Verified answer

Assume that y = Σ(n = 0 to ∞) a(n) x^n.

Substituting this into the DE yields

(2x - 3)Σ(n = 2 to ∞) n(n-1) a(n) x^(n-2) - x Σ(n = 1 to ∞) n a(n) x^(n-1) + Σ(n = 0 to ∞) a(n) x^n = 0

Simplifying:

Σ(n = 2 to ∞) 2n(n-1) a(n) x^(n-1) - Σ(n = 2 to ∞) 3n(n-1) a(n) x^(n-2) - Σ(n = 1 to ∞) n a(n) x^n

+ Σ(n = 0 to ∞) a(n) x^n = 0

Re-indexing to x^n:

Σ(n = 1 to ∞) 2(n+1)n a(n+1) x^n - Σ(n = 0 to ∞) 3(n+2)(n+1) a(n+2) x^n -

Σ(n = 1 to ∞) n a(n) x^n + Σ(n = 0 to ∞) a(n) x^n = 0

Collecting like terms:

(a(0) - 6 a(2)) + Σ(n = 1 to ∞) [-3(n+2)(n+1) a(n+2) + 2n(n+1) a(n+1) - (n-1) a(n)] x^n = 0

Assuming that a(0) and a(1) are arbitrary,

a(0) - 6 a(2) = 0

==> a(2) = (1/6) a(0).

For n > 0, we have -3(n+2)(n+1) a(n+2) + 2n(n+1) a(n+1) - (n-1) a(n) = 0

==> a(n+2) = [2n(n+1) a(n+1) - (n-1) a(n)] / [3(n+2)(n+1)].

a(3) = 4 a(2)/18 = (1/27) a(0)

a(4) = [12 a(3) - a(2)] / 36 = (5/648) a(0), which also does not have a(1) in it (nor do any other coefficients).

So, a general solution is

y = a(0) (1 + (1/6)x^2 + (1/27)x^3 + (5/648)x^4 ...) + a(1) x.

I hope this helps!