The differential equation for which sin^{-1} | vedclass

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(B) Given equation is $\sin^{-1} x + \sin^{-1} y = c$ ... $(i)$Differentiating both sides with respect to $x$,we get:$\frac{d}{dx}(\sin^{-1} x) + \fra

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

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(B) Given equation is $\sin^{-1} x + \sin^{-1} y = c$ ... $(i)$Differentiating both sides with respect to $x$,we get:$\frac{d}{dx}(\sin^{-1} x) + \frac{d}{dx}(\sin^{-1} y) = \frac{d}{dx}(c)$$\frac{1}{\sqrt{1 - x^2}} + \frac{1}{\sqrt{1 - y^2}} \cdot \frac{dy}{dx} = 0$Multiplying by $\sqrt{1 - x^2} \sqrt{1 - y^2}$,we get:$\sqrt{1 - y^2} + \sqrt{1 - x^2} \frac{dy}{dx} = 0$Rearranging the terms,we get:$\sqrt{1 - x^2} \, dy + \sqrt{1 - y^2} \, dx = 0$.

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

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