y(1-x^2)dy+x(1+y^2)dx=0 求微分方程的通解。

Question

如题所述

Answer 1

x(y^2+1)dx=-y(1-x^2)dy
x(y^2+1)dx=y(x^2-1)dy
两边同除以 (y²＋1)(x²-1)即可得：y/(1+y²)dy=x/(x²-1)dx
两边积分，得
ln(1+y²)=ln(x²-1)+lnc
微分方程的通解为：
1+y²=c(x²-1)

即y²=c(x²-1)-1

Answer 2

y(1-x^2)y'+x(1+y^2)=0

[-(x^2-1)^2/2]* [(y^2+1)/(x^2-1)]'=0
即
[(y^2+1)/(x^2-1)]'=0
(y^2+1)/(x^2-1)=C [x!=1]
结果，

(y^2+1)=C* (x^2-1) [x!=1]