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
    $(1 - {x^2}){{dy} \over {dx}} + 2xy = 0$

Answer

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

$\frac{{dy}}{{dx}} = \frac{{ - x}}{y} = \frac{{ - xy}}{{(1 - {x^2})}}$ or 

$(1 - {x^2})\frac{{dy}}{{dx}} + xy = 0$.

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