求微分方程 (xy∧2+x)dx+(y-x∧2y)dx=0的通解

(xy∧2+x)dx+(y-x∧2y)dy=0(y^2+1)xdx=(x^2-1)ydyxdx/(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)+lncx^2-1=c(y^2+1)厉害，我打错了后面的是dy，谢谢啊不客气，及时采纳呵