If dy dx + 1 1 - x^2 = 0, then - Vidyadip

MCQ

If $\frac{{dy}}{{dx}} + \frac{1}{{\sqrt {1 - {x^2}} }} = 0$, then

✓
$y + {\sin ^{ - 1}}x = c$
B
${y^2} + 2{\sin ^{ - 1}}x + c = 0$
C
$x + {\sin ^{ - 1}}y = 0$
D
${x^2} + 2{\sin ^{ - 1}}y = 1$

✓

Answer

Correct option: A.

$y + {\sin ^{ - 1}}x = c$

a
(a) $\frac{{dy}}{{dx}} = - \frac{1}{{\sqrt {1 - {x^2}} }}$==> $dy = - \frac{1}{{\sqrt {1 - {x^2}} }}dx$

On integrating, we get $y = {\cos ^{ - 1}}x + c$

==> $y = \frac{\pi }{2} - {\sin ^{ - 1}}x + c$ ==> $y + {\sin ^{ - 1}}x = c$.

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