∵xy'-y-y²=0 ==>xdy/dx=y(y+1)
==>dy/[y(y+1)]=dx/x
==>[1/y-1/(y+1)]dy=dx/x
==>ln│y│-ln│y+1│=ln│x│+ln│C│ (C≠0是积分常数)
==>y/(y+1)=Cx
∴原微分方程的通解是y/(y+1)=Cx (C≠0是积分常数).
xy'-y-y*y=0的通解
xdy/dx=y(y+1)==>dy/[y(y+1)]=dx/x==>[1/y-1/(y+1)]dy=dx/x==>ln"}}}'>