JEE Main 202329 Jan, Evening shiftMathematicsMediumDifferential Equations163× since 2002Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation xlogexdydx+y=x2logex,(x>1)x{\log _e}x{{dy} \over {dx}} + y = {x^2}{\log _e}x,(x > 1)xlogexdxdy+y=x2logex,(x>1). If y(2)=2y(2) = 2y(2)=2, then y(e)y(e)y(e) is equal toAOption A. 1+e22{{1 + {e^2}} \over 2}21+e2BOption B. 1+e24{{1 + {e^2}} \over 4}41+e2COption C. 2+e22{{2 + {e^2}} \over 2}22+e2DOption D. 4+e24{{4 + {e^2}} \over 4}44+e2CheckSkip · show solution
