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

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(D) Given $y = a \cos(\log x) + b \sin(\log x)$.Differentiating with respect to $x$:$y_1 = \frac{dy}{dx} = -a \sin(\log x) \cdot \frac{1}{x} + b \cos(

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$-y$

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(D) Given $y = a \cos(\log x) + b \sin(\log x)$.Differentiating with respect to $x$:$y_1 = \frac{dy}{dx} = -a \sin(\log x) \cdot \frac{1}{x} + b \cos(\log x) \cdot \frac{1}{x}$$x y_1 = -a \sin(\log x) + b \cos(\log x)$.Differentiating again with respect to $x$:$x y_2 + y_1 = -a \cos(\log x) \cdot \frac{1}{x} - b \sin(\log x) \cdot \frac{1}{x}$Multiply by $x$:$x^2 y_2 + x y_1 = -[a \cos(\log x) + b \sin(\log x)]$Since $y = a \cos(\log x) + b \sin(\log x)$,we have:$x^2 y_2 + x y_1 = -y$.

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

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