y = a cos x + (b + 2x) sin x Rightarrow y^{pr | vedclass

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(D) Given $y = a \cos x + (b + 2x) \sin x$. First,find the first derivative $y^{\prime}$ using the product rule: $y^{\prime} = -a \sin x + (b + 2x) \c

Vedclass · Accepted Answer

$4 \cos x$

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(D) Given $y = a \cos x + (b + 2x) \sin x$. First,find the first derivative $y^{\prime}$ using the product rule: $y^{\prime} = -a \sin x + (b + 2x) \cos x + 2 \sin x$. Now,find the second derivative $y^{\prime \prime}$: $y^{\prime \prime} = -a \cos x - (b + 2x) \sin x + 2 \cos x + 2 \cos x$. $y^{\prime \prime} = -(a \cos x + (b + 2x) \sin x) + 4 \cos x$. Since $y = a \cos x + (b + 2x) \sin x$,we substitute $y$ into the expression: $y^{\prime \prime} = -y + 4 \cos x$. Therefore,$y^{\prime \prime} + y = 4 \cos x$.

$y = a \cos x + (b + 2x) \sin x \Rightarrow y^{\prime \prime} + y = $

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