Let \(y = e^{x}\sin x\). Consider the following statements:
(A) \(\displaystyle\frac{dy}{dx} = e^{x}(\sin x - \cos x)\)
(B) \(\displaystyle\frac{d^{2}y}{dx^{2}} = 2e^{x}\cos x\)
(C) \(\displaystyle\frac{d^{2}y}{dx^{2}} - 2\frac{dy}{dx} + 2y = 0\)
(D) The value of \(\displaystyle\frac{d^{2}y}{dx^{2}}\) at \(x = 0\) is 0.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (B) and (C) only
By the product rule:
\( \displaystyle \frac{dy}{dx} = e^{x}\sin x + e^{x}\cos x = e^{x}(\sin x + \cos x) \)
\( \displaystyle \frac{d^{2}y}{dx^{2}} = e^{x}(\sin x + \cos x) + e^{x}(\cos x - \sin x) = 2e^{x}\cos x \)
Then
\(y'' - 2y' + 2y = 2e^{x}\cos x - 2e^{x}(\sin x + \cos x) + 2e^{x}\sin x = 0\)
So y′ = ex(sin x + cos x), y″ = 2ex cos x, y″ − 2y′ + 2y = 0, and y″(0) = 2.
Checking the statements
(A) \(\frac{dy}{dx} = e^{x}(\sin x - \cos x)\) — Incorrect. The derivative of sin x is +cos x, so y′ = \(e^{x}(\sin x + \cos x)\).
(B) \(\frac{d^{2}y}{dx^{2}} = 2e^{x}\cos x\) — Correct. The sin x terms cancel in the second derivative.
(C) \(\frac{d^{2}y}{dx^{2}} - 2\frac{dy}{dx} + 2y = 0\) — Correct. \(2e^{x}\cos x - 2e^{x}\sin x - 2e^{x}\cos x + 2e^{x}\sin x = 0\).
(D) The value of \(\frac{d^{2}y}{dx^{2}}\) at \(x = 0\) is 0. — Incorrect. \(2e^{0}\cos 0 = 2\).