The solution of the equation frac{dy}{dx} + s | vedclass

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(B) Given the differential equation: $\frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1 - x^2}} = 0$Separating the variables,we get: $\frac{dy}{\sqrt{1 - y^2}} =

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

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(B) Given the differential equation: $\frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1 - x^2}} = 0$Separating the variables,we get: $\frac{dy}{\sqrt{1 - y^2}} = -\frac{dx}{\sqrt{1 - x^2}}$Integrating both sides: $\int \frac{dy}{\sqrt{1 - y^2}} = -\int \frac{dx}{\sqrt{1 - x^2}}$This yields: $\sin^{-1}(y) = -\sin^{-1}(x) + C$Let the constant $C = \sin^{-1}(c)$,so: $\sin^{-1}(y) + \sin^{-1}(x) = \sin^{-1}(c)$Using the identity $\sin^{-1}(x) + \sin^{-1}(y) = \sin^{-1}(x\sqrt{1 - y^2} + y\sqrt{1 - x^2})$,we get:$\sin^{-1}(x\sqrt{1 - y^2} + y\sqrt{1 - x^2}) = \sin^{-1}(c)$Taking the sine of both sides: $x\sqrt{1 - y^2} + y\sqrt{1 - x^2} = c$.

The solution of the equation $\frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1 - x^2}} = 0$ is

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