∵dy/dx=(1+y)/(xy) ==>ydy/(1+y)=dx/x==>[1-1/(1+y)]dy=dx/x==>y-ln│1+y│=ln│x│-ln│C│ (C是积分常数)==>ln│x(1+y)│=y+ln│C│==>x(1+y)=Ce^y∴原微分方程的通解是x(1+y)=Ce^y (C是积分常数)注：这是可分离...