If y=∑aᵣxʳ | integer r∊[0,∞)
xy=∑aᵣxʳ⁺¹ | integer r∊[0,∞) = a₀x+a₁x²+∑aᵣxʳ⁺¹ | integer r∊[2,∞)
+
3xy'=3a₁x+3∑raᵣxʳ | integer r∊[2,∞)
+ (x²-1)y''=∑r(r-1)aᵣxʳ-∑r(r-1)aᵣxʳ⁻² | integer r∊[2,∞)
=a₀x+a₁x²+∑aᵣxʳ⁺¹+3a₁x+∑(3r+r(r-1))aᵣxʳ-∑r(r-1)aᵣxʳ⁻² | integer r∊[2,∞)
=a₀x+3a₁x+a₁x²+∑aᵣxʳ⁺¹+∑r(r+2)aᵣxʳ-∑r(r-1)aᵣxʳ⁻² | integer r∊[2,∞) =0.
We can expand this for a few terms:
a₀x+3a₁x+a₁x² + a₂x³+8a₂x²-2a₂ + a₃x⁴+15a₃x³-6a₃x + a₄x⁵+24a₄x⁴-12a₄x² + a₅x⁶+35a₅x⁵-20a₅x³+...=0
Now regroup by powers of x: -2a₂ +x(a₀+3a₁-6a₃) +x²(a₁+8a₂-12a₄) +x³(a₂+15a₃-20a₅)+...=0
Now a pattern emerges for xʳ where r>0: xʳ(aᵣ₋₁+r(r+2)aᵣ-(r+1)(r+2)aᵣ₊₂), giving the series:
-2a₂+∑xʳ(aᵣ₋₁+r(r+2)aᵣ-(r+1)(r+2)aᵣ₊₂)=0.
Equating coefficients:
-2a₂=0, so a₂=0 and:
aᵣ₊₂=(aᵣ₋₁+r(r+2)aᵣ)/((r+1)(r+2)).
Therefore, a₃=(a₀+3a₁)/6, a₄=(a₁+8a₂)/12=a₁/12, a₅=(a₂+15a₃)/20=3a₃/4=(a₀+3a₁)/8, a₆=(a₃+24a₄)/30=((a₀+3a₁)/6+2a₁)/30=(a₀+15a₁)/180, a₇=(a₄+35a₅)/42=(a₁/12+(a₀+3a₁)/8)/42=(3a₀+11a₁)/1008,...
If A=a₀ and B=a₁,
y=A+Bx+(A+3B)x³/6+Bx⁴/12+(A+3B)x⁵/8+(A+15B)x⁶/180+... y=A(1+x³/6+x⁵/8+x⁶/180+...)+B(x+x³/2+x⁴/12+3x⁵/8+x⁶/12+...)