Find power series solutions of the equation (x^3-1)y″+3xy′-y=0 , about the point x=1

Question

ordinary differential equations

Answer 1

y=∑a_n(x-1)ⁿ, y'=∑na_n(x-1)^n-1, y"=∑n(n-1)a_n(x-1)^n-2, if we build the series around x=1.

Expand the first few terms:

y=a₀+a₁(x-1)+a₂(x-1)²+a₃(x-1)³+a₄(x-1)⁴+a₅(x-1)⁵+...

y'=a₁+2a₂(x-1)+3a₃(x-1)²+4a₄(x-1)³+5a₅(x-1)⁴+6a₆(x-1)⁵+...

y"=2a₂+6a₃(x-1)+12a₄(x-1)²+20a₅(x-1)³+30a₆(x-1)⁴+42a₇(x-1)⁵+...

Note that x³-1=(x-1)(x²+x+1), so the DE becomes:

(x³-1)y"+3xy'-y=

(x²+x+1)(2a₂+6a₃(x-1)²+12a₄(x-1)³+20a₅(x-1)⁴+30a₆(x-1)⁵+...

+3x(a₁+2a₂(x-1)+3a₃(x-1)²+4a₄(x-1)³+5a₅(x-1)⁴+6a₆(x-1)⁵+...)

-(a₀+a₁(x-1)+a₂(x-1)²+a₃(x-1)³+a₄(x-1)⁴+a₅(x-1)⁵+...)=0.

Consider the constant terms:

2a₂-a₀=0, a₂=a₀/2, because the DE must be zero for all x.

Consider (x-1) terms:

(x³-1)y"+3xy'-y=0=(x²+x+1)∑n(n-1)a_n(x-1)^n-1+3x∑na_n(x-1)^n-1-∑a_n(x-1)ⁿ=0,

∑na_n(x-1)^n-1[(n-1)(x²+x+1)+3x]-∑a_n(x-1)ⁿ=0.

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