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If $y = x + {1 \over x}$, then
A${x^2}{{dy} \over {dx}} + xy = 0$
B${x^2}{{dy} \over {dx}} + xy + 2 = 0$
C${x^2}{{dy} \over {dx}} - xy + 2 = 0$
DNone of these
Solution
(c) $y = x + \frac{1}{x}$==> $\frac{{dy}}{{dx}} = 1 – \frac{1}{{{x^2}}}$
Therefore,${x^2}.\frac{{dy}}{{dx\left( {1 – \frac{1}{{}}} \right)}} – xy + 2 = {x^2}$
Therefore,${x^2}.\frac{{dy}}{{dx\left( {1 – \frac{1}{{}}} \right)}} – xy + 2 = {x^2}$
Standard 11
Physics
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