`y(1+logx)(dx)/(dy)-xlogx=0, y=e^(2)" if "x=w`

x(dy)/(dx)+y=xlogx

The particular solution of the differential equation y (1+logx)(dx)/(dy)-xlog x =0 ," when" x = e, y = e^(2) is

x(dy)/(dx)+2y-x^(2)logx=0

Solve: xlogx(dy)/(dx)+y=2logx

Solve: xlogx(dy)/(dx)+y=2logx

Solve: (xlogx)(dy)/(dx)+y=2/xlogx

Let y(x) be the solution of the differential equation (xlogx)(dy)/(dx)+y=2xlogx, (x ge1) Then y(e) is equal to-