∵dx+(x+y^2)dy=0
==>e^ydx+xe^ydy+y^2e^ydy=0 (等式两端同乘e^y)
==>e^ydx+xd(e^y)+y^2e^ydy=0
==>d(xe^y)+d((y^2-2y+2)e^y)=0
==>xe^y+(y^2-2y+2)e^y=C (C是常数)
==>x=Ce^(-y)-y^2+2y-2
∴原方程的通解是x=Ce^(-y)-y^2+2y-2.