求微分方程 (xy∧2+x)dx+(y-x∧2y)dx=0的通解

2025-06-26 13:48:59
推荐回答(1个)
回答1:

(xy∧2+x)dx+(y-x∧2y)dy=0
(y^2+1)xdx=(x^2-1)ydy
xdx/(x^2-1)=ydy/(y^2+1)
两边积分得:
Sxdx/(x^2-1)=Sydy/(y^2+1)
1/2*Sd(x^2-1)/(x^2-1)=1/2*Sd(y^2+1)/(y^2+1)
ln(x^2-1)=ln(y^2+1)+lnc
x^2-1=c(y^2+1)