MathematicsMedium38×since 2002Q3610Let 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 :A1 −-− 4e³B1 −-− 4e⁶C1 + 4e³D1 + 4e⁶Check answerSkip