令p=y'
则y"=dp/dx=dp/dy*dy/dx=pdp/dy
代入原方程：ypdp/dy-p^2+p=0
得：p=0或ydp/dy-p+1=0
p=0得：dy/dx=0,即：y=c
ydp/dy-p+1=0,得：dp/(p-1)=dy/y,得：ln(p-1)=lny+c1,得：p-1=cy
得：dy/dx=cy+1,
得：dy/(cy+1)=cx,
得：ln(cy+1)=cx^2/2+c2
cy+1=e^(cx^2/2+c2)
y=[e^(cx^2/2+c2)-1]/c