5.1 — Power Series Solutions Around Ordinary Points

Finding analytic solutions to differential equations using power series expansions

Learning Objectives
  • Distinguish between ordinary and singular points of a differential equation
  • Represent solutions as power series around an ordinary point
  • Derive and solve recurrence relations for power series coefficients
  • Determine the radius of convergence for power series solutions

Why Power Series?

Many important differential equations arising in physics and engineering have no closed-form solutions using the methods of Chapters 1–4. Power series solutions provide a systematic way to find analytic solutions in the form of infinite series. This approach is particularly valuable for:

Theory

Ordinary vs. Singular Points

Consider the second-order linear differential equation:

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

Divide by $P(x)$ (where $P(x) \neq 0$) to obtain the standard form:

$$y'' + p(x)y' + q(x)y = 0$$

where $p(x) = \frac{Q(x)}{P(x)}$ and $q(x) = \frac{R(x)}{P(x)}$.

Definition: Ordinary and Singular Points

A point $x_0$ is called an ordinary point of the differential equation if both $p(x)$ and $q(x)$ are analytic (have convergent Taylor series) at $x_0$. Otherwise, $x_0$ is a singular point.

Existence of Power Series Solutions

Theorem: Power Series Solution at an Ordinary Point

If $x_0$ is an ordinary point of $y'' + p(x)y' + q(x)y = 0$, then the general solution can be expressed as:

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

This series converges at least for $|x - x_0| < R$, where $R$ is the distance from $x_0$ to the nearest singular point.

Power Series Review

Key facts about power series that we'll use:

Recurrence Relations

When we substitute the power series $y = \sum a_n(x-x_0)^n$ into the differential equation, we obtain an identity in powers of $(x-x_0)$. Equating coefficients of like powers yields a recurrence relation — a formula expressing each coefficient in terms of earlier coefficients. The first few coefficients $a_0, a_1, \ldots$ are determined by initial or boundary conditions, and the recurrence relation then determines all subsequent coefficients uniquely.

Step-by-Step Method

  1. Verify $x_0$ is ordinary: Check that $p(x)$ and $q(x)$ are analytic at $x_0$.
  2. Assume a power series solution: Let $y = \sum_{n=0}^{\infty} a_n(x-x_0)^n$
  3. Compute derivatives:
    $$y' = \sum_{n=1}^{\infty} n a_n(x-x_0)^{n-1}, \quad y'' = \sum_{n=2}^{\infty} n(n-1)a_n(x-x_0)^{n-2}$$
  4. Substitute into the ODE: Replace $y, y', y''$ in the differential equation.
  5. Align powers of $(x-x_0)$: Shift indices so all sums have the same power $(x-x_0)^n$.
  6. Extract the recurrence relation: Set the coefficient of each power to zero. This gives a formula for $a_n$ in terms of earlier coefficients.
  7. Express general solution: Use initial conditions to determine $a_0$ and $a_1$. The recurrence then generates all other coefficients.
  8. Write the general solution: Combine the two linearly independent solutions (one with arbitrary $a_0, a_1=0$ and one with $a_0=0, a_1$ arbitrary).

Worked Examples

Setup: We solve $y'' - y = 0$ using power series around $x_0 = 0$. (Note: This has the known solution $y = c_1 e^x + c_2 e^{-x}$, so we can verify our result.)

Point type: $x_0 = 0$ is an ordinary point (no singular points).

Assume: $y = \sum_{n=0}^{\infty} a_n x^n$

Derivatives:

$$y' = \sum_{n=1}^{\infty} n a_n x^{n-1}, \quad y'' = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}$$

Substitute into $y'' - y = 0$:

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

Shift index: Let $m = n-2$ in the first sum:

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

Combine:

$$\sum_{n=0}^{\infty} \left[(n+2)(n+1) a_{n+2} - a_n\right] x^n = 0$$

Recurrence relation: Each coefficient must vanish:

$$a_{n+2} = \frac{a_n}{(n+2)(n+1)} \quad \text{for } n \geq 0$$

Two families of solutions:

  • Even indices ($a_1 = 0$): $a_2 = \frac{a_0}{2!}, a_4 = \frac{a_2}{4 \cdot 3} = \frac{a_0}{4!}, a_6 = \frac{a_0}{6!}, \ldots$ gives $y_1 = a_0 \cosh x$
  • Odd indices ($a_0 = 0$): $a_3 = \frac{a_1}{3 \cdot 2}, a_5 = \frac{a_1}{5!}, a_7 = \frac{a_1}{7!}, \ldots$ gives $y_2 = a_1 \sinh x$

General solution:

$$y = c_1 \cosh x + c_2 \sinh x$$

This matches the known solution $c_1 e^x + c_2 e^{-x}$ (with different constants).

Setup: The Airy equation $y'' - xy = 0$ has no solution in terms of elementary functions. This classic example demonstrates the power of series methods.

Assume: $y = \sum_{n=0}^{\infty} a_n x^n$

Derivatives:

$$y'' = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}$$

Substitute into $y'' - xy = 0$:

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

Shift indices: First sum: $m = n-2$; second sum: $m = n+1$

$$\sum_{m=0}^{\infty} (m+2)(m+1) a_{m+2} x^m - \sum_{m=1}^{\infty} a_{m-1} x^m = 0$$

Separate by power:

  • Coefficient of $x^0$: $2a_2 = 0 \Rightarrow a_2 = 0$
  • Coefficient of $x^n$ ($n \geq 1$): $(n+2)(n+1)a_{n+2} - a_{n-1} = 0$

Recurrence relation:

$$a_{n+2} = \frac{a_{n-1}}{(n+2)(n+1)} \quad \text{for } n \geq 1$$

First few terms:

$$y_1 = a_0\left(1 + \frac{x^3}{6} + \frac{x^6}{180} + \cdots\right), \quad y_2 = a_1\left(x + \frac{x^4}{12} + \frac{x^7}{504} + \cdots\right)$$

The solutions are called Airy functions $\text{Ai}(x)$ and $\text{Bi}(x)$, which are essential in quantum mechanics and wave theory.

Setup: This is Legendre's equation with $n=2$. It's important in electrostatics and gravitational theory.

Singular points: $P(x) = 1-x^2 = 0$ at $x = \pm 1$, so $R = 1$.

Assume: $y = \sum_{n=0}^{\infty} a_n x^n$

Compute derivatives:

$$y' = \sum_{n=1}^{\infty} n a_n x^{n-1}, \quad y'' = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}$$

Substitute:

$$(1-x^2)\sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} - 2x\sum_{n=1}^{\infty} n a_n x^{n-1} + 6\sum_{n=0}^{\infty} a_n x^n = 0$$

Expand and align powers: (After shifting indices)

$$\sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^n - \sum_{n=0}^{\infty} n(n-1)a_n x^n - \sum_{n=0}^{\infty} 2n a_n x^n + 6\sum_{n=0}^{\infty} a_n x^n = 0$$

Recurrence:

$$a_{n+2} = \frac{n(n+1) - 6}{(n+2)(n+1)}a_n = \frac{n^2 + n - 6}{(n+2)(n+1)}a_n = \frac{(n+3)(n-2)}{(n+2)(n+1)}a_n$$

Key observation: For $n=2$, we get $a_4 = \frac{(5)(0)}{(4)(3)}a_2 = 0$, and all subsequent even terms vanish. The solution terminates to a polynomial:

$$P_2(x) = \frac{1}{2}(3x^2 - 1)$$

This is the Legendre polynomial of degree 2, a terminating power series (polynomial).

Assume: $y = \sum_{n=0}^{\infty} a_n x^n$

Substitute:

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

Shift and align (first sum: $m=n-2$, second: $m=n+2$):

$$\sum_{m=0}^{\infty} (m+2)(m+1) a_{m+2} x^m + \sum_{m=2}^{\infty} a_{m-2} x^m = 0$$

Recurrence:

  • $2a_2 = 0 \Rightarrow a_2 = 0$
  • $6a_3 = 0 \Rightarrow a_3 = 0$
  • For $n \geq 2$: $(n+2)(n+1)a_{n+2} + a_{n-2} = 0 \Rightarrow a_{n+2} = -\frac{a_{n-2}}{(n+2)(n+1)}$

Solution forms:

$$y_1 = a_0\left(1 - \frac{x^4}{3 \cdot 4} + \frac{x^8}{3 \cdot 4 \cdot 7 \cdot 8} - \cdots\right)$$
$$y_2 = a_1\left(x - \frac{x^5}{4 \cdot 5} + \frac{x^9}{4 \cdot 5 \cdot 8 \cdot 9} - \cdots\right)$$

Setup: This is Hermite's equation with $n=2$. The initial conditions will force the solution to be a polynomial.

Assume: $y = \sum_{n=0}^{\infty} a_n x^n$

From ICs: $y(0) = a_0 = 1$ and $y'(0) = a_1 = 0$

Compute derivatives:

$$y' = \sum_{n=1}^{\infty} n a_n x^{n-1}, \quad y'' = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}$$

Substitute into $y'' - 2xy' + 4y = 0$:

$$\sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} - 2x\sum_{n=1}^{\infty} n a_n x^{n-1} + 4\sum_{n=0}^{\infty} a_n x^n = 0$$

Align powers (after shifting):

$$\sum_{n=0}^{\infty} \left[(n+2)(n+1)a_{n+2} - 2na_n + 4a_n\right] x^n = 0$$

Recurrence:

$$a_{n+2} = \frac{2n - 4}{(n+2)(n+1)} a_n = \frac{2(n-2)}{(n+2)(n+1)} a_n$$

Compute coefficients:

  • $a_0 = 1$, $a_1 = 0$
  • $a_2 = \frac{2(0-2)}{2 \cdot 1} \cdot 1 = \frac{-4}{2} = -2$
  • $a_3 = \frac{2(1-2)}{3 \cdot 2} \cdot 0 = 0$
  • $a_4 = \frac{2(2-2)}{4 \cdot 3} \cdot (-2) = 0$
  • All higher terms: $a_n = 0$ for $n > 2$

Solution:

$$y = 1 - 2x^2$$

The solution is exactly the Hermite polynomial $H_2(x) = 4x^2 - 2$ (up to scaling), proving that Hermite polynomials are solutions to Hermite's equation.

📝 Exam-Style Practice Problems

These problems match the style and difficulty of past exam questions on series solutions about an ordinary point. Work each one on paper first, then click to reveal the full solution.

Practice 1: Classify the points of $(x^2+4)y'' + xy' - y = 0$ and give the guaranteed radius of convergence about $x_0 = 0$ and $x_0 = 1$ ▼

Solution

Step 1: Put the equation in standard form

$$y'' + \underbrace{\frac{x}{x^2+4}}_{p(x)}y' + \underbrace{\frac{-1}{x^2+4}}_{q(x)}y = 0$$

Both $p$ and $q$ are rational functions, so they fail to be analytic only where the denominator vanishes.

Step 2: Locate the singular points

$$x^2 + 4 = 0 \quad\Longrightarrow\quad x = \pm 2i$$

There are no real singular points, so every real number is an ordinary point. But the radius of convergence is governed by the nearest singularity in the complex plane, and $\pm 2i$ certainly counts.

Step 3: Radius about $x_0 = 0$

$$R \ge |0 - 2i| = 2$$

Step 4: Radius about $x_0 = 1$

$$R \ge |1 - 2i| = \sqrt{1^2 + 2^2} = \sqrt{5} \approx 2.236$$ $$\boxed{\text{Ordinary everywhere on }\mathbb{R};\qquad R_{x_0=0} = 2,\qquad R_{x_0=1} = \sqrt{5}}$$
Exam tip: “No real singular points” does not mean $R = \infty$. Always factor the leading coefficient over $\mathbb{C}$.
Practice 2: Find the recurrence relation and the first four nonzero terms of two linearly independent solutions of $y'' + xy' + 2y = 0$ about $x_0 = 0$ ▼

Solution

Step 1: Check the point and substitute the series

Here $p(x) = x$ and $q(x) = 2$ are polynomials, so $x_0 = 0$ is an ordinary point and $R = \infty$. Put $y = \sum_{n\ge0} a_n x^n$:

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

Step 2: Shift the first sum so every term carries $x^n$

$$\sum_{n=0}^{\infty}\Big[(n+2)(n+1)a_{n+2} + n a_n + 2a_n\Big]x^n = 0$$

(The $n a_n$ term contributes nothing at $n = 0$, so the bracket is valid for all $n \ge 0$.)

Step 3: Simplify the recurrence

$$(n+2)(n+1)a_{n+2} + (n+2)a_n = 0 \quad\Longrightarrow\quad \boxed{a_{n+2} = -\frac{a_n}{n+1}},\qquad n \ge 0$$

The factor $(n+2)$ cancels — a very common place to lose a step.

Step 4: Even family ($a_0 = 1,\ a_1 = 0$)

$$a_2 = -\frac{a_0}{1} = -1,\qquad a_4 = -\frac{a_2}{3} = \frac{1}{3},\qquad a_6 = -\frac{a_4}{5} = -\frac{1}{15},\qquad a_8 = -\frac{a_6}{7} = \frac{1}{105}$$ $$y_1(x) = 1 - x^2 + \frac{x^4}{3} - \frac{x^6}{15} + \frac{x^8}{105} - \cdots$$

Step 5: Odd family ($a_0 = 0,\ a_1 = 1$)

$$a_3 = -\frac{a_1}{2} = -\frac{1}{2},\qquad a_5 = -\frac{a_3}{4} = \frac{1}{8},\qquad a_7 = -\frac{a_5}{6} = -\frac{1}{48}$$ $$y_2(x) = x - \frac{x^3}{2} + \frac{x^5}{8} - \frac{x^7}{48} + \cdots = x\,e^{-x^2/2}$$

Check: with $y = xe^{-x^2/2}$ we get $y' = (1-x^2)e^{-x^2/2}$ and $y'' = (x^3-3x)e^{-x^2/2}$, so

$$y'' + xy' + 2y = \big[(x^3-3x) + x(1-x^2) + 2x\big]e^{-x^2/2} = 0. \;\checkmark$$ $$\boxed{y = c_1\left(1 - x^2 + \frac{x^4}{3} - \frac{x^6}{15} + \cdots\right) + c_2\left(x - \frac{x^3}{2} + \frac{x^5}{8} - \frac{x^7}{48} + \cdots\right)}$$
Practice 3: Show that $y'' - 2xy' + 4y = 0$ has a polynomial solution, and find the second (non-terminating) solution ▼

Solution

Step 1: Derive the recurrence

This is Hermite's equation with $2\nu = 4$, i.e. $\nu = 2$. Substituting $y = \sum a_n x^n$ and collecting $x^n$:

$$(n+2)(n+1)a_{n+2} - 2n a_n + 4a_n = 0 \quad\Longrightarrow\quad a_{n+2} = \frac{2(n-2)}{(n+2)(n+1)}\,a_n$$

Step 2: Spot the termination

The numerator $2(n-2)$ vanishes at $n = 2$, so $a_4 = 0$ — and then $a_6 = a_8 = \cdots = 0$. The even branch terminates.

$$a_2 = \frac{2(0-2)}{2\cdot 1}a_0 = -2a_0,\qquad a_4 = 0$$ $$y_1(x) = 1 - 2x^2 \qquad (\text{taking } a_0 = 1)$$

Check: $y_1'' - 2xy_1' + 4y_1 = -4 - 2x(-4x) + 4(1-2x^2) = -4 + 8x^2 + 4 - 8x^2 = 0.\;\checkmark$

Step 3: The odd branch never terminates

With $a_1 = 1$ the numerator $2(n-2)$ is never zero for odd $n$:

$$a_3 = \frac{2(1-2)}{3\cdot 2}a_1 = -\frac{1}{3},\qquad a_5 = \frac{2(3-2)}{5\cdot 4}a_3 = \frac{a_3}{10} = -\frac{1}{30},\qquad a_7 = \frac{2(5-2)}{7\cdot 6}a_5 = \frac{a_5}{7} = -\frac{1}{210}$$ $$y_2(x) = x - \frac{x^3}{3} - \frac{x^5}{30} - \frac{x^7}{210} - \cdots$$

Step 4: General solution

$$\boxed{y = c_1\left(1 - 2x^2\right) + c_2\left(x - \frac{x^3}{3} - \frac{x^5}{30} - \frac{x^7}{210} - \cdots\right)}$$
Exam tip: a series solution terminates exactly when the recurrence numerator hits zero. For $y'' - 2xy' + 2\nu y = 0$ that happens iff $\nu$ is a non-negative integer — this is precisely how the Hermite polynomials arise.

Practice Problems

Test your understanding with these multiple-choice questions. Each correct answer earns one point.

0 / 6 correct

Q1: Which of the following is a singular point of $(1-x^2)y'' - 2xy' + n(n+1)y = 0$?

Q2: A series solution obeys the recurrence $a_{n+2} = \dfrac{a_n}{(n+2)(n+1)}$ for $n \ge 0$. What is $a_4$ in terms of $a_0$?

Q3: For the Airy equation $y'' - xy = 0$ the even-index solution begins $y_1 = a_0\left(1 + \dfrac{x^3}{6} + \cdots\right)$. What is the coefficient of $x^6$?

Q4: What radius of convergence is guaranteed for a series solution of $(x^2+4)y'' + 3xy' + y = 0$ about $x_0 = 0$?

Q5: For $(x^2+9)y'' + xy' + y = 0$, what radius of convergence is guaranteed for a series expanded about $x_0 = 2$?

Q6: Substituting $y = \sum_{n\ge0} a_n x^n$ into $y'' + xy = 0$ and shifting so every sum carries $x^n$ gives which relation?

Quick Reference Card

Power Series Definition

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

Converges for $|x-x_0| < R$

First Derivative

$y' = \sum_{n=1}^{\infty} n a_n(x-x_0)^{n-1}$

Term-by-term differentiation

Second Derivative

$y'' = \sum_{n=2}^{\infty} n(n-1)a_n(x-x_0)^{n-2}$

Valid in the interval of convergence

Index Shift Formula

$\sum_{n=k}^{\infty} a_n x^{n-r} = \sum_{m=k-r}^{\infty} a_{m+r} x^m$

Use to align powers

Radius of Convergence

$R = $ distance to nearest singular point

Find singular points from $p(x), q(x)$

Recurrence Relation

Solve: $a_{n} = f(a_{n-1}, a_{n-2}, \ldots)$

Determined by equating coefficients