2 added 582 characters in body edited Nov 13 '15 at 18:48 Pillsy 12.8k11 gold badge3535 silver badges7979 bronze badges Here, In[4]:= Array[Derivative[#1][f][0] &, 3, 2] Out[4]= {Derivative[2][f][0], Derivative[3][f][0], Derivative[4][f][0]}  is just a compact way of describing the derivatives we want to solve for, Reduce attempts to eliminate the listed variables from the equation, and Refine applies the assumptions (set by Assuming) that $$a \ne 0$$ and $$L \ne 0$$. Which comes out to Which comes out to Here, In[4]:= Array[Derivative[#1][f][0] &, 3, 2] Out[4]= {Derivative[2][f][0], Derivative[3][f][0], Derivative[4][f][0]}  is just a compact way of describing the derivatives we want to solve for, Reduce attempts to eliminate the listed variables from the equation, and Refine applies the assumptions (set by Assuming) that $$a \ne 0$$ and $$L \ne 0$$. Which comes out to 1 answered Nov 13 '15 at 18:25 Pillsy 12.8k11 gold badge3535 silver badges7979 bronze badges I think there are two issues here. The first is that LogicalExpand is for expanding logical expressions, like so: In[1]:= LogicalExpand[(a || b) && (b || c)] Out[1]= (a || b) && (b || c)  Second, you can actually work directly with power series and expand them about $$\infty$$, as follows: In[2]:= series = Series[f[(a L /x)], {x, Infinity, 4}];  In Mathematica, this is a SeriesData object that represents the series in a compact way, but you can do many normal operations on it. Using TeXForm to get nice output for SE yields: $$f(0)+\frac{a L f'(0)}{x}+\frac{a^2 L^2 f''(0)}{2 x^2}+\frac{a^3 f^{(3)}(0) L^3}{6 x^3}+\frac{a^4 f^{(4)}(0) L^4}{24 x^4}+O\left(\left(\frac{1}{x}\right)^5\right)$$ However, for the rest of the problem, I'll use your definition in terms of $$w = \frac{1}{x}$$. Let's use your DE definition and substitute in the series for f[a L w] before taking the derivatives using Block: ode = Block[{F = Series[f[a*L*w], {w, 0, 4}]}, Simplify[(1/16)*((16*a^2*L^2*(1 + a*L*w)^4*W^2)/((1 + 2*a*L*w)^2* (1 + 2*a*L*w*(1 + a*L*w))^2) - (a^2*L^2*W*(4*I + W))/ (1 + a*L*w)^2 + (8*(-1 - 2*a*L*w + 2*a^3*L^3*w^3)*(I*a*L*W))/ (w*(1 + a*L*w)*(1 + 2*a*L*w)*(1 + 2*a*L*w*(1 + a*L*w))))* D[F, {w, 0}] + (1/16)*((8*I*a*L*W)/(1 + a*L*w) + (8*(-1 - 2*a*L*w + 2*a^3*L^3*w^3)*(4*(1 + a*L*w)))/ (w*(1 + a*L*w)*(1 + 2*a*L*w)*(1 + 2*a*L*w*(1 + a*L*w))))* D[F, {w, 1}] + D[F, {w, 2}] == 0]];  The Mathematica output is kinda gross, so I'll TeXForm it again: $$\frac{a L \left(-2 f'(0)-\frac{1}{2} i f(0) W\right)}{w}+\frac{1}{16} a^2 L^2 \left(-16 f''(0)+64 f'(0)+15 f(0) W^2+20 i f(0) W\right)+\frac{1}{16} a^3 L^3 w \left(W^2 \left(15 f'(0)-62 f(0)\right)-4 i W \left(-f''(0)-3 f'(0)+8 f(0)\right)+64 \left(f''(0)-f'(0)\right)\right)+\frac{1}{96} a^4 L^4 w^2 \left(W^2 \left(45 f''(0)-372 f'(0)+942 f(0)\right)+4 i W \left(4 f^{(3)}(0)+3 f''(0)-36 f'(0)+66 f(0)\right)+16 \left(f^{(4)}(0)+12 f^{(3)}(0)-24 f''(0)+24 f'(0)\right)\right)+O\left(w^3\right)=0$$ As you can see, this equation is given in terms of $$\left\{f(0),f'(0),f''(0),f^{(3)}(0),f^{(4)}(0)\right\}$$. Assuming $$f(0)$$ and $$f'(0)$$ are fixed (since it's a second-order ODE), we can reduce ode for non-zero $$a$$ and $$L$$: reduced = Assuming[{a != 0, L != 0}, Refine[Reduce[ode, Array[Derivative[#1][f][0] & , 3, 2]]]];  Which comes out to $$\left(f'(0)=0\land f(0)=0\land f''(0)=0\land f^{(4)}(0)=-i f^{(3)}(0) (W-12 i)\right)\lor \left(f(0)\neq 0\land W=\frac{4 i f'(0)}{f(0)}\land f''(0)=\frac{1}{4} i (15 W+4 i) f'(0)\land f^{(4)}(0)=-\frac{1}{64} i \left(64 f^{(3)}(0) W-768 i f^{(3)}(0)+675 W^3 f'(0)+1848 i W^2 f'(0)+8688 W f'(0)+1152 i f'(0)\right)\right)$$