JEE Main 2004Annual, Paper 1MathematicsMediumDifferential Equations163× since 2002Solution of the differential equation ydx+(x+x2y)dy=0ydx + \left( {x + {x^2}y} \right)dy = 0ydx+(x+x2y)dy=0 isAOption A. logloglog y=Cxy=Cxy=CxBOption B. −1xy+logy=C- {1 \over {xy}} + \log y = C−xy1+logy=CCOption C. 1xy+logy=C{1 \over {xy}} + \log y = Cxy1+logy=CDOption D. −1xy=C- {1 \over {xy}} = C−xy1=CCheckSkip · show solution
