---
title: Power Series Solutions Near Ordinary Points
module: Series Solutions and Special Functions
moduleNumber: 4
lessonNumber: 1
order: 401
summary: >
  A linear equation with variable coefficients has no characteristic equation.
  A power series substituted into the equation matches coefficients to a
  recurrence relation, which near an ordinary point yields two independent
  analytic solutions. The radius of convergence is at least the distance from
  the expansion point to the nearest singular point in the complex plane.
topics: [Series Solutions and Special Functions]
sources:
  - book: Boyce
    ref: "Ch. 5 — Series Solutions of Second-Order Linear Equations; §5.1 Review of Power Series; §5.2 Series Solutions Near an Ordinary Point, Part I; §5.3 Part II"
  - book: Simmons
    ref: "Ch. 5 §26 Introduction. A Review of Power Series; §27 Series Solutions of First Order Equations; §28 Second Order Linear Equations. Ordinary Points"
draft: false
---

The [constant-coefficient methods](/differential-equations/second-order-linear/homogeneous-constant-coefficients) reduce $ay'' + by' + cy = 0$ to algebra through the substitution $y = e^{rt}$. Most equations of mathematical physics have variable coefficients, and no exponential guess works. Two recurring examples are

$$
\underbrace{x^2 y'' + x y' + (x^2 - \nu^2)\,y = 0}_{\text{Bessel's equation}},
\qquad
\underbrace{(1 - x^2)\,y'' - 2x y' + \alpha(\alpha + 1)\,y = 0}_{\text{Legendre's equation}},
$$

which arise from heat flow and wave motion in cylinders and from potential theory in spheres.[^boyce-intro] Their coefficients are polynomials in $x$, and no exponential guess works. The replacement strategy is older than either equation: assume the solution is a power series

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

substitute it into the equation, and solve for the coefficients $a_n$ one at a time. The method is guaranteed to work near an **ordinary point** of the equation, and the interval on which the resulting series converges can be read off in advance.

## Power series preliminaries

A power series about $x_0$ is an expression $\sum_{n=0}^{\infty} a_n (x - x_0)^n$. The full theory belongs to [calculus](/calculus/sequences-and-series/power-series); the working facts are these.[^boyce-review]

- **Convergence set.** Every power series has a radius of convergence $\rho \in [0, \infty]$: the series converges absolutely for $|x - x_0| < \rho$ and diverges for $|x - x_0| > \rho$. The ratio test computes $\rho$ when the limit $\lim_{n\to\infty} |a_{n+1}/a_n|$ exists.
- **Termwise calculus.** Inside $|x - x_0| < \rho$ the series may be differentiated and integrated term by term; the derived series have the same radius of convergence.
- **Algebra.** Two series about the same point may be added, multiplied, and (where the denominator is nonzero) divided, with the results valid on the common interval of convergence.
- **Identity theorem.** If $\sum a_n (x - x_0)^n = \sum b_n (x - x_0)^n$ for all $x$ in some open interval, then $a_n = b_n$ for every $n$. In particular, a power series that sums to zero on an interval has every coefficient equal to zero. The whole method rests on this fact.

> **Definition (Analytic function).** A function $f$ is **analytic at $x_0$** if it equals the sum of a power series $\sum a_n (x - x_0)^n$ with positive radius of convergence on a neighborhood of $x_0$. In that case the series is the Taylor series of $f$, so $a_n = f^{(n)}(x_0)/n!$. Polynomials, $e^x$, $\sin x$, $\cos x$ are analytic everywhere; a ratio of polynomials is analytic wherever its denominator is nonzero.

### Shifting the index of summation

The method requires rewriting sums so that every series displays the same generic power of $x$. The rule: replacing $n$ by $n + k$ inside the sum, while starting the count $k$ lower, leaves the series unchanged. For example,

$$
\sum_{n=2}^{\infty} n(n-1)\,a_n\,x^{n-2}
\;=\;
\sum_{n=0}^{\infty} (n+2)(n+1)\,a_{n+2}\,x^{n},
$$

as writing out the first few terms of each side confirms.[^boyce-shift] Combined with the identity theorem, index shifting already solves equations. Suppose a function satisfies $y' = y$ with $y = \sum a_n x^n$. Then

$$
\sum_{n=1}^{\infty} n a_n x^{n-1} = \sum_{n=0}^{\infty} a_n x^n
\quad\Longrightarrow\quad
\sum_{n=0}^{\infty} (n+1)\,a_{n+1} x^n = \sum_{n=0}^{\infty} a_n x^n,
$$

so $(n+1)a_{n+1} = a_n$ for every $n$, giving $a_n = a_0/n!$ and $y = a_0 \sum x^n/n! = a_0 e^x$. The same procedure works for any first-order linear equation at a point where $Q/P$ is analytic.[^simmons-first]

## Ordinary and singular points

Fix the homogeneous second-order equation

$$
P(x)\,y'' + Q(x)\,y' + R(x)\,y = 0,
$$

where $P$, $Q$, $R$ are polynomials with no factor common to all three (if there is one, divide it out first).

> **Definition (Ordinary and singular points).** A point $x_0$ is an **ordinary point** of the equation if $P(x_0) \neq 0$ — equivalently, if $p = Q/P$ and $q = R/P$ are analytic at $x_0$. If $P(x_0) = 0$, then $x_0$ is a **singular point**: at least one of $p$, $q$ becomes unbounded as $x \to x_0$, and the basic existence theorem no longer applies there.

Near an ordinary point the equation divides through by $P$ to give $y'' + p(x)y' + q(x)y = 0$ with continuous coefficients, so the [existence and uniqueness theorem](/differential-equations/first-order/existence-uniqueness-euler) supplies a unique solution for any initial data $y(x_0) = y_0$, $y'(x_0) = y_0'$. This solution is analytic, not merely continuous, and can be computed term by term.

Which points are singular is decided in the complex plane, not just on the real line, and this matters for convergence. The equation $(1 + x^2)y'' + 2xy' + 4x^2 y = 0$ has no real singular points at all, yet $P(x) = 1 + x^2$ vanishes at $x = \pm i$, and those complex zeros limit every real series solution centered at the origin to radius $1$.[^boyce-part2]

$$
% caption: Singular points live in the complex plane. For $(1+x^2)y'' + 2xy' + 4x^2y = 0$
% the zeros of $P$ sit at $\pm i$; a series solution centered at the origin is guaranteed
% only inside the disk reaching the nearest one, so its real interval is $|x| < 1$.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (-2.6,0) -- (2.9,0) node[below] {Re};
\draw[->, black] (0,-2.0) -- (0,2.3) node[left] {Im};
\draw[acc, dashed, thick] (0,0) circle (1.5);
\fill[acc] (0,0) circle (2.2pt);
\node[acc, anchor=north east] at (-0.08,-0.08) {$x_0 = 0$};
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\draw[black, thick] (-0.09,1.59) -- (0.09,1.41);
\draw[black, thick] (-0.09,-1.41) -- (0.09,-1.59);
\draw[black, thick] (-0.09,-1.59) -- (0.09,-1.41);
\node[anchor=west] at (0.2,1.62) {singular point $x = i$};
\node[anchor=west] at (0.2,-1.62) {singular point $x = -i$};
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\node[acc, anchor=west, align=left] at (0.72,0.52) {distance 1};
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\draw[->, acc] (1.7,-0.9) .. controls (1.55,-0.82) .. (1.34,-0.79);
\end{tikzpicture}
$$

## The method at an ordinary point

Let $x_0$ be an ordinary point. Assume a solution

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

convergent on some interval $|x - x_0| < \rho$, differentiate termwise, substitute into the equation, expand the coefficients $P, Q, R$ in powers of $x - x_0$ where needed, shift indices until one generic power remains, and set each coefficient to zero. The result is a **recurrence relation** expressing later coefficients in terms of earlier ones, while $a_0$ and $a_1$ stay free and fix the initial data $y(x_0)$ and $y'(x_0)$.

```algorithm
caption: $\textsc{SeriesSolve}(P, Q, R, x_0)$ — power-series solution at an ordinary point
require $P(x_0) \neq 0$
assume $y = \sum_{n=0}^{\infty} a_n (x - x_0)^n$ with $\rho > 0$
compute $y' = \sum_{n \ge 1} n\,a_n (x - x_0)^{n-1}$ and $y'' = \sum_{n \ge 2} n(n-1)\,a_n (x - x_0)^{n-2}$
substitute into $P y'' + Q y' + R y = 0$, writing $P, Q, R$ in powers of $x - x_0$
shift indices until every sum displays the generic power $(x - x_0)^n$
for each $n \ge 0$ do
  set the coefficient of $(x - x_0)^n$ to zero, solving for the highest-index $a$
solve the resulting recurrence along the chain seeded by $a_0$ and the chain seeded by $a_1$
return $y = a_0\,y_1(x) + a_1\,y_2(x)$
```

Take the equation $y'' + y = 0$, where every point is ordinary ($P = 1$) and the answer is known in advance, as a first test of the method.

> **Worked example.** Solve $y'' + y = 0$ by series about $x_0 = 0$.[^boyce-ex1] Substitute the series and its second derivative:
>
> $$
> \sum_{n=2}^{\infty} n(n-1)a_n x^{n-2} + \sum_{n=0}^{\infty} a_n x^n = 0
> \quad\Longrightarrow\quad
> \sum_{n=0}^{\infty} \Big[ (n+2)(n+1)\,a_{n+2} + a_n \Big] x^n = 0.
> $$
>
> By the identity theorem every bracket vanishes:
>
> $$
> a_{n+2} = -\frac{a_n}{(n+2)(n+1)}, \qquad n = 0, 1, 2, \ldots
> $$
>
> The recurrence steps by two, so the even and odd coefficients form independent chains, one seeded by each free constant. Running the even chain, $a_2 = -a_0/2!$, $a_4 = +a_0/4!$, $a_6 = -a_0/6!$, and in general $a_{2k} = (-1)^k a_0/(2k)!$; the odd chain gives $a_{2k+1} = (-1)^k a_1/(2k+1)!$. Assembling,
>
> $$
> y = a_0 \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}\,x^{2n}
>   + a_1 \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}\,x^{2n+1}
>   = a_0 \cos x + a_1 \sin x .
> $$

$$
% caption: The recurrence $a_{n+2} = -a_n/((n+2)(n+1))$ splits the coefficients into two
% independent chains; the seeds $a_0$ and $a_1$ are the initial data $y(0)$ and $y'(0)$,
% and each chain assembles one member of the fundamental set.
\begin{tikzpicture}[>=stealth, font=\footnotesize,
  c/.style={draw, black, minimum width=11mm, minimum height=8mm, align=center},
  seed/.style={draw, thick, minimum width=11mm, minimum height=8mm, align=center}]
\definecolor{acc}{HTML}{4A6FA5}
\node[seed, draw=acc, text=acc] (a0) at (0,1.2) {$a_0$};
\node[c] (a2) at (2.4,1.2) {$a_2$};
\node[c] (a4) at (4.8,1.2) {$a_4$};
\node[c] (a6) at (7.2,1.2) {$a_6$};
\node at (8.7,1.2) {$\cdots$};
\node[seed, draw=acc, text=acc] (a1) at (0,-1.2) {$a_1$};
\node[c] (a3) at (2.4,-1.2) {$a_3$};
\node[c] (a5) at (4.8,-1.2) {$a_5$};
\node[c] (a7) at (7.2,-1.2) {$a_7$};
\node at (8.7,-1.2) {$\cdots$};
\draw[->, acc] (a0) -- (a2) node[midway, above] {$n = 0$};
\draw[->, acc] (a2) -- (a4) node[midway, above] {$n = 2$};
\draw[->, acc] (a4) -- (a6) node[midway, above] {$n = 4$};
\draw[->, acc] (a1) -- (a3) node[midway, above] {$n = 1$};
\draw[->, acc] (a3) -- (a5) node[midway, above] {$n = 3$};
\draw[->, acc] (a5) -- (a7) node[midway, above] {$n = 5$};
\node[anchor=west, text=black] at (-0.7,2.2) {even chain: builds $y_1(x) = \cos x$};
\node[anchor=west, text=black] at (-0.7,-2.2) {odd chain: builds $y_2(x) = \sin x$};
\end{tikzpicture}
$$

Both series pass the ratio test for every $x$, which justifies the termwise operations. The two bracketed series satisfy $y_1(0) = 1, y_1'(0) = 0$ and $y_2(0) = 0, y_2'(0) = 1$, so their Wronskian at $0$ is $1$ and they form a fundamental set. Nothing in the computation used prior knowledge of sine and cosine; had these functions been unknown, the series would _define_ them, and their standard properties ($S' = C$, $C' = -S$, parity) all follow from the series.[^boyce-define] Several classical special functions have no more elementary definition than this.

Airy's equation $y'' - xy = 0$ has non-constant coefficients and solutions that are not elementary functions, so the series is the only way in.

> **Worked example.** Solve Airy's equation $y'' - xy = 0$ by series about $x_0 = 0$.[^boyce-airy] Every point is ordinary. Substituting the series and shifting the index,
>
> $$
> 2 a_2 + \sum_{n=1}^{\infty} \Big[ (n+2)(n+1)\,a_{n+2} - a_{n-1} \Big] x^n = 0,
> $$
>
> so $a_2 = 0$ and
>
> $$
> a_{n+2} = \frac{a_{n-1}}{(n+2)(n+1)}, \qquad n \ge 1.
> $$
>
> The recurrence reaches back three indices, so the coefficients split into three chains: $a_0$ feeds $a_3, a_6, a_9, \ldots$; $a_1$ feeds $a_4, a_7, a_{10}, \ldots$; and $a_2 = 0$ forces $a_5 = a_8 = \cdots = 0$. Running the $a_0$ chain from the recurrence,
>
> $$
> a_3 = \frac{a_0}{2 \cdot 3}, \qquad
> a_6 = \frac{a_0}{2 \cdot 3 \cdot 5 \cdot 6}, \qquad
> a_9 = \frac{a_0}{2 \cdot 3 \cdot 5 \cdot 6 \cdot 8 \cdot 9},
> $$
>
> and the $a_1$ chain gives $a_4 = a_1/(3 \cdot 4)$, $a_7 = a_1/(3 \cdot 4 \cdot 6 \cdot 7)$, $a_{10} = a_1/(3 \cdot 4 \cdot 6 \cdot 7 \cdot 9 \cdot 10)$. Assembling both,
>
> $$
> y(x) = a_0 \left[ 1 + \frac{x^3}{2 \cdot 3} + \frac{x^6}{2 \cdot 3 \cdot 5 \cdot 6} + \cdots \right]
>      + a_1 \left[ x + \frac{x^4}{3 \cdot 4} + \frac{x^7}{3 \cdot 4 \cdot 6 \cdot 7} + \cdots \right]
>      = a_0\,y_1(x) + a_1\,y_2(x),
> $$
>
> with both series convergent for all $x$ by the ratio test.

The solutions oscillate with growing frequency and shrinking amplitude for $x < 0$ and grow monotonically for $x > 0$. The behavior is visible from the equation itself: $y'' = xy$ makes solutions curve away from the axis where $x > 0$ (exponential character) and toward it where $x < 0$ (oscillatory character). Airy functions are not expressible in elementary terms, but as solutions of an initial-value problem they are concrete computational objects.

> **Worked example.** Solve the same equation $y'' - xy = 0$ by series about the ordinary point $x_0 = 1$.[^boyce-airy-shift] Write $y = \sum a_n (x - 1)^n$ and expand the coefficient as $x = 1 + (x - 1)$ so that both sides run in powers of $x - 1$. Substituting and shifting,
>
> $$
> \sum_{n=0}^{\infty} (n+2)(n+1)\,a_{n+2}(x-1)^n = \sum_{n=0}^{\infty} a_n (x-1)^n + \sum_{n=1}^{\infty} a_{n-1}(x-1)^n,
> $$
>
> which gives a three-term recurrence,
>
> $$
> 2a_2 = a_0, \qquad (n+2)(n+1)\,a_{n+2} = a_n + a_{n-1}, \quad n \ge 1.
> $$
>
> Solving in order,
>
> $$
> a_2 = \frac{a_0}{2}, \quad
> a_3 = \frac{a_0}{6} + \frac{a_1}{6}, \quad
> a_4 = \frac{a_0}{24} + \frac{a_1}{12}, \quad
> a_5 = \frac{a_0}{30} + \frac{a_1}{120},
> $$
>
> so
>
> $$
> y = a_0\left[ 1 + \frac{(x-1)^2}{2} + \frac{(x-1)^3}{6} + \frac{(x-1)^4}{24} + \cdots \right]
>   + a_1\left[ (x-1) + \frac{(x-1)^3}{6} + \frac{(x-1)^4}{12} + \cdots \right].
> $$
>
> A three-term recurrence rarely has a closed-form general coefficient, but as many terms as needed follow in order. Since $P = 1$ never vanishes, both series converge for all $x$.

## Accuracy of truncated series

The series is exact; any finite partial sum is a local approximation, accurate near $x_0$ and worthless far away. The polynomial approximations to $\cos x$ show the pattern: each added pair of terms extends the interval of useful accuracy, but every polynomial eventually leaves the bounded oscillation behind.

$$
% caption: Partial sums of the series solution of $y'' + y = 0$. Degree 4 tracks
% $\cos x$ to about $x = 2$, degree 8 to about $x = 4.5$; every truncation eventually
% peels away, so a partial sum is only a local approximation near the expansion point.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (-0.3,0) -- (8.9,0) node[below] {$x$};
\draw[->, black] (0,-1.7) -- (0,1.7) node[left] {$y$};
\foreach \x/\l in {1.6/1, 3.2/2, 4.8/3, 6.4/4, 8.0/5}
  \draw[black] (\x,0.05) -- (\x,-0.05) node[below] {\l};
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$$

A practical error estimate needs no theory: plot two consecutive partial sums and trust the approximation only where their graphs have not yet separated.[^boyce-airy]

## Radius of convergence

Fuchs settled the convergence question in 1866.[^boyce-fuchs]

> **Theorem (Series solutions at an ordinary point).** Let $x_0$ be an ordinary point of $P(x)y'' + Q(x)y' + R(x)y = 0$, so that $p = Q/P$ and $q = R/P$ are analytic at $x_0$. Then the general solution is
> $$
> y = \sum_{n=0}^{\infty} a_n (x - x_0)^n = a_0\,y_1(x) + a_1\,y_2(x),
> $$
> where $a_0, a_1$ are arbitrary and $y_1, y_2$ form a fundamental set of solutions, each analytic at $x_0$. The radius of convergence of each of $y_1, y_2$ is at least as large as the minimum of the radii of convergence of the series for $p$ and $q$.

For polynomial coefficients the lower bound becomes geometric. If $Q/P$ is in lowest terms, its Taylor series about $x_0$ converges out to the nearest zero of $P$ in the **complex plane**. So:

> **Corollary (Radius from the singular points).** If $P, Q, R$ are polynomials and $P(x_0) \neq 0$, every series solution about $x_0$ converges at least for
> $$
> |x - x_0| < \rho = \min \{ |x_0 - s| : P(s) = 0,\ s \in \mathbb{C} \},
> $$
> the distance from $x_0$ to the nearest complex zero of $P$.

The bound is computed by inspection, before any coefficients are found. Worked instances:[^boyce-part2]

| Equation | $P(x)$ | Zeros of $P$ | Center $x_0$ | Guaranteed radius |
| --- | --- | --- | --- | --- |
| $y'' + y = 0$ | $1$ | none | any | $\infty$ |
| $y'' - xy = 0$ (Airy) | $1$ | none | any | $\infty$ |
| $(1 - x^2)y'' - 2xy' + \alpha(\alpha+1)y = 0$ (Legendre) | $1 - x^2$ | $\pm 1$ | $0$ | $1$ |
| $(1 + x^2)y'' + 2xy' + 4x^2 y = 0$ | $1 + x^2$ | $\pm i$ | $0$ | $1$ |
| same | $1 + x^2$ | $\pm i$ | $-\tfrac12$ | $\sqrt{5}/2$ |
| $(x^2 - 2x + 2)^{-1}$ as a coefficient | $x^2 - 2x + 2$ | $1 \pm i$ | $0$ | $\sqrt{2}$ |

Two remarks:

- **It is only a lower bound.** For Legendre's equation with $\alpha$ a nonnegative integer, one of the two series terminates: the solution is a polynomial and converges for all $x$, far beyond the guaranteed $|x| < 1$. These polynomial solutions are the Legendre polynomials of the [special-functions lesson](/differential-equations/series-solutions/bessel-and-special-functions).
- **Analytic, not just polynomial, coefficients work.** For $y'' + (\sin x)y' + (1 + x^2)y = 0$, both coefficients are entire (their Taylor series converge everywhere), so series solutions about $0$ converge for all $x$ even though the coefficients are not polynomials.

$$
% caption: For Legendre's equation the zeros of $P(x) = 1 - x^2$ sit at $x = 1$ and
% $x = -1$; a series about the origin is guaranteed on $|x| < 1$ only. When the
% parameter is a nonnegative integer one solution terminates and outruns the bound.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (-4.2,0) -- (4.5,0) node[below] {$x$};
\foreach \x/\l in {-2.6/-1, 0/0, 2.6/1}
  \draw[black] (\x,0.07) -- (\x,-0.07) node[below=1pt] {$\l$};
\draw[acc, very thick] (-2.6,0) -- (2.6,0);
\fill[acc] (0,0) circle (2.4pt);
\node[acc, anchor=south] at (0,0.42) {$x_0 = 0$};
\draw[black, thick] (-2.72,-0.12) -- (-2.48,0.12);
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\draw[black, thick] (2.48,-0.12) -- (2.72,0.12);
\draw[black, thick] (2.48,0.12) -- (2.72,-0.12);
\node[anchor=south, align=center] at (-2.6,0.35) {singular point};
\node[anchor=south, align=center] at (2.6,0.35) {singular point};
\draw[->, acc] (0,-0.75) -- (2.55,-0.75) node[midway, below] {guaranteed radius $= 1$};
\draw[acc] (0,-0.62) -- (0,-0.88);
\end{tikzpicture}
$$

> **Worked example.** Find a lower bound for the radius of convergence of series solutions of $(1 + x^2)y'' + 2xy' + 4x^2 y = 0$ about $x_0 = 0$ and about $x_0 = -\tfrac12$.[^boyce-radius] Here $P(x) = 1 + x^2$ has zeros $x = \pm i$; no series work is needed, only two distances in the complex plane:
>
> $$
> \rho_0 = |0 - i| = 1,
> \qquad
> \rho_{-1/2} = \left| -\tfrac12 - i \right| = \sqrt{\tfrac14 + 1} = \frac{\sqrt 5}{2}.
> $$
>
> A series about $0$ converges at least on $|x| < 1$; a series about $-\tfrac12$ converges at least on $\left| x + \tfrac12 \right| < \sqrt 5/2$.

The equation has no real singular points, so the basic existence theorem gives a unique solution on all of $(-\infty, \infty)$; the series theorem only guarantees a power-series representation of that solution on the disk above. The solution exists globally, but its Taylor series about a point need not converge globally. Existence of the solution and convergence of one particular representation of it are different questions.

## Failure at singular points

At a singular point $P(x_0) = 0$, the coefficient functions $p, q$ blow up, Fuchs' theorem gives no guarantee, and the solutions themselves typically fail to be analytic: near a singular point they may grow without bound, oscillate infinitely fast, or involve fractional powers and logarithms that no Taylor series can represent. Singular points cannot simply be avoided. In applications they are usually where the geometry concentrates (the center of the membrane, the ends of the interval), and the physically relevant question is how solutions behave as $x \to x_0$. The singular points mild enough to handle are the **regular singular points**, and the power series generalizes there to the Frobenius form $x^r \sum a_n x^n$, treated in [the Frobenius method](/differential-equations/series-solutions/regular-singular-frobenius).

[^boyce-intro]: **Boyce**, _Elementary Differential Equations and Boundary Value Problems_, §5.2 — the Bessel and Legendre equations as the motivating examples of polynomial-coefficient equations from mathematical physics.
[^boyce-review]: **Boyce**, §5.1 — Review of Power Series: radius of convergence, termwise differentiation, and equality of coefficients; **Simmons**, _Differential Equations with Applications and Historical Notes_, §26.
[^boyce-shift]: **Boyce**, §5.1, Examples 3–6 — shifting the index of summation and the $y' = y$ coefficient computation.
[^simmons-first]: **Simmons**, §27 — Series Solutions of First Order Equations: the ordinary-point condition and radius bound for $P(x)y' + Q(x)y = 0$.
[^boyce-ex1]: **Boyce**, §5.2, Example 1 — the full series computation for $y'' + y = 0$, the recurrence relation, and the induction argument for the closed-form coefficients.
[^boyce-define]: **Boyce**, §5.2 — defining $C(x)$ and $S(x)$ by their series and recovering the sine/cosine properties from the differential equation alone.
[^boyce-airy]: **Boyce**, §5.2, Example 2 — Airy's equation: the step-three recurrence, the coefficient chains, and convergence for all $x$; the graphical rule of comparing consecutive partial sums.
[^boyce-airy-shift]: **Boyce**, §5.2, Example 3 — Airy's equation solved in powers of $x - 1$: the coefficient $x = 1 + (x-1)$, the three-term recurrence $(n+2)(n+1)a_{n+2} = a_n + a_{n-1}$, and the first coefficients.
[^boyce-radius]: **Boyce**, §5.3, Example 5 — the radius-of-convergence bound for $(1 + x^2)y'' + 2xy' + 4x^2 y = 0$ about $0$ and about $-\tfrac12$ from the complex zeros of $P$.
[^boyce-fuchs]: **Boyce**, §5.3, Theorem 5.3.1 — analyticity of the fundamental set at an ordinary point and the radius-of-convergence bound; the complex-plane distance computations of Examples 2–5; **Simmons**, §28.