Power Series Solutions at an Ordinary Point

Assume the solution is a power series with unknown coefficients, substitute it into the equation, and collect like powers of $x$ . Each power gives one equation, and together they form a recurrence relation that determines every coefficient from the first two. Those two stay free, and they are the two arbitrary constants a second-order equation must have.

Quick Reference

Item	Statement
Equation in standard form	y'' + P(x)y' + Q(x)y = 0
Ordinary point	$x_{0}$ where $P$ and $Q$ are both analytic
Singular point	Any point that is not ordinary
Existence theorem	At an ordinary point, two independent solutions $\sum a_{k} {(x - x_{0})}^{k}$ exist
Guaranteed radius	At least the distance from $x_{0}$ to the nearest singular point, in the complex plane
Series for $y$	$y = \sum_{k = 0}^{\infty} a_{k} x^{k}$
Series for y'	y' = \displaystyle\sum_{k=1}^{\infty}ka_kx^{k-1}
Series for y''	y'' = \displaystyle\sum_{k=2}^{\infty}k(k-1)a_kx^{k-2}
Free constants	$a_{0} = y (0)$ and a_1 = y'(0)
Simplification	Take $x_{0} = 0$ after the substitution $X = x - x_{0}$

Ordinary and Singular Points

Consider y'' + P(x)y' + Q(x)y = 0. A point $x_{0}$ is an ordinary point of the equation if both $P (x)$ and $Q (x)$ are analytic at $x_{0}$ , meaning each has a power series expansion valid in some interval around $x_{0}$ . A point that is not ordinary is a singular point.

For an equation written as a_0(x)y'' + a_1(x)y' + a_2(x)y = 0 with polynomial coefficients, the singular points are exactly the zeros of the leading coefficient $a_{0} (x)$ , after any common factors have been cancelled.

Existence theorem. If $x_{0}$ is an ordinary point, then the equation has two linearly independent solutions of the form

y = \sum_{k = 0}^{\infty} a_{k} {(x - x_{0})}^{k},

and each series converges at least on the interval $| x - x_{0} | < ρ$ , where $ρ$ is the distance from $x_{0}$ to the nearest singular point in the complex plane.

The words "in the complex plane" are not decoration. For \left(1+x^{2}\right)y'' + y = 0, the coefficient $1 + x^{2}$ never vanishes for real $x$ , yet the singular points $x = \pm i$ are at distance $1$ from the origin. So the series about $x = 0$ is guaranteed only for $| x | < 1$ , even though nothing visible goes wrong on the real line.

From now on we take $x_{0}$ as the origin, that is, $x_{0} = 0$ . If $x_{0}$ is not already zero, the substitution $X = x - x_{0}$ moves the ordinary point to the origin without changing anything essential.

The Method

Write 
  y
  =
  
      ∑
      
        k
        =
        0
      
        ∞
      
      a
      k
    
      x
      
        k
      
 and differentiate term by term to obtain y' and y''.
Substitute into the differential equation.
Shift indices so that every series carries the same power 
  
    x
    
      n
    
.
Combine into a single series and set the coefficient of each 
  
    x
    
      n
    
 to zero. This is the recurrence relation.
Generate the coefficients from 
  
    a
    0
  
 and 
  
    a
    1
  
, which remain arbitrary.
Split the answer into the part multiplying 
  
    a
    0
  
 and the part multiplying 
  
    a
    1
  
; these are the two independent solutions.

An Introductory Example

Consider the nonlinear differential equation

y'' = x^{2} + y^{2} ,

and find the series solution with $y = 0$ and y' = 1 at $x = 0$ .

Solution

The method is not restricted to linear equations, though the recurrence relation is messier when squaring is involved. Let

y = \sum_{k = 0}^{\infty} a_{k} x^{k} = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + \dots

Then

y' = \sum_{k=1}^{\infty}ka_kx^{k-1} = a_1 + 2a_2x + 3a_3x^{2} + \cdots

and

y'' = \sum_{k=2}^{\infty}k(k-1)a_kx^{k-2} = 2a_2 + 6a_3x + 12a_4x^{2} + \cdots

Substituting $y$ and y'' into the equation,

2 a_{2} + 6 a_{3} x + 12 a_{4} x^{2} + \dots = x^{2} + {(a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + \dots)}^{2}

Expanding the square and collecting terms up to $x^{2}$ ,

2 a_{2} + 6 a_{3} x + 12 a_{4} x^{2} + \dots = x^{2} + a_{0}^{2} + 2 a_{0} a_{1} x + (a_{1}^{2} + 2 a_{0} a_{2}) x^{2} + \dots

Equating coefficients of like powers of $x$ ,

\begin{aligned} a_2 &= \frac{1}{2}a_0^{2} \\ a_3 &= \frac{1}{3}a_0a_1 \\ a_4 &= \frac{1}{12}\left(1 + a_1^{2} + 2a_0a_2\right) = \frac{1}{12}\left(1 + a_0^{3} + a_1^{2}\right) \\ &\ \ \vdots \end{aligned}

where the last simplification used $2 a_{0} a_{2} = 2 a_{0} \cdot \frac{1}{2} a_{0}^{2} = a_{0}^{3}$ . The constant $1$ inside the parentheses comes from the $x^{2}$ on the right-hand side of the differential equation.

Looking for the solution with $y = 0$ and y' = 1 at $x = 0$ means $a_{0} = 0$ and $a_{1} = 1$ . Consequently

a_{2} = 0, a_{3} = 0, a_{4} = \frac{1}{12}, \dots

and the solution with those initial conditions is

y = x + \frac{1}{12} x^{4} + \dots .

For $x$ near $0$ even the two-term truncation $y \approx x + \frac{1}{12} x^{4}$ tracks the true solution closely, since the omitted terms carry higher powers of a small number.

Notice the general pattern already visible here: $a_{0}$ and $a_{1}$ are free, and every later coefficient is determined by them. That is the series-method version of the fact that an initial value problem needs exactly $y (0)$ and y'(0).

Linear Examples

Solve y'' + y = 0 by power series, and identify the two solutions.

Solution

Every point is ordinary, since $P = 0$ and $Q = 1$ are analytic everywhere. Substituting the series,

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

Shift the first index with $n = k - 2$ :

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

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

By the identity theorem every bracket vanishes, giving the recurrence relation

a_{n + 2} = - \frac{a_{n}}{(n + 2) (n + 1)}, n \geq 0 .

The recurrence links coefficients two apart, so the even ones descend from $a_{0}$ and the odd ones from $a_{1}$ , independently.

Even coefficients, from $a_{0}$ :

a_{2} = - \frac{a_{0}}{2!}, a_{4} = - \frac{a_{2}}{4 \cdot 3} = \frac{a_{0}}{4!}, a_{6} = - \frac{a_{0}}{6!}, \dots

giving the pattern $a_{2 m} = \frac{(- 1)^{m} a_{0}}{(2 m)!}$ .

Odd coefficients, from $a_{1}$ :

a_{3} = - \frac{a_{1}}{3!}, a_{5} = \frac{a_{1}}{5!}, \dots

giving $a_{2 m + 1} = \frac{(- 1)^{m} a_{1}}{(2 m + 1)!}$ .

Therefore

y = a_{0} (1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \dots) + a_{1} (x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \dots) = a_{0} \cos x + a_{1} \sin x .

The method recovers the answer we already knew, which is a reassuring check. Note that both series converge for all $x$ , consistent with the equation having no singular points at all.

Solve Airy's equation y'' - xy = 0 by power series.

Solution

This equation has no elementary solutions, so the series is the answer rather than a rediscovery.

Substituting,

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

Shift both indices to the common exponent $n$ . In the first, $n = k - 2$ ; in the second, $n = k + 1$ , so $k = n - 1$ and the sum starts at $n = 1$ :

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

The $n = 0$ term appears only in the first sum:

2 a_{2} = 0 ⟹ a_{2} = 0 .

For $n \geq 1$ ,

(n + 2) (n + 1) a_{n + 2} = a_{n - 1} ⟹ a_{n + 2} = \frac{a_{n - 1}}{(n + 2) (n + 1)} .

This recurrence links coefficients three apart, so it produces three chains: one from $a_{0}$ , one from $a_{1}$ , and one from $a_{2} = 0$ which therefore vanishes entirely.

From $a_{0}$ : $a_{3} = \frac{a_{0}}{3 \cdot 2} = \frac{a_{0}}{6}$ , then $a_{6} = \frac{a_{3}}{6 \cdot 5} = \frac{a_{0}}{180}$ , and so on.

From $a_{1}$ : $a_{4} = \frac{a_{1}}{4 \cdot 3} = \frac{a_{1}}{12}$ , then $a_{7} = \frac{a_{4}}{7 \cdot 6} = \frac{a_{1}}{504}$ .

From $a_{2} = 0$ : $a_{5} = \frac{a_{2}}{5 \cdot 4} = 0$ , and likewise $a_{8} = a_{11} = \dots = 0$ .

Therefore

y = a_{0} (1 + \frac{x^{3}}{6} + \frac{x^{6}}{180} + \dots) + a_{1} (x + \frac{x^{4}}{12} + \frac{x^{7}}{504} + \dots) .

These two series define the Airy functions, which appear in optics and quantum mechanics. Since the equation has no singular points, both converge for every $x$ .

Solve y' = 2xy by power series and confirm the answer independently.

Solution

Substituting,

\sum_{k = 1}^{\infty} k a_{k} x^{k - 1} = 2 \sum_{k = 0}^{\infty} a_{k} x^{k + 1} .

Shifting to the common power $x^{n}$ , with $n = k - 1$ on the left and $n = k + 1$ on the right,

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

The $n = 0$ term gives $a_{1} = 0$ . For $n \geq 1$ ,

a_{n + 1} = \frac{2 a_{n - 1}}{n + 1} .

Since $a_{1} = 0$ , all odd coefficients vanish. For the even ones,

a_{2} = \frac{2 a_{0}}{2} = a_{0}, a_{4} = \frac{2 a_{2}}{4} = \frac{a_{0}}{2}, a_{6} = \frac{2 a_{4}}{6} = \frac{a_{0}}{6},

suggesting $a_{2 m} = \frac{a_{0}}{m!}$ .

Therefore

y = a_{0} (1 + x^{2} + \frac{x^{4}}{2!} + \frac{x^{6}}{3!} + \dots) = a_{0} \sum_{m = 0}^{\infty} \frac{{(x^{2})}^{m}}{m!} = a_{0} e^{x^{2}} .

Independent check. The equation is separable: $\frac{d y}{y} = 2 x d x$ gives $\ln | y | = x^{2} + c$ , so $y = C e^{x^{2}}$ . ✓

Only one arbitrary constant appears, as it must for a first-order equation.

Exercises

Find the singular points of \left(x^{2}-4\right)y'' + xy' + y = 0 and state the guaranteed radius of convergence for a series about $x_{0} = 0$ and about $x_{0} = 1$ .

Solution

The leading coefficient vanishes at $x = 2$ and $x = - 2$ , so those are the singular points.

About $x_{0} = 0$ , the distances to the singular points are $2$ and $2$ , so $ρ \geq 2$ and the series converges at least on $- 2 < x < 2$ .

About $x_{0} = 1$ , the distances are $| 2 - 1 | = 1$ and $| - 2 - 1 | = 3$ , so $ρ \geq 1$ and the series converges at least on $0 < x < 2$ .

Find the recurrence relation for y'' - 2xy' + 2y = 0 and compute the first few terms of both solutions.

Solution

Substituting and shifting to the common power $x^{n}$ ,

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

Note that the middle sum may start at $n = 0$ as well, since its $n = 0$ term is zero. Collecting,

(n + 2) (n + 1) a_{n + 2} = 2 (n - 1) a_{n} ⟹ a_{n + 2} = \frac{2 (n - 1) a_{n}}{(n + 2) (n + 1)} .

From $a_{0}$ : $a_{2} = \frac{2 (- 1) a_{0}}{2} = - a_{0}$ , then $a_{4} = \frac{2 (1) a_{2}}{12} = - \frac{a_{0}}{6}$ , then $a_{6} = \frac{2 (3) a_{4}}{30} = - \frac{a_{0}}{30}$ .

From $a_{1}$ : $a_{3} = \frac{2 (0) a_{1}}{6} = 0$ , and every later odd coefficient is then zero as well.

y = a_{0} (1 - x^{2} - \frac{x^{4}}{6} - \frac{x^{6}}{30} - \dots) + a_{1} x .

The odd chain terminates, giving the polynomial solution $y = x$ . This is Hermite's equation with parameter $1$ , and terminating series are exactly how the Hermite polynomials arise.

Solve the initial value problem y'' + xy' + y = 0, $y (0) = 1$ , y'(0) = 0, through the $x^{6}$ term.

Solution

Substituting and shifting,

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

so $(n + 2) (n + 1) a_{n + 2} + (n + 1) a_{n} = 0$ , and cancelling the common factor $n + 1$ ,

a_{n + 2} = - \frac{a_{n}}{n + 2} .

The conditions give $a_{0} = 1$ and $a_{1} = 0$ , so all odd coefficients vanish. For the even ones,

a_{2} = - \frac{1}{2}, a_{4} = - \frac{a_{2}}{4} = \frac{1}{8}, a_{6} = - \frac{a_{4}}{6} = - \frac{1}{48} .

Therefore

y = 1 - \frac{x^{2}}{2} + \frac{x^{4}}{8} - \frac{x^{6}}{48} + \dots

The pattern $a_{2 m} = \frac{(- 1)^{m}}{2^{m} m!}$ identifies the sum as $e^{- x^{2} / 2}$ , which you can verify satisfies the equation directly.

Frequently Asked Questions

What is an ordinary point?

A point where the coefficients $P$ and $Q$ of the standard form y'' + Py' + Qy = 0 are both analytic, that is, both have convergent power series expansions there. For polynomial coefficients, the ordinary points are everywhere the leading coefficient is nonzero.

Why do

a_{0}

and

a_{1}

stay arbitrary?

Because $a_{0} = y (0)$ and a_1 = y'(0), the two pieces of initial data a second-order equation requires. The recurrence relation determines every other coefficient in terms of these two, which is exactly the statement that the solution space is two-dimensional.

How do I get the two independent solutions out of one series?

Group the terms. Every coefficient is a multiple of $a_{0}$ or of $a_{1}$ , so the series splits as $a_{0} y_{1} (x) + a_{1} y_{2} (x)$ . Setting $a_{0} = 1$ , $a_{1} = 0$ gives $y_{1}$ ; setting $a_{0} = 0$ , $a_{1} = 1$ gives $y_{2}$ . They are independent because $y_{1} (0) = 1$ , y_1'(0) = 0 while $y_{2} (0) = 0$ , y_2'(0) = 1, so their Wronskian at $0$ is $1$ .

What if I cannot find a closed form for the coefficients?

That is common and perfectly acceptable. The recurrence relation itself is the answer: it generates as many terms as you want. Bessel's and Legendre's equations have tidy patterns, but many equations do not, and the recurrence is then the practical description of the solution.

How far can I trust the series?

At least out to the nearest singular point, measured in the complex plane. That distance is a guaranteed lower bound for the radius; the true radius may be larger. Convergence also slows dramatically as you approach the boundary, so many terms are needed there for a given accuracy.

Does this work for nonlinear equations?

Sometimes, as the introductory example shows. The obstacle is that squaring or multiplying series produces convolutions of the coefficients rather than simple linear relations, so the recurrence is nonlinear and rarely has a closed form. There is also no existence theorem guaranteeing that the resulting series converges.

Power Series Solutions

Quick Reference

Ordinary and Singular Points

The Method

An Introductory Example

Linear Examples

Exercises

Frequently Asked Questions

Quick Links

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