Discussion
Eliminating the arbitrary constants from \( y = Ae^{x} + Be^{-x} \) gives
Answer: C
Differentiating twice returns the function itself:
\[ y' \;=\; Ae^{x} - Be^{-x}, \qquad y'' \;=\; Ae^{x} + Be^{-x} \;=\; y. \]
So \( y'' - y = 0 \). Two constants call for two differentiations and yield an equation of order two. The equation \( y'' + y = 0 \) belongs to \( \sin x \) and \( \cos x \) instead
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