求解高数dy/dx==(2y/x)-y^2.我做了一晚上都想不出来

dy/dx=(2y/x)-y^2
xdy=2ydx-xy^2dx
设x=u/y
dx=du/y-udy/y^2
(u/y)dy=(2y-uy)*(du/y-udy/y^2)
[(u/y)+(2-u)u/y]dy=(2-u)du
dy/y=(2-u)du/(3u-u^2)
dy/y=(u-2)du/(u^2-3u)
3lny=2lnu+ln(u-3)+lnC
y^3=Cu^2 *(u-3)
通解y^3=C(xy)^2 *(xy-3)
∫(u-2)du/(u^2-3u)=(2/3)∫du/u+(1/3)∫du/(u-3)=(2/3)lnu+(1/3)ln(u-3)
再问： 感谢你耐心回答，，但你给的答案是错得，，答案是y=(3x^2)/[(x^3)+C]
再答： y^3=C(x^2y^2)*(xy-3) y^3=Cx^3y^3-3Cx^2y^2 y=Cx^3y-3Cx^2 3Cx^2=Cx^3y-y y=3Cx^2/(Cx^3-1) =3x^2/(x^3-1/C) 答案是一样的 只是y=3x^2/(x^3+C0)中的C0= 1/C吧