JEE Main 2004Annual, Paper 1MathematicsEasyDifferential Equations163× since 2002The differential equation for the family of circle x2+y2−2ay=0,{x^2} + {y^2} - 2ay = 0,x2+y2−2ay=0, where a is an arbitrary constant is :AOption A. (x2+y2)y′=2xy\left( {{x^2} + {y^2}} \right)y' = 2xy(x2+y2)y′=2xyBOption B. 2(x2+y2)y′=xy2\left( {{x^2} + {y^2}} \right)y' = xy2(x2+y2)y′=xyCOption C. (x2−y2)y′=2xy\left( {{x^2} - {y^2}} \right)y' =2 xy(x2−y2)y′=2xyDOption D. 2(x2−y2)y′=xy2\left( {{x^2} - {y^2}} \right)y' = xy2(x2−y2)y′=xyCheckSkip · show solution
