(dy/dx)(1-x)=a(y^2+y)
dy/(y^2+y)=adx/(1-x)
1/2*(1/y-1/(y+2))dy=-ad(1-x)/(1-x)
ln y-ln(y-1)=-2aln(1-x)+ln c
y/(y-1)=c(1-x)^(-2a)
c为任意常数
这样好了
求解微分方程 y'-xy'=a(y^2+y)
(dy/dx)(1-x)=a(y^2+y)
dy/(y^2+y)=adx/(1-x)
1/2*(1/y-1/(y+2))dy=-ad(1-x)/(1-x)
ln y-ln(y-1)=-2aln(1-x)+ln c
y/(y-1)=c(1-x)^(-2a)
c为任意常数
这样好了