If y = x + frac{1}{x},then | vedclass

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(C) Given $y = x + \frac{1}{x}$.Differentiating both sides with respect to $x$,we get:$\frac{dy}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(x^{-1})$$\frac{d

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$x^2 \frac{dy}{dx} - xy + 2 = 0$

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(C) Given $y = x + \frac{1}{x}$.Differentiating both sides with respect to $x$,we get:$\frac{dy}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(x^{-1})$$\frac{dy}{dx} = 1 - x^{-2} = 1 - \frac{1}{x^2}$.Now,consider the expression $x^2 \frac{dy}{dx} - xy + 2$:Substitute $\frac{dy}{dx} = 1 - \frac{1}{x^2}$ and $y = x + \frac{1}{x}$ into the expression:$= x^2 \left( 1 - \frac{1}{x^2} \right) - x \left( x + \frac{1}{x} \right) + 2$$= (x^2 - 1) - (x^2 + 1) + 2$$= x^2 - 1 - x^2 - 1 + 2$$= 0$.Thus,$x^2 \frac{dy}{dx} - xy + 2 = 0$.

If $y = x + \frac{1}{x}$,then

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