The solution of the differential equation 1−y2dx+xdy−sin−1ydy=0, is

dxdy+x1−y2=sin−1y1−y2 IF=∫edy1−y2=esin−1ySolution of given diff eq isxesin−1y=∫esin−1y sin−1y1−y2.dyxesin−1y=esin−1y(sin−1y−1)+cx=sin−1y−1+ce−sin−1y

The solution of the differential equation 1−y2dx+xdy−sin−1ydy=0, is

dxdy+x1−y2=sin−1y1−y2 IF=∫edy1−y2=esin−1ySolution of given diff eq isxesin−1y=∫esin−1y sin−1y1−y2.dyxesin−1y=esin−1y(sin−1y−1)+cx=sin−1y−1+ce−sin−1y

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The solution of the differential equation $\sqrt{1 - y^{2}} d x + x d y - \sin^{- 1} y d y = 0,$ is

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

$y = x \sqrt{1 - y^{2}} + \sin^{- 1} y + c$

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

$y = \sin^{- 1} y - 1 + x \sqrt{1 - y^{2}} + c$

answer is A.

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Detailed Solution

$\frac{d x}{d y} + \frac{x}{\sqrt{1 - y^{2}}} = \frac{\sin^{- 1} y}{\sqrt{1 - y^{2}}} I F = \int_{e} \frac{d y}{\sqrt{1 - y^{2}}} = e^{\sin^{- 1} y}$
Solution of given diff eq is
$\begin{array}{l} x e^{\sin^{- 1}} y = \int e^{\sin^{- 1} y} \frac{\sin^{- 1} y}{\sqrt{1 - y^{2}}} . d y \\ x e^{\sin^{- 1}} y = e^{\sin^{- 1}} y (\sin^{- 1} y - 1) + c \\ x = \sin^{- 1} y - 1 + c {e^{-}}^{\sin^{- 1} y} \end{array}$