JEE Main 202120 Jul, Morning shiftMathematicsMediumDifferential Equations163× since 2002Let y = y(x) be the solution of the differential equation ex1−y2dx+(yx)dy=0{e^x}\sqrt {1 - {y^2}} dx + \left( {{y \over x}} \right)dy = 0ex1−y2dx+(xy)dy=0, y(1) = −-−1. Then the value of (y(3))² is equal to :AOption A. 1 −-− 4e³BOption B. 1 −-− 4e⁶COption C. 1 + 4e³DOption D. 1 + 4e⁶CheckSkip · show solution
