If y=a cos (log x)+b sin (log x), where a, b | vedclass

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(B) Given $y=a \cos (\log x)+b \sin (\log x)$.First, differentiate with respect to $x$:$y^{\prime} = \frac{d}{dx} [a \cos (\log x)+b \sin (\log x)] =

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$-y$

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(B) Given $y=a \cos (\log x)+b \sin (\log x)$.First, differentiate with respect to $x$:$y^{\prime} = \frac{d}{dx} [a \cos (\log x)+b \sin (\log x)] = -a \sin (\log x) \cdot \frac{1}{x} + b \cos (\log x) \cdot \frac{1}{x} = \frac{-a \sin (\log x) + b \cos (\log x)}{x}$.So, $x y^{\prime} = -a \sin (\log x) + b \cos (\log x)$.Now, differentiate again with respect to $x$:$\frac{d}{dx} (x y^{\prime}) = \frac{d}{dx} [-a \sin (\log x) + b \cos (\log x)]$.Using the product rule on the left side: $x y^{\prime \prime} + y^{\prime} = -a \cos (\log x) \cdot \frac{1}{x} - b \sin (\log x) \cdot \frac{1}{x}$.Multiply both sides by $x$:$x^2 y^{\prime \prime} + x y^{\prime} = -[a \cos (\log x) + b \sin (\log x)]$.Since $y = a \cos (\log x) + b \sin (\log x)$, we have:$x^2 y^{\prime \prime} + x y^{\prime} = -y$.

If $y=a \cos (\log x)+b \sin (\log x)$, where $a, b$ are parameters, then $x^2 y^{\prime \prime}+x y^{\prime}$ is equal to

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