求解常微分方程 y+2xy'+(x^2)y'' 坐等……呃，是y+2xy'+(x^2)y''=0

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不是已经解了吗?
y+2xy'+(x^2)y''=0
设x=e^t, t=lnx
y'(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)+C2sin(t√3/2))
=x^(-1/2)(C1cos(√3lnx/2)+C2sin(√3lnx/2))