If y=2 cos (2 log x)+3 sin (2 log x), then x^ | vedclass

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(A) Given $y = 2 \cos (2 \log x) + 3 \sin (2 \log x)$.First, differentiate with respect to $x$:$y^{\prime} = -2 \sin (2 \log x) \cdot \frac{2}{x} + 3

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$-2 y$

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(A) Given $y = 2 \cos (2 \log x) + 3 \sin (2 \log x)$.First, differentiate with respect to $x$:$y^{\prime} = -2 \sin (2 \log x) \cdot \frac{2}{x} + 3 \cos (2 \log x) \cdot \frac{2}{x}$$x y^{\prime} = -4 \sin (2 \log x) + 6 \cos (2 \log x)$Now, differentiate again with respect to $x$:$x y^{\prime \prime} + y^{\prime} = -4 \cos (2 \log x) \cdot \frac{2}{x} - 6 \sin (2 \log x) \cdot \frac{2}{x}$Multiply by $x$:$x^2 y^{\prime \prime} + x y^{\prime} = -8 \cos (2 \log x) - 12 \sin (2 \log x)$$x^2 y^{\prime \prime} + x y^{\prime} = -4 [2 \cos (2 \log x) + 3 \sin (2 \log x)]$Since $y = 2 \cos (2 \log x) + 3 \sin (2 \log x)$, we have:$x^2 y^{\prime \prime} + x y^{\prime} = -4 y$Therefore, $x^2 y^{\prime \prime} + x y^{\prime} + 2 y = -4 y + 2 y = -2 y$.

If $y=2 \cos (2 \log x)+3 \sin (2 \log x)$, then $x^2 y^{\prime \prime}+x y^{\prime}+2 y=$

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