求微分方程x^2y撇+xy=y^3满足初始条件y(1)=1的特解

y'+y/x=(y/x)^2
令u=y/x,则y'=u+xu'
u+xu'+u=u^2
xdu/dx=u^2-2u
du/(u(u-2))=dx/x
1/2*(1/(u-2)-1/u)du=dx/x
1/2*(ln|u-2|-ln|u|)=ln|x|+C
(u-2)/u=Cx^2
u=y/x=2/(1-Cx^2)
y=2x/(1-Cx^2)
令x=1：1=2/(1-C),C=-1
所以y=2x/(1+x^2)