1.级数法：
y^(n)=[1/(1+x^2)]^((n-1))=[∑{0≤k
(1+x^2)y^(n)+(n-1)(1+x^2)'y^((n-1))+(n-1)(n-2)/2(1+x^2)''y^((n-2))=
=0==>
(1+x^2)y^(n)+2(n-1)x*y^((n-1))+(n-1)(n-2)y^((n-2))=0.
答：看图````````````````````````````````详情>>
答：详情>>
