求文档:求全微分(x^2+2xy)dx+xydy=0的通解

(x^2+2xy)dx+xydy=0
x^2dx+2xydx+xydy=0
(x/y)dx+2dx+dy=0
x/y=u dx=ydu+udy
uydu+u^2dy+2ydu+2udy+dy=0
(uy+2y)du+(u^2+2u+1)dy=0
-(u+2)du/(u+1)^2=dy/y
-du/(u+1)-du/(u+1)^2=dlny
d-ln(u+1)+1/(u+1)=dlny
dlny+ln(u+1)-1/(u+1)=0
lny+ln(u+1)-1/(u+1)=C
lny(u+1)-1/(u+1)=C
ln(x+y)-y/(x+y)=C