Solution of the differential equationdxdy−xlogx1+logx=ey1+logx′ if y(1)=0, is

[a] (1+logx)dxdy−xlogx=ey putting x log xlogx=t⇒(1+logx)dx=dt ∴dtdy−t=ey Now, I.F. =e∫−1dy=e−y ⇒te−y=∫e−yeydy+C ⇒t=Cey+yey⇒xlogx=(C+y)ey, Since, y(1)=0, Then 0=(C+0)1⇒C=0 ∴yey=xlogx⇒xx=eyey