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(C) Given,$y = e^x(A \cos x + B \sin x)$ ...$(i)$Differentiating with respect to $x$ using the product rule:$y^{\prime} = e^x(A \cos x + B \sin x) + e

Vedclass · Accepted Answer

$y^{\prime \prime}-2 y^{\prime}+2 y=0$

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(C) Given,$y = e^x(A \cos x + B \sin x)$ ...$(i)$Differentiating with respect to $x$ using the product rule:$y^{\prime} = e^x(A \cos x + B \sin x) + e^x(-A \sin x + B \cos x)$$y^{\prime} = y + e^x(-A \sin x + B \cos x)$$y^{\prime} - y = e^x(-A \sin x + B \cos x)$ ...(ii)Differentiating again with respect to $x$:$y^{\prime \prime} - y^{\prime} = e^x(-A \sin x + B \cos x) + e^x(-A \cos x - B \sin x)$Substitute $e^x(-A \sin x + B \cos x) = y^{\prime} - y$ from (ii) and $e^x(-A \cos x - B \sin x) = -y$ from $(i)$:$y^{\prime \prime} - y^{\prime} = (y^{\prime} - y) - y$$y^{\prime \prime} - y^{\prime} = y^{\prime} - 2y$$y^{\prime \prime} - 2y^{\prime} + 2y = 0$Thus,the correct option is $C$.

The differential equation corresponding to the family of curves $y=e^x(A \cos x+B \sin x)$ is

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