JEE Main 202225 Jun, Evening shiftMathematicsEasyDifferential Equations163× since 2002If y=y(x)y = y(x)y=y(x) is the solution of the differential equation 2x2dydx−2xy+3y2=02{x^2}{{dy} \over {dx}} - 2xy + 3{y^2} = 02x2dxdy−2xy+3y2=0 such that y(e)=e3y(e) = {e \over 3}y(e)=3e, then y(1) is equal to :AOption A. 13{1 \over 3}31BOption B. 23{2 \over 3}32COption C. 32{3 \over 2}23DOption D. 3CheckSkip · show solution
