(x-y+1)dx+(2y-x)dy=0
xdx-ydx+dx+2ydy-xdy=0
xdx+dx+2ydy=ydx+xdy
而d(xy)=ydx+xdy故
xdx+dx+2ydy=d(xy)
两边积分得:
(1/2)x^2+x+y^2=xy+C (C为常数)
(x-y+1)dx+(2y-x)dy=0的通解
(x-y+1)dx+(2y-x)dy=0
xdx-ydx+dx+2ydy-xdy=0
xdx+dx+2ydy=ydx+xdy
而d(xy)=ydx+xdy故
xdx+dx+2ydy=d(xy)
两边积分得:
(1/2)x^2+x+y^2=xy+C (C为常数)