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

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Question

(c) $y = x + \frac{1}{x}$==> $\frac{{dy}}{{dx}} = 1 - \frac{1}{{{x^2}}}$
Therefore,${x^2}.\frac{{dy}}{{dx\left( {1 - \frac{1}{{}}} \right)}} - xy

Vedclass · Accepted Answer

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

Vedclass · Answer

(c) $y = x + \frac{1}{x}$==> $\frac{{dy}}{{dx}} = 1 - \frac{1}{{{x^2}}}$
Therefore,${x^2}.\frac{{dy}}{{dx\left( {1 - \frac{1}{{}}} \right)}} - xy + 2 = {x^2}$

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

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