Power series solution of y''+x²y=0

Equation: y″ + x²y = 0. x = 0 is an ordinary point; take y = Σ aₙxⁿ.

Substitution

y″ = Σ (n+2)(n+1)aₙ₊₂ xⁿ and x²y = Σ aₙ₋₂ xⁿ (n ≥ 2).

• x⁰: 2a₂ = 0, so a₂ = 0.
• x¹: 6a₃ = 0, so a₃ = 0.
• xⁿ (n ≥ 2): (n+2)(n+1)aₙ₊₂ + aₙ₋₂ = 0, so

aₙ₊₂ = − aₙ₋₂ / ((n+1)(n+2))

Coefficients (a₀, a₁ arbitrary)

n=2:  a₄ = −a₀/(3·4)  = −a₀/12
n=3:  a₅ = −a₁/(4·5)  = −a₁/20
n=4:  a₆ = −a₂/30 = 0
n=5:  a₇ = −a₃/42 = 0
n=6:  a₈ = −a₄/(7·8) = a₀/672
n=7:  a₉ = −a₅/(8·9) = a₁/1440

Only powers x⁴ᵐ and x⁴ᵐ⁺¹ survive.

Solution

y = a₀(1 − x⁴/12 + x⁸/672 − …) + a₁(x − x⁵/20 + x⁹/1440 − …)

a₀ and a₁ are arbitrary constants.