求微分方程(1-x^2)dy+(2xy-cosx)dx=0满足初始条件y(0)=1的特解

(1-x^2)dy+(2xy-cosx)dx=0
(1-x^2)dy+yd(x^2-1)=cosxdx
dy/(1-x^2)+yd(1/(1-x^2)）=cosxdx/(1-x^2)^2
d(y/(1-x^2))=cosxdx/(1-x^2)^2
通解y/(1-x^2)=∫cosxdx/(1-x^2)^2