不是已经解了吗?y+2xy'+(x^2)y''=0设x=e^t,t=lnxy'(x)=y'(t)/x . xy'(x)=y'(t)y''(x)=(y''(t)-y'(t))/x^2 . x^2y''(x)=y''(t)-y'(t)y''(t)-y'(t)+2y'(t)+y=0 y''(t)+y'(t)+y=0 解得：y=e^(-t/2)(C1cos(t√3/2)+C...