MCQ
If $y = x + {1 \over x}$, then
  • A
    ${x^2}{{dy} \over {dx}} + xy = 0$
  • B
    ${x^2}{{dy} \over {dx}} + xy + 2 = 0$
  • ${x^2}{{dy} \over {dx}} - xy + 2 = 0$
  • D
    None of these

Answer

Correct option: C.
${x^2}{{dy} \over {dx}} - xy + 2 = 0$
c
(c) $y = x + \frac{1}{x}$==> $\frac{{dy}}{{dx}} = 1 - \frac{1}{{{x^2}}}$

Therefore, ${{x}^{2}}.\frac{dy}{dx}-xy+2$

$={{x}^{2}} \left( {1 - \frac{1}{{{x^2}}}} \right) - x(x + \frac{1}{x}) + 2 = 0$

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