JEE Main 202228 Jun, Morning shiftMathematicsMediumDifferential Equations163× since 2002Let y = y(x) be the solution of the differential equation x(1−x2)dydx+(3x2y−y−4x3)=0x(1 - {x^2}){{dy} \over {dx}} + (3{x^2}y - y - 4{x^3}) = 0x(1−x2)dxdy+(3x2y−y−4x3)=0, x>1x > 1x>1, with y(2)=−2y(2) = - 2y(2)=−2. Then y(3) is equal to :AOption A. −-−18BOption B. −-−12COption C. −-−6DOption D. −-−3CheckSkip · show solution
