求微分方程（x-1）y’=y（1+2xy）的通解
为凑微分,两边同除以y²,得
y'(x-1)/y²=(1+2xy)/y
(dy/dx)(x-1)/y²=(1+2xy)/y
(x-1)dy/y²=(1+2xy)dx/y
[(x/y²)-(1/y²)]dy=[(1/y)+2x]dx
(x/y²)dy-(1/y²)dy=(1/y)dx+2xdx
-(1/y²)dy-2xdx=(1/y)dx-(x/y²)dy
因为 -(1/y²)dy=1/y,2xdx=x²,
[x*(1/y)]的微分=dx*1/y+x(-1/y²)dy=(1/y)dx-(x/y²)dy,
所以,有 1/y-x²=x/y+C
所以通解是：x²+x/y-1/y+C=0