(y(0) = 0\), \(y'(0) = 1\), \(y(\pi/4) = y'(\pi/4) = \dfrac{1}{\sqrt{2}}\) is :
Answer: D
The extremal is \(y(x) = \sin x\).
For \(F(x, y, y'') = y''^{2} - y^{2} + x^{2}\), the Euler-Poisson equation is \(F_{y} + \dfrac{d^{2}}{dx^{2}}F_{y''} = 0\):
\[ -2y + 2y^{(4)} = 0 \Rightarrow y^{(4)} = y \]
So \(y = c_{1}e^{x} + c_{2}e^{-x} + c_{3}\cos x + c_{4}\sin x\). The four conditions pick out \(\sin x\): \(\sin 0 = 0\), \(\cos 0 = 1\), and \(\sin(\pi/4) = \cos(\pi/4) = 1/\sqrt{2}\), and the four conditions determine the constants uniquely.
\(1 - \cos x\) has \(y'(0) = 0\); \(\cos x\) has \(y(0) = 1\); \(\tan x\) has \(y(\pi/4) = 1\). None meets the conditions.
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