MCQ
If $y = \sqrt {(1 - x)(1 + x)} $, then
- A$(1 - {x^2}){{dy} \over {dx}} - xy = 0$
- ✓$(1 - {x^2}){{dy} \over {dx}} + xy = 0$
- C$(1 - {x^2}){{dy} \over {dx}} - 2xy = 0$
- D$(1 - {x^2}){{dy} \over {dx}} + 2xy = 0$
$\frac{{dy}}{{dx}} = \frac{{ - x}}{y} = \frac{{ - xy}}{{(1 - {x^2})}}$ or
$(1 - {x^2})\frac{{dy}}{{dx}} + xy = 0$.
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