原式等价于 ((x^2-1)y)'=cosx(x^2-1)y=sinx+C(常数)y=(sinx+C)/(x^2-1)
dy/dx=y'=(-2xy+cosx)/（x^2-1)（x^2-1)dy=(-2xy+cosx)dxx^2dy+2xydx-dy-cosxdx=0因为2xydx＝yd(x^2)所以x^2dy+yd(x^2)-dy-cosxdx=0x^2*y-y-sinx=c
