MathematicsEasy38×since 2002Q3508Let y(x)y(x)y(x) be the solution of the differential equation (x logx)dydx+y=2x logx,(x≥1).\left( {x\,\log x} \right){{dy} \over {dx}} + y = 2x\,\log x,\left( {x \ge 1} \right).(xlogx)dxdy+y=2xlogx,(x≥1). Then y(e)y(e)y(e) is equal to :A222B2e2e2eCeeeD000Check answerSkip