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Define power series; solve y'+2xy=0

Short answer

A power series is an infinite sum Σ a_n (x-x_0)^n. Substituting y = Σ a_n x^n into y'+2xy=0 yields a_1=0 and a_{n+1} = -2 a_{n-1}/(n+1), giving y = a_0 e^{-x^2}.

Define power series. Solve by power series method of differential equation, y' + 2xy = 0.

Power series

An infinite series of the form

Σ aₙ(x − x₀)ⁿ = a₀ + a₁(x − x₀) + a₂(x − x₀)² + …

(with constant coefficients aₙ) is called a power series about the point x₀. About x₀ = 0 it is Σ aₙxⁿ. It converges for |x − x₀| < R, where R is the radius of convergence.

Solve y′ + 2xy = 0

x = 0 is an ordinary point. Let y = Σ aₙxⁿ, so y′ = Σ (n+1)aₙ₊₁ xⁿ and 2xy = Σ 2aₙ₋₁ xⁿ (n ≥ 1).

Coefficient of x⁰: a₁ = 0. Coefficient of xⁿ (n ≥ 1): (n+1)aₙ₊₁ + 2aₙ₋₁ = 0, so

aₙ₊₁ = − 2aₙ₋₁ / (n + 1)

a₁ = 0  →  all odd coefficients a₃ = a₅ = … = 0
a₂ = −2a₀/2 = −a₀
a₄ = −2a₂/4 = a₀/2
a₆ = −2a₄/6 = −a₀/6

Pattern: a₂ₘ = (−1)^m a₀ / m!.

y = a₀(1 − x² + x⁴/2! − x⁶/3! + …) = a₀ e^(−x²)

Answer: y = a₀e^(−x²). (Check by separation of variables: dy/y = −2x dx gives y = Ce^(−x²).)

✓ Answer checked by Claude Opus 5.5 (high effort)

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