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Power series solution of y''+xy'+y=0

Short answer

Assume y = Σ a_n x^n. Recurrence a_{n+2} = -a_n/(n+2) yields even terms a_{2m}=(-1)^m a_0/(2^m m!), forming a_0 e^(-x²/2), and odd terms a_{2m+1}=(-1)^m a_1/(1·3·…·(2m+1)). The general solution is y = a_0 e^(-x²/2) + a_1 (x - x³/3 + x⁵/15 - …).

Solve by power series method the differential equation y'' + xy' + y = 0.

Equation: y″ + xy′ + y = 0. x = 0 is an ordinary point, so take y = Σ aₙxⁿ.

Substitution

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

Coefficient of xⁿ: (n+2)(n+1)aₙ₊₂ + (n + 1)aₙ = 0, so

aₙ₊₂ = − aₙ / (n + 2)

Coefficients

a₀ and a₁ arbitrary.

Even:  a₂ = −a₀/2,  a₄ = −a₂/4 = a₀/(2·4),  a₆ = −a₀/(2·4·6) …
       a₂ₘ = (−1)^m a₀ / (2^m m!)
Odd:   a₃ = −a₁/3,  a₅ = −a₃/5 = a₁/(3·5),  a₇ = −a₁/(3·5·7) …
       a₂ₘ₊₁ = (−1)^m a₁ / (1·3·5···(2m+1))

Solution

y = a₀(1 − x²/2 + x⁴/8 − x⁶/48 + …) + a₁(x − x³/3 + x⁵/15 − x⁷/105 + …)

The first series is e^(−x²/2), because Σ (−x²/2)^m / m! = e^(−x²/2). So

y = a₀ e^(−x²/2) + a₁ (x − x³/3 + x⁵/15 − …)

Check: y = e^(−x²/2) gives y′ = −x e^(−x²/2), y″ = (x² − 1)e^(−x²/2); then y″ + xy′ + y = (x² − 1 − x² + 1)e^(−x²/2) = 0.

✓ Answer checked by Claude Opus 5.5 (high effort)

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