Discussion
The differential equation \(\frac{d^{2}y}{dx^{2}} + 16y = 0\) for \(y(x)\) with the two boundary conditions \(\left.\frac{dy}{dx}\right|_{x=0} = 1\) and \(\left.\frac{dy}{dx}\right|_{x=\frac{\pi}{2}} = -1\) has
Answer: A
The problem has no solution.
The general solution is \(y = A\cos 4x + B\sin 4x\), so \(y' = -4A\sin 4x + 4B\cos 4x\).
\[ \begin{aligned} y'(0) &= 4B = 1 \Rightarrow B = \tfrac{1}{4} \\ y'(\pi/2) &= -4A\sin 2\pi + 4B\cos 2\pi = 4B = 1 \end{aligned} \]
The second condition demands -1, but every solution gives y'(\(\pi\)/2) = 1, so the two conditions contradict each other.
If the second condition had been +1, A would be free and there would be infinitely many solutions; a linear problem like this can never have exactly two.
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