现在我对这个微分方程的解猜出了一个形式解，如图所示，现在我想要级数展开然后求解前面的常系数，但是使用Series函数只能得到幂级数的展开解，带log z的这些项体现不出来，我要怎么做才能得到我想要的形式展开呢？
以下是代码：
In[4]:= SetAttributes[{u1, u0, zh, v4}, Constant]
In[2]:= eq1 =
y''[z] + (-3/z + 2*u1^2*z + 4*z^3/(z^4 - zh^4)) y'[z] +
1/z^2*zh^4/(zh^4 - z^4) (3*y[z] - 4*v4*y[z]^3) == 0
Out[2]= (zh^4 (3 y[z] - 4 v4 y[z]^3))/(
z^2 (-z^4 + zh^4)) + (-(3/z) + 2 u1^2 z + (4 z^3)/(
z^4 - zh^4)) Derivative[1][y][z] + (y^\[Prime]\[Prime])[z] == 0
In[6]:= Series[eq1, {z, 0, 5}]
Out[6]= (3 y(0)-(4 v4) y(0)^3)/z^2-(((12 v4) y(0)^2) y^\[Prime](0))/z+(-((6 v4) y(0)) (y(0) y^\[Prime]\[Prime](0)+2 y^\[Prime](0)^2)-1/2 y^\[Prime]\[Prime](0))+z ((2 u1^2) y^\[Prime](0)-((4 v4) y(0)^3) (y^(3)(0)/(2 y(0))+((2 y^\[Prime](0)) y^\[Prime]\[Prime](0))/y(0)^2+(y^\[Prime](0) (y^\[Prime]\[Prime](0)/y(0)+y^\[Prime](0)^2/y(0)^2))/y(0)))+z^2 ((2 u1^2) y^\[Prime]\[Prime](0)+zh^4 ((1/8 y^(4)(0)-((4 v4) y(0)^3) (y^(4)(0)/(8 y(0))+y^\[Prime]\[Prime](0)^2/(4 y(0)^2)+(y^(3)(0) (2 y^\[Prime](0)))/(3 y(0)^2)+(y^\[Prime]\[Prime](0) (y^\[Prime]\[Prime](0)/y(0)+y^\[Prime](0)^2/y(0)^2))/(2 y(0))+(y^\[Prime](0) (y^(3)(0)/(3 y(0))+(y^\[Prime](0) y^\[Prime]\[Prime](0))/y(0)^2))/y(0)))/zh^4+(3 y(0)-(4 v4) y(0)^3)/zh^8))+z^3 (u1^2 y^(3)(0)+zh^4 ((3 y^\[Prime](0)-((12 v4) y(0)^2) y^\[Prime](0))/zh^8+1/zh^4 (1/40 y^(5)(0)-((4 v4) y(0)^3) (y^(5)(0)/(40 y(0))+(y^(4)(0) y^\[Prime](0))/(6 y(0)^2)+(y^(3)(0) y^\[Prime]\[Prime](0))/(6 y(0)^2)+(y^(3)(0) (y^\[Prime]\[Prime](0)/y(0)+y^\[Prime](0)^2/y(0)^2))/(6 y(0))+(y^\[Prime]\[Prime](0) (y^(3)(0)/(3 y(0))+(y^\[Prime](0) y^\[Prime]\[Prime](0))/y(0)^2))/(2 y(0))+(y^\[Prime](0) (y^(4)(0)/(12 y(0))+y^\[Prime]\[Prime](0)^2/(4 y(0)^2)+(y^(3)(0) y^\[Prime](0))/(3 y(0)^2)))/y(0))))+1/24 y^(5)(0)-(4 y^\[Prime](0))/zh^4)+z^4 (1/3 u1^2 y^(4)(0)+zh^4 (((3 y^\[Prime]\[Prime](0))/2-((6 v4) y(0)) (y(0) y^\[Prime]\[Prime](0)+2 y^\[Prime](0)^2))/zh^8+1/zh^4 (1/240 y^(6)(0)-((4 v4) y(0)^3) (y^(6)(0)/(240 y(0))+y^(3)(0)^2/(36 y(0)^2)+(y^(5)(0) y^\[Prime](0))/(30 y(0)^2)+(y^(4)(0) y^\[Prime]\[Prime](0))/(24 y(0)^2)+(y^(4)(0) (y^\[Prime]\[Prime](0)/y(0)+y^\[Prime](0)^2/y(0)^2))/(24 y(0))+(y^(3)(0) (y^(3)(0)/(3 y(0))+(y^\[Prime](0) y^\[Prime]\[Prime](0))/y(0)^2))/(6 y(0))+(y^\[Prime]\[Prime](0) (y^(4)(0)/(12 y(0))+y^\[Prime]\[Prime](0)^2/(4 y(0)^2)+(y^(3)(0) y^\[Prime](0))/(3 y(0)^2)))/(2 y(0))+(y^\[Prime](0) (y^(5)(0)/(60 y(0))+(y^(4)(0) y^\[Prime](0))/(12 y(0)^2)+(y^(3)(0) y^\[Prime]\[Prime](0))/(6 y(0)^2)))/y(0))))+1/60 y^(6)(0)-(4 y^\[Prime]\[Prime](0))/zh^4)+z^5 (1/12 u1^2 y^(5)(0)+zh^4 ((1/2 y^(3)(0)-((4 v4) y(0)^3) (y^(3)(0)/(2 y(0))+((2 y^\[Prime](0)) y^\[Prime]\[Prime](0))/y(0)^2+(y^\[Prime](0) (y^\[Prime]\[Prime](0)/y(0)+y^\[Prime](0)^2/y(0)^2))/y(0)))/zh^8+1/zh^4 (y^(7)(0)/1680-((4 v4) y(0)^3) (y^(7)(0)/(1680 y(0))+(y^(6)(0) y^\[Prime](0))/(180 y(0)^2)+(y^(5)(0) y^\[Prime]\[Prime](0))/(120 y(0)^2)+(y^(3)(0) y^(4)(0))/(72 y(0)^2)+(y^(5)(0) (y^\[Prime]\[Prime](0)/y(0)+y^\[Prime](0)^2/y(0)^2))/(120 y(0))+(y^(4)(0) (y^(3)(0)/(3 y(0))+(y^\[Prime](0) y^\[Prime]\[Prime](0))/y(0)^2))/(24 y(0))+(y^(3)(0) (y^(4)(0)/(12 y(0))+y^\[Prime]\[Prime](0)^2/(4 y(0)^2)+(y^(3)(0) y^\[Prime](0))/(3 y(0)^2)))/(6 y(0))+(y^\[Prime]\[Prime](0) (y^(5)(0)/(60 y(0))+(y^(4)(0) y^\[Prime](0))/(12 y(0)^2)+(y^(3)(0) y^\[Prime]\[Prime](0))/(6 y(0)^2)))/(2 y(0))+(y^\[Prime](0) (y^(6)(0)/(360 y(0))+y^(3)(0)^2/(36 y(0)^2)+(y^(5)(0) y^\[Prime](0))/(60 y(0)^2)+(y^(4)(0) y^\[Prime]\[Prime](0))/(24 y(0)^2)))/y(0))))+1/240 y^(7)(0)-(2 y^(3)(0))/zh^4)+O(z^6)==0
In[3]:= Series[eq1, {z, 0, 5}] // FullSimplify
Out[3]= (3 y(0)-4 v4 y(0)^3)/z^2-(12 (v4 y(0)^2 y^\[Prime](0)))/z+(-6 v4 y(0) (y(0) y^\[Prime]\[Prime](0)+2 y^\[Prime](0)^2)-1/2 y^\[Prime]\[Prime](0))+z (2 y^\[Prime](0) (u1^2-2 v4 (3 y(0) y^\[Prime]\[Prime](0)+y^\[Prime](0)^2))-2 v4 y(0)^2 y^(3)(0))+z^2 (2 u1^2 y^\[Prime]\[Prime](0)+1/8 y^(4)(0) (1-4 v4 y(0)^2)-v4 (3 y(0) y^\[Prime]\[Prime](0)^2+4 y(0) y^(3)(0) y^\[Prime](0)+6 y^\[Prime](0)^2 y^\[Prime]\[Prime](0))+(3 y(0)-4 v4 y(0)^3)/zh^4)+z^3 (u1^2 y^(3)(0)+1/30 y^(5)(0) (2-3 v4 y(0)^2)-2 v4 y(0) y^(3)(0) y^\[Prime]\[Prime](0)-2 v4 y^(3)(0) y^\[Prime](0)^2-(y^\[Prime](0) (v4 (y(0) y^(4)(0) zh^4+3 zh^4 y^\[Prime]\[Prime](0)^2+12 y(0)^2)+1))/zh^4)+1/(240 zh^4) z^4 (5 zh^4 (16 u1^2 y^(4)(0)+y^(6)(0))-4 v4 (y(0)^2 (y^(6)(0) zh^4+360 y^\[Prime]\[Prime](0))+30 zh^4 (y^\[Prime]\[Prime](0)^3+y^(4)(0) y^\[Prime](0)^2+4 y^(3)(0) y^\[Prime](0) y^\[Prime]\[Prime](0))+2 y(0) (6 y^\[Prime](0) (y^(5)(0) zh^4+60 y^\[Prime](0))+5 zh^4 (2 y^(3)(0)^2+3 y^(4)(0) y^\[Prime]\[Prime](0))))-600 y^\[Prime]\[Prime](0))+1/420 z^5 (y^(7)(0) (2-v4 y(0)^2)-1/zh^4 7 (-5 u1^2 y^(5)(0) zh^4+2 v4 (120 y^\[Prime](0)^3+3 y(0) y^(5)(0) zh^4 y^\[Prime]\[Prime](0)+3 y^(5)(0) zh^4 y^\[Prime](0)^2+5 y^(3)(0) (y(0) y^(4)(0) zh^4+3 zh^4 y^\[Prime]\[Prime](0)^2+12 y(0)^2)+y^\[Prime](0) (y(0) y^(6)(0) zh^4+10 y^(3)(0)^2 zh^4+15 (y^(4)(0) zh^4+24 y(0)) y^\[Prime]\[Prime](0)))+90 y^(3)(0)))+O(z^6)==0

Map[# == 0 &,Rest@CoefficientList[你图上的结果, z]] 可以得到常系数的方程

抱歉，没有看懂你的问题。在eq1中有z^-2，z^-1，z^0，z^1，…，在黑色图片里面没有z^-1，我猜是要用 z^-1 = Log[z+1]/z^2 +1/2 - z/3 + z^2/4 - z^3/5 + 将z^-1消去并且重算各项系数？不然我不太理解怎么出现Log函数的