JEE Main 202325 Jan, Morning shiftMathematicsMediumDifferential Equations163× since 2002Let y=y(x)y = y(x)y=y(x) be the solution curve of the differential equation dydx=yx(1+xy2(1+logex)),x>0,y(1)=3{{dy} \over {dx}} = {y \over x}\left( {1 + x{y^2}(1 + {{\log }_e}x)} \right),x > 0,y(1) = 3dxdy=xy(1+xy2(1+logex)),x>0,y(1)=3. Then y2(x)9{{{y^2}(x)} \over 9}9y2(x) is equal to :AOption A. x25−2x3(2+logex3){{{x^2}} \over {5 - 2{x^3}(2 + {{\log }_e}{x^3})}}5−2x3(2+logex3)x2BOption B. x23x3(1+logex2)−2{{{x^2}} \over {3{x^3}(1 + {{\log }_e}{x^2}) - 2}}3x3(1+logex2)−2x2COption C. x27−3x3(2+logex2){{{x^2}} \over {7 - 3{x^3}(2 + {{\log }_e}{x^2})}}7−3x3(2+logex2)x2DOption D. x22x3(2+logex3)−3{{{x^2}} \over {2{x^3}(2 + {{\log }_e}{x^3}) - 3}}2x3(2+logex3)−3x2CheckSkip · show solution
