The solution of frac{dy}{dx} = sqrt{1-y^2} wi | vedclass

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(B) Given the differential equation $\frac{dy}{dx} = \sqrt{1-y^2}$.Separating the variables,we get $\frac{dy}{\sqrt{1-y^2}} = dx$.Integrating both sid

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

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(B) Given the differential equation $\frac{dy}{dx} = \sqrt{1-y^2}$.Separating the variables,we get $\frac{dy}{\sqrt{1-y^2}} = dx$.Integrating both sides,we have $\int \frac{dy}{\sqrt{1-y^2}} = \int dx$.This results in $\sin^{-1}(y) = x + C$.Using the initial condition $y(0) = 1$,we substitute $x = 0$ and $y = 1$ into the equation:$\sin^{-1}(1) = 0 + C$,which gives $C = \sin^{-1}(1)$.Substituting the value of $C$ back into the general solution,we get $\sin^{-1}(y) = x + \sin^{-1}(1)$.

The solution of $\frac{dy}{dx} = \sqrt{1-y^2}$ with the initial condition $y(0) = 1$ is:

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