The differential equation having y = (sin–1x)2 + Acos–1x + B where A and B are arbitary constants is

y=sin−1⁡x2+Acos−1⁡x+B y'=2sin-1x1-x2-A1-x2 1-x2 y'=2sin-1x-A 1-x2 y''-x1-x2y'=21-x2 1-x2 y''-xy'=2

The differential equation having y = (sin–1x)2 + Acos–1x + B where A and B are arbitary constants is

y=sin−1⁡x2+Acos−1⁡x+B y'=2sin-1x1-x2-A1-x2 1-x2 y'=2sin-1x-A 1-x2 y''-x1-x2y'=21-x2 1-x2 y''-xy'=2

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The differential equation having y = (sin^–1x)² + Acos^–1x + B where A and B are arbitary constants is

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$(1 - x^{2}) \frac{d^{2} y}{{dx}^{2}} - x \frac{dy}{dx} = 0$

$(1 + x^{2}) \frac{d^{2} y}{{dx}^{2}} + 3 y = 0$

$(1 - x^{2}) \frac{d^{2} y}{{dx}^{2}} - x \frac{dy}{dx} = 2$

$(1 + x^{2}) \frac{dy}{dx} + x \frac{d^{2} y}{{dx}^{2}} = 0$

answer is B.

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Detailed Solution

$y = {(\sin^{- 1} x)}^{2} + {Acos}^{- 1} x + B y' = \frac{2 \sin^{- 1} x}{\sqrt{1 - x^{2}}} - \frac{A}{\sqrt{1 - x^{2}}} \sqrt{1 - x^{2}} y' = 2 \sin^{- 1} x - A \sqrt{1 - x^{2}} y'' - \frac{x}{\sqrt{1 - x^{2}}} y' = \frac{2}{\sqrt{1 - x^{2}}} (1 - x^{2}) y'' - xy' = 2$