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Q)

Solve: $\large\frac{dy}{dx}=\frac{-\sqrt {1-y^2}}{x-e^{\Large\sin ^{-1}y}}$

$(a)\;e^{\sin^{-1}}y=x+c \\ (b)\;2xe^{\sin^{-1}}y=e^{\sin ^{-1}y}+c \\ (c)\;2e^{\sin^{-1}}y=e^{3\sin ^{-1}y}+c \\ (d)\;e^{\sin^{-1}}=e^{3\sin ^{-1}y}+c$

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A)
$\sqrt {1-y^2}+(x- e^{\sin ^{-1}y}) \large\frac{dy}{dx}=0$
$\large\frac{dx}{dy}+\frac{x}{\sqrt {1-y^2}}=\frac{e^{\sin ^{-1}y}}{\sqrt {1-y^2}}$
$I.f= e^{\int \large\frac{dy}{\sqrt {1-y^2}}}$
$\qquad= e^{\sin ^{-1}}y$
$x= e^{\sin ^{-1}y} =\int \large\frac{ e^{\sin ^{-1}y}. e^{\sin ^{-1}y}}{\sqrt {1-y^2}}$$dy$
$e^{\sin ^{-1}y} =t$
$x. e^{\sin ^{-1}y} =\int t.dt$
$xe^{\sin ^{-1}y}=\large\frac{t^2}{2}+\frac{c}{2}$
$2xe^{\sin^{-1}}y=e^{\sin ^{-1}y}+c$
Hence b is the correct answer.
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