Question

求微分方程（x^2-1)y'+2xy-cosx=0的通解

我解到u'(x^2+1)+4ux=cosx/(x^2-1)这一步就不会做了，谁帮忙详细解一下

Answer 1

原式等价于 ((x^2-1)y)'=cosx
(x^2-1)y=sinx+C(常数)
y=(sinx+C)/(x^2-1)

Answer 2

dy/dx=y'=(-2xy+cosx)/（x^2-1)
（x^2-1)dy=(-2xy+cosx)dx
x^2dy+2xydx-dy-cosxdx=0
因为2xydx＝yd(x^2)
所以x^2dy+yd(x^2)-dy-cosxdx=0
x^2*y-y-sinx=c