MathematicsMedium38×since 2002Q3555Let 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 toA1+e22{{1 + {e^2}} \over 2}21+e2B1+e24{{1 + {e^2}} \over 4}41+e2C2+e22{{2 + {e^2}} \over 2}22+e2D4+e24{{4 + {e^2}} \over 4}44+e2Check answerSkip