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(A) The given differential equation is: $\frac{dy}{dx} = \sin^{-1} x$ Integrating both sides with respect to $x$: $y = \int \sin^{-1} x \, dx$ Using i

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

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(A) The given differential equation is: $\frac{dy}{dx} = \sin^{-1} x$ Integrating both sides with respect to $x$: $y = \int \sin^{-1} x \, dx$ Using integration by parts,let $u = \sin^{-1} x$ and $dv = dx$. Then $du = \frac{1}{\sqrt{1-x^2}} dx$ and $v = x$. $y = x \sin^{-1} x - \int x \cdot \frac{1}{\sqrt{1-x^2}} dx$ Let $t = 1 - x^2$,then $dt = -2x \, dx$,which implies $x \, dx = -\frac{1}{2} dt$. $y = x \sin^{-1} x - \int \frac{-1/2}{\sqrt{t}} dt$ $y = x \sin^{-1} x + \frac{1}{2} \int t^{-1/2} dt$ $y = x \sin^{-1} x + \frac{1}{2} \cdot \frac{t^{1/2}}{1/2} + C$ $y = x \sin^{-1} x + \sqrt{t} + C$ $y = x \sin^{-1} x + \sqrt{1 - x^2} + C$

Find the general solution of the differential equation: $\frac{dy}{dx} = \sin^{-1} x$

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