Question

求(1+x^2)y''-2xy'=0 的通解

Answer 1

(1+x^2)y''-2xy'=0

凑积分得

[y'/(1+x^2)]'=0

即

y'/(1+x^2)=C1

y'=C1(1+x^2)

两边再积分得

y=C1x+C1x^3/3+C2