The differential equation for which ^ - 1 x + ^ - 1 y = c is give… - Vidyadip

MCQ

The differential equation for which ${\sin ^{ - 1}}x + {\sin ^{ - 1}}y = c$ is given by

A
$\sqrt {1 - {x^2}} \,\,dx\,\, + \sqrt {1 - {y^2}} \,\,dy = 0$
✓
$\sqrt {1 - {x^2}} \,\,dy\,\, + \sqrt {1 - {y^2}} \,\,dx = 0$
C
$\sqrt {1 - {x^2}} \,\,dy\,\, - \sqrt {1 - {y^2}} \,\,dx = 0$
D
$\sqrt {1 - {x^2}} \,\,dx\, - \sqrt {1 - {y^2}} \,\,dy = 0$

✓

Answer

Correct option: B.

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

b
(b) Given equation is ${\sin ^{ - 1}}x + {\sin ^{ - 1}}y = c .....(i)$

On differentiating w.r.t. to $x$, we get

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

==>$\frac{{dy}}{{dx}} = - \frac{{\sqrt {1 - {y^2}} }}{{\sqrt {1 - {x^2}} }}$

==> $\sqrt {1 - {x^2}} \,\,dy + \sqrt {1 - {y^2}} \,\,dx = 0$.

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