If y=a cos 3 x+b e^{-x}, then y^{prime prime} | vedclass

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(B) Given $y = a \cos 3x + b e^{-x}$.First derivative: $y' = -3a \sin 3x - b e^{-x}$.Second derivative: $y'' = -9a \cos 3x + b e^{-x}$.Now, calculate

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$10 y^{\prime} \cos 3 x+3 y(\sin 3 x+3 \cos 3 x)$

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(B) Given $y = a \cos 3x + b e^{-x}$.First derivative: $y' = -3a \sin 3x - b e^{-x}$.Second derivative: $y'' = -9a \cos 3x + b e^{-x}$.Now, calculate $y''(3 \sin 3x - \cos 3x)$:$y''(3 \sin 3x - \cos 3x) = (-9a \cos 3x + b e^{-x})(3 \sin 3x - \cos 3x)$$= -27a \cos 3x \sin 3x + 9a \cos^2 3x + 3b e^{-x} \sin 3x - b e^{-x} \cos 3x$.We can rewrite this expression by manipulating the terms to match the form $10y' \cos 3x + 3y(\sin 3x + 3 \cos 3x)$:$= 10(-3a \sin 3x - b e^{-x}) \cos 3x + 3a \cos 3x(\cos 3x + 3 \sin 3x) + 3b e^{-x}(\sin 3x + 3 \cos 3x)$$= 10y' \cos 3x + 3(a \cos 3x + b e^{-x})(\sin 3x + 3 \cos 3x)$$= 10y' \cos 3x + 3y(\sin 3x + 3 \cos 3x)$.

If $y=a \cos 3 x+b e^{-x}$, then $y^{\prime \prime}(3 \sin 3 x-\cos 3 x)=$

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