Power Series Solutions to ODEs — Mathematics Knowledge System

Overview

When a second-order linear ODE has analytic (power-series expandable) coefficients, its solutions near an ordinary point are themselves analytic and can be found by substituting a power series ansatz, differentiating term-by-term, and matching coefficients to obtain a recurrence relation. Two linearly independent solutions emerge from the two free parameters a₀ and a₁, corresponding to the two-dimensional solution space guaranteed by the existence-uniqueness theorem.

Intuition

The idea is to 'guess' that the solution is a power series (which it must be, since the coefficients are analytic) and then determine the coefficients by substitution. Differentiating a power series term-by-term is easy, and the ODE forces all the coefficients of each power xⁿ to satisfy a simple algebraic recurrence. The two free parameters a₀ and a₁ (for a second-order ODE) generate the two linearly independent solutions.

Formal Definition

Definition

Consider a second-order linear ODE in standard form with analytic coefficient functions P, Q near x₀:

P(x)y'' + Q(x)y' + R(x)y = 0, \quad P(x_0) \neq 0

Standard second-order linear ODE; x₀ is an ordinary point if P(x₀) ≠ 0

standard-form

y = \sum_{n=0}^{\infty} a_n (x - x_0)^n

Power series ansatz centered at x₀

power-series-ansatz

a_{n+2} = f(n, a_n, a_{n-1}, \ldots), \quad n \geq 0

Recurrence relation obtained by matching coefficients of (x−x₀)ⁿ

recurrence

\rho \geq \min(\rho_P, \rho_Q, \rho_R)

Radius of convergence ≥ distance from x₀ to the nearest singularity of Q/P or R/P

convergence

Worked Examples

Substitute the ansatz y = ∑_{n=0}^∞ aₙxⁿ. Compute y' and y''.
$y'' = \sum_{n=2}^{\infty} n(n-1)a_n x^{n-2} = \sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n$
Compute xy' = ∑_{n=0}^∞ n aₙ xⁿ (the n=0 term vanishes).
$xy' = \sum_{n=0}^{\infty} n a_n x^n$
Substitute into the ODE and collect powers of xⁿ:
$\sum_{n=0}^{\infty}\left[(n+2)(n+1)a_{n+2} - n\,a_n - a_n\right]x^n = 0$
Setting each coefficient to zero gives the recurrence:
$a_{n+2} = \frac{a_n}{n+2}, \quad n \geq 0$
Even terms (from a₀): a₂ = a₀/2, a₄ = a₀/8, a_{2k} = a₀/(2^k k!). Odd terms (from a₁): a₃ = a₁/3, a₅ = a₁/15, a_{2k+1} = a₁/(2k+1)!!.
$a_{2k} = \frac{a_0}{2^k k!}, \qquad a_{2k+1} = \frac{a_1}{(2k+1)!!}$

Answer: y = a₀ ∑_{k=0}^∞ x^{2k}/(2^k k!) + a₁ ∑_{k=0}^∞ x^{2k+1}/((2k+1)!!), where (2k+1)!! = 1·3·5⋯(2k+1). Both series converge for all x.

Practice Problems

Difficulty 6/10

Use the power series method to find the recurrence relation for y'' + xy = 0 (the Airy equation) about x = 0.

Difficulty 7/10

What is the radius of convergence guaranteed for solutions of (x²−4)y'' + 3xy' + y = 0 found by the power series method about x = 0?

Difficulty 8/10

Show that the power series solution to y'' − y = 0 about x = 0 yields y₁ = cosh(x) and y₂ = sinh(x).

Common Mistakes

Common Mistake

Assuming the power series method always works at singular points.

At a regular singular point (where P(x₀)=0), the Frobenius method (with a factor (x−x₀)^r) is needed. At an irregular singular point, power series solutions may not exist.

Common Mistake

Forgetting to re-index the series before collecting like powers.

After differentiating ∑ aₙxⁿ twice, shift the summation index so all series run from n=0 with the same power xⁿ before summing coefficients.

Quiz

Historical Background

The method of series solutions was pioneered by Frobenius (1873) in his study of ODEs near singular points, generalising earlier work by Euler and Gauss on the hypergeometric equation. Many classical special functions — Legendre polynomials, Bessel functions, Hermite polynomials — were discovered as power-series solutions to specific ODEs arising in mathematical physics.

1748
Euler solves differential equations with power series
Leonhard Euler
1812
Gauss studies the hypergeometric ODE and its series solutions
Carl Friedrich Gauss
1873
Frobenius extends the method to regular singular points
Ferdinand Georg Frobenius

Summary

At an ordinary point x₀, a second-order linear ODE with analytic coefficients has two linearly independent analytic solutions.
Substitute y = ∑ aₙ(x−x₀)ⁿ, differentiate term-by-term, substitute into the ODE, and collect powers to get a recurrence for aₙ.
The two free parameters a₀ and a₁ generate the two independent solutions; all other coefficients are determined by the recurrence.
The radius of convergence is at least the distance from x₀ to the nearest singularity of Q/P or R/P.
Many classical special functions (Legendre polynomials, Hermite polynomials, Airy functions) arise as series solutions to specific ODEs.

References

WebsiteWikipedia — Power series solution of differential equations