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

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(C) Given $y = a \cos (\ln x) + b \sin (\ln x)$.First,differentiate with respect to $x$:$\frac{dy}{dx} = -a \sin (\ln x) \cdot \frac{1}{x} + b \cos (\

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$-y$

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(C) Given $y = a \cos (\ln x) + b \sin (\ln x)$.First,differentiate with respect to $x$:$\frac{dy}{dx} = -a \sin (\ln x) \cdot \frac{1}{x} + b \cos (\ln x) \cdot \frac{1}{x}$Multiply by $x$:$x \frac{dy}{dx} = -a \sin (\ln x) + b \cos (\ln x)$Now,differentiate again with respect to $x$ using the product rule on the left side:$\frac{d}{dx} (x \frac{dy}{dx}) = \frac{d}{dx} (-a \sin (\ln x) + b \cos (\ln x))$$x \frac{d^2y}{dx^2} + \frac{dy}{dx} = -a \cos (\ln x) \cdot \frac{1}{x} - b \sin (\ln x) \cdot \frac{1}{x}$Multiply the entire equation by $x$:$x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} = -a \cos (\ln x) - b \sin (\ln x)$$x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} = -(a \cos (\ln x) + b \sin (\ln x))$Since $y = a \cos (\ln x) + b \sin (\ln x)$,we get:$x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} = -y$

If $y = a \cos (\ln x) + b \sin (\ln x)$,then $x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx}$ is equal to

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