---
title: "Series Solutions About an Ordinary Point"
description: "When the coefficients are not constant, assume a power series, match coefficients and generate a recurrence. Named functions arise this way, such as the Airy functions, which solve y'' = xy."
canonical: https://lightmysky.com/learn/mathematics/series-solutions-about-an-ordinary-point-mt_E6NiSd7MED
source: https://lightmysky.com/learn/mathematics/series-solutions-about-an-ordinary-point-mt_E6NiSd7MED.md
retrieved: 2026-11-12
---

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# Series Solutions About an Ordinary Point

When the coefficients are not constant, assume a power series, match coefficients and generate a recurrence. Named functions arise this way, such as the Airy functions, which solve y'' = xy.

Subject: Mathematics · Area: Differential Equations · Ages 20 to 21
Page: https://lightmysky.com/learn/mathematics/series-solutions-about-an-ordinary-point-mt_E6NiSd7MED

## Ready when they can

- Substitute a power series and derive the recurrence relation
- Generate the first several coefficients from initial conditions
- Say what the radius of convergence means for the solution's range

## Lesson: Guessing a power series for the answer

When the coefficients are not constant, assume a power series and let it tell you its own terms. Write y as the sum of a sub n times x to the n, differentiate term by term, shift the indices so like powers line up, and match coefficients. That matching hands you a recurrence relation linking each coefficient to the earlier ones. Named functions come out of this: y double prime minus x y equals 0, solved below, is Airy's equation, and its solutions are the Airy functions.

**Example.** Solve y prime equals 3 y with a0 equals 2. Term by term differentiation turns the equation into n plus 1 times a sub n plus 1 equals 3 a sub n. At n equals 0 that reads 1 times a1 equals 3 times 2, so a1 is 6. At n equals 1 it reads 2 times a2 equals 3 times 6, so a2 is 9. Feed each answer into the next step and the series builds itself: 2 plus 6 x plus 9 x squared onward.

Check your series against expansions you already know. The series above is 2 times e to the 3 x expanded, and matching a known expansion catches algebra slips fast. Also mind where the series can be trusted: the radius of convergence says how far from the centre the sum still represents the solution. At an ordinary point the method runs smoothly, and when the coefficients are polynomials, as in y double prime minus x y equals 0, the series converges for every x.

**Example.** Now a variable coefficient. For y'' - xy = 0, write y = Σ a_n x^n. Then y'' = Σ (n + 2)(n + 1) a_(n+2) x^n, and xy = Σ a_(n-1) x^n starting at n = 1. The x^0 terms give 2a_2 = 0, so a_2 = 0. For n ≥ 1, (n + 2)(n + 1) a_(n+2) = a_(n-1). The initial conditions set the seeds: a_0 = y(0) and a_1 = y'(0). With y(0) = 1 and y'(0) = 0, a_3 = 1/6, a_4 = a_5 = 0 and a_6 = a_3/30 = 1/180, so y = 1 + x³/6 + x⁶/180 + ...

**Recap.** Substitute the series, shift indices, read the recurrence, and grow the coefficients step by step.

## Practice

14 questions on this page, each with its working shown.

## Needs first

- [Fourier Series](https://lightmysky.com/learn/mathematics/fourier-series-mt_DJ3iwfj7NK)

## Optional connections

- [Taylor and Maclaurin Series](https://lightmysky.com/learn/mathematics/taylor-and-maclaurin-series-mt_SgI9Pn9RIO)

## Opens up

- [The Laplace Transform](https://lightmysky.com/learn/mathematics/the-laplace-transform-mt_8j8GXzITN0)
