Rearrange the given equation to:\[\sqrt{1 - y^2} \, dx = (\sin^{-1} y - x) \, dy.\]
Divide both sides by \( \sqrt{1 - y^2} \):\[dx = \frac{\sin^{-1} y - x}{\sqrt{1 - y^2}} \, dy.\]
Define \( P(y) = \frac{1}{\sqrt{1 - y^2}} \) and \( Q(y) = \frac{\sin^{-1} y}{\sqrt{1 - y^2}} \).Step 1: Compute the integrating factor (IF)
The integrating factor is:\[\text{IF} = e^{\int P(y) \, dy} = e^{\sin^{-1} y}.\]Step 2: Solve the differential equation
Multiply by the integrating factor \( \text{IF} \):\[e^{\sin^{-1} y} \, dx + e^{\sin^{-1} y} \frac{x}{\sqrt{1 - y^2}} \, dy = e^{\sin^{-1} y} \frac{\sin^{-1} y}{\sqrt{1 - y^2}} \, dy.\]
This simplifies to:\[\frac{d}{dy} \left( x e^{\sin^{-1} y} \right) = e^{\sin^{-1} y} \frac{\sin^{-1} y}{\sqrt{1 - y^2}}.\]
Integrate both sides:\[x e^{\sin^{-1} y} = \int e^{\sin^{-1} y} \frac{\sin^{-1} y}{\sqrt{1 - y^2}} \, dy + c.\]
The simplified solution is:\[x = \sin^{-1} y - 1 + c e^{-\sin^{-1} y}.\] Final Answer:\[\boxed{x = \sin^{-1} y - 1 + c e^{-\sin^{-1} y}}.\]