求黎卡提微分方程(x^2-1)y'+y^2-2xy+1=0的通解.

Question

求黎卡提微分方程(x^2-1)y'+y^2-2xy+1=0的通解.

Answer 1

令y-x=p,dy/dx--1=dp/dx,代入得
(x^2--1)(1+dp/dx)+p^2+1--x^2=0,即(x^2--1)dp/dx=-p^2,
--dp/p^2=dx/(x^2--1)
d(1/p)=0.5d(ln|(x--1)/(x+1)|),
1/(y--x)=0.5ln|(x--1)/(x+1)|+C
解出y即可.