The general solution of the differential equa | vedclass

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Question

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

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

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(C) Given the differential equation: $\frac{dy}{dx} + \frac{1}{\sqrt{1-x^2}} = 0$ Rearranging the terms,we get: $\frac{dy}{dx} = -\frac{1}{\sqrt{1-x^2}}$ Integrating both sides with respect to $x$: $\int dy = -\int \frac{1}{\sqrt{1-x^2}} dx$ We know that $\int \frac{1}{\sqrt{1-x^2}} dx = \sin^{-1} x + c$ Therefore,the solution is: $y = -\sin^{-1} x + c$ This can be rewritten as: $y + \sin^{-1} x = c$

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

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