∵dy/dx=y(lny-lnx+1)/x ==>dy/dx=y(ln(y/x)+1)/x.(1)
∴令z=y/x,则代入(1),得xz'+z=z(lnz+1)
==>xz'=zlnz
==>dz/(zlnz)=dx/x
==>d(lnz)/lnz=dx/x
==>ln│lnz│=ln│x│+ln│C│ (C是积分常数)
==>lnz=Cx
==>z=e^(Cx)
==>y/x=e^(Cx)
==>y=xe^(Cx)
故原方程的通解是y=xe^(Cx) (C是积分常数).