凡数y'=f(y/x)的,都可设u=y/x
dy/dx=(2+(y/x))/(y/x)
u=y/xy=uxy'=u'x+u
u 'x+u=(2+u)/u
u'x=(2+u-u^2)/u
u/(-u^2+u+2)du=1/xdx
u/[(u-2)(u+1)]du=1/xdx
[(1/3)/(u+1)+(2/3)/(u-2)]du=1/xdx
1/3ln(u+1)+2/3ln(u-2)=lnx+c
再将y/x=u代入即可