求微分方程(x^2-1)y'+2xy-cosx=0的通解
问题描述：
求微分方程(x^2-1)y'+2xy-cosx=0的通解
答
(x^2-1)y'+2xy-cosx=0
(x^2-1)dy+2xydx-cosxdx=0
d(yx^2-y-sinx)=0
yx^2-y-sinx=C
y=(sinc+C)/(x^2-1)