MCQ
જો $y = x + {1 \over x}$, તો
- A${x^2}{{dy} \over {dx}} + xy = 0$
- B${x^2}{{dy} \over {dx}} + xy + 2 = 0$
- ✓${x^2}{{dy} \over {dx}} - xy + 2 = 0$
- Dએકપણ નહીં
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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