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Question: If \(\frac { d y } { d x } + \frac { 1 } { \sqrt { 1 - x ^ { 2 } } } = 0\), then...

If dydx+11−x2=0\frac { d y } { d x } + \frac { 1 } { \sqrt { 1 - x ^ { 2 } } } = 0, then

A

y+sin⁡−1x=cy + \sin ^ { - 1 } x = c

B

y2+2sin⁡−1x+c=0y ^ { 2 } + 2 \sin ^ { - 1 } x + c = 0

C

x+sin⁡−1y=0x + \sin ^ { - 1 } y = 0

D

x2+2sin⁡−1y=1x ^ { 2 } + 2 \sin ^ { - 1 } y = 1

Answer

y+sin⁡−1x=cy + \sin ^ { - 1 } x = c

Explanation

Solution

dydx=−11−x2\frac { d y } { d x } = - \frac { 1 } { \sqrt { 1 - x ^ { 2 } } }⇒ dy=−11−x2dxd y = - \frac { 1 } { \sqrt { 1 - x ^ { 2 } } } d x

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

⇒ y=π2−sin⁡−1x+cy = \frac { \pi } { 2 } - \sin ^ { - 1 } x + c ⇒ y+sin⁡−1x=cy + \sin ^ { - 1 } x = c.