Series solution of (1+x²)y''+xy'−y=0

Equation: (1 + x²)y″ + xy′ − y = 0. x = 0 is an ordinary point; let y = Σ aₙxⁿ.

Substitution

(1+x²)y″ = Σ (n+2)(n+1)aₙ₊₂ xⁿ + Σ n(n−1)aₙ xⁿ
xy′ = Σ n aₙ xⁿ,    −y = −Σ aₙ xⁿ

Coefficient of xⁿ: (n+2)(n+1)aₙ₊₂ + [n(n−1) + n − 1]aₙ = 0. Since n(n−1) + n − 1 = n² − 1,

aₙ₊₂ = − (n² − 1) aₙ / ((n+1)(n+2)) = (1 − n)aₙ/(n+2)

(because n² − 1 = (n − 1)(n + 1)).

Coefficients (a₀, a₁ arbitrary)

n=0:  a₂ = a₀/2
n=1:  a₃ = 0   (so a₅ = a₇ = … = 0)
n=2:  a₄ = −a₂/4 = −a₀/8
n=4:  a₆ = −3a₄/6 = a₀/16

Solution

y = a₀(1 + x²/2 − x⁴/8 + x⁶/16 − …) + a₁x

The odd solution terminates: y = x satisfies the equation exactly (0 + x − x = 0). The even series is the expansion of √(1 + x²).