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

Correct Answer is : x = sin−1y−1+ce−sin−1y

x = \sin^{-1}y-1+ce^{-\sin^{-1}y}

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

x = 1 + \sin^{-1} y + ce^{\sin^{-1}y}

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

Correct Answer is : x = sin−1y−1+ce−sin−1y

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TS EAMCET 03 May 2019 Shift 1 Solved Paper

Section: Mathematics

Question : 80 of 160

Marks: +1, -0

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