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

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(B) Given $y = a \cos(\log x) + b \sin(\log x)$.Differentiating with respect to $x$ using the chain rule:$y' = -a \sin(\log x) \cdot \frac{1}{x} + b \

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$-y$

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(B) Given $y = a \cos(\log x) + b \sin(\log x)$.Differentiating with respect to $x$ using the chain rule:$y' = -a \sin(\log x) \cdot \frac{1}{x} + b \cos(\log x) \cdot \frac{1}{x}$Multiplying both sides by $x$:$xy' = -a \sin(\log x) + b \cos(\log x)$Differentiating again with respect to $x$ using the product rule on the left side:$x y'' + y' = -a \cos(\log x) \cdot \frac{1}{x} - b \sin(\log x) \cdot \frac{1}{x}$Multiplying both sides by $x$ again:$x^2 y'' + xy' = -a \cos(\log x) - b \sin(\log x)$Factoring out the negative sign:$x^2 y'' + xy' = -(a \cos(\log x) + b \sin(\log x))$Since $y = a \cos(\log x) + b \sin(\log x)$,we substitute $y$ back into the equation:$x^2 y'' + xy' = -y$.

If $y = a \cos(\log x) + b \sin(\log x)$ where $a, b$ are parameters,then ${x^2}y'' + xy' = $

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