If y=e^{-x} cos 2x, then which of the followi | vedclass

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(A) Given, $y = e^{-x} \cos 2x$. Taking the first derivative with respect to $x$ using the product rule: $\frac{dy}{dx} = e^{-x}(-2 \sin 2x) + \cos 2x

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$\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+5 y=0$

Vedclass · Answer

(A) Given, $y = e^{-x} \cos 2x$. Taking the first derivative with respect to $x$ using the product rule: $\frac{dy}{dx} = e^{-x}(-2 \sin 2x) + \cos 2x(-e^{-x}) = -2e^{-x} \sin 2x - y$. Rearranging gives: $\frac{dy}{dx} + y = -2e^{-x} \sin 2x$. Taking the derivative again with respect to $x$: $\frac{d^2y}{dx^2} + \frac{dy}{dx} = -2[e^{-x}(2 \cos 2x) + \sin 2x(-e^{-x})] = -4(e^{-x} \cos 2x) + 2(e^{-x} \sin 2x)$. Substituting $y = e^{-x} \cos 2x$ and $-2e^{-x} \sin 2x = \frac{dy}{dx} + y$: $\frac{d^2y}{dx^2} + \frac{dy}{dx} = -4y - (\frac{dy}{dx} + y)$. $\frac{d^2y}{dx^2} + \frac{dy}{dx} = -4y - \frac{dy}{dx} - y$. $\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 5y = 0$.

If $y=e^{-x} \cos 2x$, then which of the following differential equations is satisfied?

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