(1-x²)y''-2xy'+4y=0, given y=∑aᵣxʳ, a power series in x for r∈[0,n], so for r∈[2,n]:
We can expect the power series to be infinite, that is, n→∞.
The DE can now be written:
4a₀+2a₁x+∑aᵣxʳ(4-r-r²)+∑r(r-1)aᵣxʳ⁻²=0 for r∈[2,n].
This has to be true for all x, so we need to find the constant terms, that is, x⁰ coefficients: 4a₀+2a₂=0 (r-2=0 implies r=2). So a₂=-2a₀.
If we now expand the resulting polynomial a little we get:
Because 4a₀+2a₂=0, this becomes:
We can divide through by x:
Therefore, 2a₁+6a₃=0, and a₃=-⅓a₁.
Progressively, the coefficients can be found. Let A=a₀ and B=a₁ (constants of integration), so a₂=-2A and a₃=-B/3. If we can find out the relation of successive coefficients and hence their relation to A and B, the polynomial can be written in the form y=Ap(x)+Bq(x).
For example, following the above method, we deduce that -2a₂+12a₄=0, a₄=a₂/6=-2A/6=-A/3, and -8a₃+20a₅=0, a₅=2a₃/5=-2B/15.
Therefore we need to find the coefficients of consecutive powers of x as a formula.
The general formula is: aᵣ(4-r-r²)+(r+2)(r+1)aᵣ₊₂=0,
If r=0, a₂=-4a₀/2=-2a₀=-2A, as discovered earlier.
If r=1, a₃=-2a₁/6=-a₁/3=-B/3 as before.
The series in parentheses are both infinite.