Power series: y''−4xy'+(4x²−2)y=0

Equation: y″ − 4xy′ + (4x² − 2)y = 0. x = 0 is an ordinary point, so let y = Σ aₙxⁿ.

Substitution

y″ = Σ (n+2)(n+1) aₙ₊₂ xⁿ, −4xy′ = −Σ 4n aₙ xⁿ, 4x²y = Σ 4aₙ₋₂ xⁿ, −2y = −Σ 2aₙ xⁿ.

Coefficient of xⁿ:

(n+2)(n+1) aₙ₊₂ − 4n aₙ − 2aₙ + 4aₙ₋₂ = 0
aₙ₊₂ = [(4n + 2) aₙ − 4 aₙ₋₂] / ((n+2)(n+1))

(with a₋₁ = a₋₂ = 0).

Coefficients

n=0:  2a₂ = 2a₀                 → a₂ = a₀
n=1:  6a₃ = 6a₁                 → a₃ = a₁
n=2:  12a₄ = 10a₂ − 4a₀ = 6a₀   → a₄ = a₀/2
n=3:  20a₅ = 14a₃ − 4a₁ = 10a₁  → a₅ = a₁/2
n=4:  30a₆ = 18a₄ − 4a₂ = 5a₀   → a₆ = a₀/6

a₀ and a₁ are arbitrary.

Solution

y = a₀(1 + x² + x⁴/2! + x⁶/3! + …) + a₁(x + x³ + x⁵/2! + x⁷/3! + …)

The first series is e^(x²), the second is x·e^(x²).

Answer: y = (a₀ + a₁x) e^(x²).

Check by substitution: for y = e^(x²), y″ = (2 + 4x²)e^(x²), y′ = 2x e^(x²), so y″ − 4xy′ + (4x² − 2)y = (2 + 4x² − 8x² + 4x² − 2)e^(x²) = 0.