y = e^x (A cos x + B sin x) is the solution o | vedclass

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(C) Given the solution $y = e^x (A \cos x + B \sin x)$.First,differentiate with respect to $x$ using the product rule:$\frac{dy}{dx} = e^x (A \cos x +

Vedclass · Accepted Answer

$\frac{d^2 y}{dx^2} - 2 \frac{dy}{dx} + 2y = 0$

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(C) Given the solution $y = e^x (A \cos x + B \sin x)$.First,differentiate with respect to $x$ using the product rule:$\frac{dy}{dx} = e^x (A \cos x + B \sin x) + e^x (-A \sin x + B \cos x)$$\frac{dy}{dx} = e^x ((A + B) \cos x + (B - A) \sin x)$Now,differentiate again with respect to $x$:$\frac{d^2 y}{dx^2} = e^x ((A + B) \cos x + (B - A) \sin x) + e^x (-(A + B) \sin x + (B - A) \cos x)$$\frac{d^2 y}{dx^2} = e^x ((A + B + B - A) \cos x + (B - A - A - B) \sin x)$$\frac{d^2 y}{dx^2} = e^x (2B \cos x - 2A \sin x)$Now,substitute $\frac{dy}{dx}$ and $y$ into the equation $\frac{d^2 y}{dx^2} - 2 \frac{dy}{dx} + 2y = 0$:$e^x (2B \cos x - 2A \sin x) - 2e^x ((A + B) \cos x + (B - A) \sin x) + 2e^x (A \cos x + B \sin x)$$= e^x [ (2B - 2A - 2B + 2A) \cos x + (-2A - 2B + 2A + 2B) \sin x ]$$= e^x [ 0 \cos x + 0 \sin x ] = 0$.Thus,the correct option is $C$.

$y = e^x (A \cos x + B \sin x)$ is the solution of the differential equation

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