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Question

(A) Given differential equation is $\frac{dy}{dx} + \sqrt{\frac{1-y^2}{1-x^2}} = 0$. Rearranging the terms,we get $\frac{dy}{dx} = -\frac{\sqrt{1-y^2}

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$\sin^{-1} x + \sin^{-1} y = C$

Vedclass · Answer

(A) Given differential equation is $\frac{dy}{dx} + \sqrt{\frac{1-y^2}{1-x^2}} = 0$. Rearranging the terms,we get $\frac{dy}{dx} = -\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$. Separating the variables,we have $\frac{dy}{\sqrt{1-y^2}} = -\frac{dx}{\sqrt{1-x^2}}$. Integrating both sides,we get $\int \frac{dy}{\sqrt{1-y^2}} = -\int \frac{dx}{\sqrt{1-x^2}}$. This yields $\sin^{-1} y = -\sin^{-1} x + C$. Therefore,the general solution is $\sin^{-1} x + \sin^{-1} y = C$.

Find the general solution of the differential equation $\frac{dy}{dx} + \sqrt{\frac{1-y^2}{1-x^2}} = 0$.

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