If $y = e^x \cos x$, then $\dfrac{d^2 y}{dx^2} - 2\dfrac{dy}{dx} + 2y$ equals:
A$2y$
B$0$
C$e^x$
D$y$
Answer & Solution
Correct answer: B. $0$
$y' = e^x(\cos x - \sin x)$. $y'' = e^x(\cos x - \sin x) + e^x(-\sin x - \cos x) = -2 e^x \sin x$. Check: $y'' - 2y' + 2y = -2e^x\sin x - 2e^x(\cos x - \sin x) + 2 e^x\cos x = -2e^x\sin x - 2e^x\cos x + 2e^x\sin x + 2e^x\cos x = 0$. ✓
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