If y=3 cos (log x)+4 sin (log x),then x^2 y_2 | vedclass

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(C) Given $y=3 \cos (\log x)+4 \sin (\log x)$.First,differentiate with respect to $x$:$y_1 = \frac{dy}{dx} = -3 \sin (\log x) \cdot \frac{1}{x} + 4 \c

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$-y$

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(C) Given $y=3 \cos (\log x)+4 \sin (\log x)$.First,differentiate with respect to $x$:$y_1 = \frac{dy}{dx} = -3 \sin (\log x) \cdot \frac{1}{x} + 4 \cos (\log x) \cdot \frac{1}{x}$.Multiply both sides by $x$:$x y_1 = -3 \sin (\log x) + 4 \cos (\log x)$.Differentiate again with respect to $x$ using the product rule on the left side:$1 \cdot y_1 + x \cdot y_2 = -3 \cos (\log x) \cdot \frac{1}{x} - 4 \sin (\log x) \cdot \frac{1}{x}$.Multiply the entire equation by $x$:$x y_1 + x^2 y_2 = -3 \cos (\log x) - 4 \sin (\log x)$.Factor out $-1$ from the right side:$x^2 y_2 + x y_1 = -(3 \cos (\log x) + 4 \sin (\log x))$.Since $y = 3 \cos (\log x) + 4 \sin (\log x)$,we have:$x^2 y_2 + x y_1 = -y$.

If $y=3 \cos (\log x)+4 \sin (\log x)$,then $x^2 y_2+x y_1=$

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