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)

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

Recurrence:

$$a_{n+2} = -\frac{(n+2)(n+1) - 6}{(n+2)(n+1)}a_n = -\frac{n^2 + 3n - 4}{(n+2)(n+1)}a_n$$

Key observation: For $n=2$, we get $a_4 = -\frac{4+6-4}{4 \cdot 3}a_2 = 0$, and all subsequent even terms vanish if we choose $a_2$ appropriately. 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.

Practice Problems

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

0 / 6 correct

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

Question 2: For the recurrence relation $a_{n+2} = \frac{a_n}{(n+2)(n+1)}$, what is $a_4$ in terms of $a_0$?

Question 3: For the Airy equation $y'' - xy = 0$, what is the coefficient of $x^6$ in the solution $y_1 = a_0(1 + \frac{x^3}{6} + \cdots)$?

Question 4: A power series solution $y = \sum_{n=0}^{\infty} a_n x^n$ to a DE with radius of convergence $R = \infty$ means:

Question 5: For the equation $y'' - xy' - y = 0$, which statement about the radius of convergence $R$ around $x_0 = 0$ is correct?

Question 6: In the power series method, we derive a recurrence relation by:

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