The solution of dy dx + , ( 1 - y^2 1 - x^2 ) , = ,0 is - Vidyadip

MCQ

The solution of $\frac{{dy}}{{dx}} + \sqrt {\,\left( {\frac{{1 - {y^2}}}{{1 - {x^2}}}} \right)} \, = \,0$ is

A
${\tan ^{ - 1}}x + {\cot ^{ - 1}}x = c$
✓
${\sin ^{ - 1}}x + {\sin ^{ - 1}}y = c$
C
${\sec ^{ - 1}}x + {\rm{cose}}{{\rm{c}}^{ - 1}}x = c$
D
None of these

✓

Answer

Correct option: B.

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

b
(b) Given equation is $\int {\frac{{dy}}{{\sqrt {1 - {y^2}} }} + \int {\frac{{dx}}{{\sqrt {1 - {x^2}} }} = 0} } $

Integrating we get, ${\sin ^{ - 1}}y + {\sin ^{ - 1}}x = c$.

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