Series Solutions of ODEs — Mathematics Knowledge System

Overview

Many linear differential equations that arise in practice — Legendre's equation, Bessel's equation, Hermite's equation — have variable (non-constant) coefficients, so the characteristic-equation trick used for constant-coefficient ODEs does not apply. The power series method instead looks for a solution directly as a series y = Σₙ aₙxⁿ around a point x₀. Substituting the series into the ODE and collecting like powers of x turns the differential equation into a recurrence relation among the coefficients aₙ, which can be solved term by term. Where the series converges (guaranteed near any 'ordinary point' where the coefficients are analytic, by a theorem essentially due to Cauchy and later made rigorous by Fuchs), it defines a genuine solution, and often the series can be recognized as the expansion of a familiar function.

Intuition

Think of the series method as 'guessing infinitely many undetermined coefficients at once.' Instead of trying a single exponential eʳˣ as with constant coefficients, you try an entire polynomial-like expansion and let the differential equation itself tell you, order by order in x, how each coefficient must relate to the earlier ones. Shifting the index of the y'' series so every sum lines up with the same power xⁿ is the key bookkeeping trick — once every series is written as Σ(...)xⁿ, the ODE becomes 'the bracketed coefficient of xⁿ must be zero for every n', which is exactly a recurrence relation.

Formal Definition

Definition

For a linear ODE with analytic coefficients, seek a solution as a power series about an ordinary point x₀ = 0 and substitute term by term:

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

Trial series solution

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

Term-by-term derivatives

P(x)y'' + Q(x)y' + R(x)y = 0 \ \implies\ \sum_n c_n(a_0,\dots,a_n) x^n = 0 \implies c_n = 0\ \forall n

Substitution forces every coefficient of x^n to vanish, giving a recurrence

Derivation

Apply the method to y'' + xy' + y = 0, an ODE with a variable coefficient (the xy' term) that has no elementary characteristic-equation shortcut. Substitute y=Σaₙxⁿ and shift indices so every sum is written in powers of x^n:

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

Substitute y, y', y'' into y'' + xy' + y = 0

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

Reindex the first sum (m=n-2) and absorb x into the middle sum (n a_n x^n, valid at n=0 too since the term is 0)

(n+2)(n+1)a_{n+2} + n a_n + a_n = 0 \quad \text{for every } n \ge 0

Coefficient of x^n must vanish for every n

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

The recurrence relation obtained by solving for a_{n+2}

Applications

Legendre's equation (Laplace's equation in spherical coordinates) and Bessel's equation (cylindrical problems, vibrating membranes) are both solved by power series, giving the Legendre polynomials and Bessel functions used throughout mathematical physics.

Worked Examples

Substituting y=Σaₙxⁿ and matching coefficients of xⁿ gives the recurrence derived above.
$a_{n+2} = \frac{-a_n}{n+2}$
Initial conditions give a₀ = y(0) = 1 and a₁ = y'(0) = 0. Since a₁=0, every odd-indexed coefficient is 0.
$a_0 = 1, \quad a_1 = 0 \implies a_3=a_5=\cdots=0$
Compute the even coefficients from a₀=1: a₂=-a₀/2, a₄=-a₂/4, a₆=-a₄/6.
$a_2 = -\tfrac{1}{2}, \quad a_4 = -\tfrac{a_2}{4} = \tfrac{1}{8}, \quad a_6 = -\tfrac{a_4}{6} = -\tfrac{1}{48}$
These match the Taylor coefficients of e^{-x²/2}: (-1)^k/(2^k k!) gives -1/2, 1/8, -1/48 for k=1,2,3.
$y = 1 - \frac{x^2}{2} + \frac{x^4}{8} - \frac{x^6}{48} + \cdots = e^{-x^2/2}$

Answer: y = e^(-x²/2) — the series solution collapses exactly to a closed form here (verify: y'=-xe^{-x²/2}, y''=(x²-1)e^{-x²/2}, so y''+xy'+y = (x²-1)e^{-x²/2} - x^2e^{-x^2/2} + e^{-x^2/2} = 0 ✓).

Practice Problems

Difficulty 6/10

For the recurrence a_{n+2} = -a_n/(n+2) with a₀=1, a₁=0 (as derived above), compute a₈.

Difficulty 6/10

Using the same recurrence a_{n+2} = -a_n/(n+2), if instead a₀=0 and a₁=1, find a₃ and a₅.

Difficulty 7/10

Verify that y = e^{-x²/2} satisfies y'' + xy' + y = 0 by direct differentiation.

Common Mistakes

Common Mistake

Forgetting to reindex (shift) the series for y'' before comparing coefficients, and instead matching coefficients of xⁿ in one sum against x^{n-2} in another.

Every series must be rewritten in the same power of x (usually xⁿ) before the coefficient-matching step is valid; substituting m=n-2 in Σn(n-1)aₙx^{n-2} is what produces the aligned Σ(n+2)(n+1)a_{n+2}xⁿ.

Common Mistake

Assuming the power series method always yields a solution expressible in elementary closed form.

Most series solutions (Legendre polynomials, Bessel functions) do NOT collapse to elementary functions; the e^{-x²/2} example is a special, unusually clean case chosen for illustration.

Quiz

Summary

For linear ODEs with variable (non-constant) coefficients, substitute y = Σaₙxⁿ and match coefficients of each power of x.
This turns the ODE into a recurrence relation among the coefficients aₙ, which is solved term by term from the initial conditions a₀, a₁.
Series solutions converge near ordinary points of the ODE and sometimes collapse to elementary closed forms, as in y'' + xy' + y = 0 giving y = e^{-x²/2}.
The method underlies Legendre's, Bessel's, and Hermite's equations, whose series solutions are named special functions used throughout mathematical physics.

References

WebsiteWikipedia — Power series solution of differential equations
BookBoyce, W. & DiPrima, R. Elementary Differential Equations, 11th ed., Ch. 5.