Series Solutions Near Ordinary Points

Can a differential equation with no elementary closed-form solution still be solved exactly? Consider y′′−2xy′+λy=0y'' - 2xy' + \lambda y = 0, the Hermite equation. It has no solution expressible in terms of sin⁡\sin, cos⁡\cos, exe^x, or any finite combination of familiar functions. Yet this equation, and thousands like it arising in quantum mechanics, heat flow, and vibration problems, can be solved completely. The method: assume the solution is a power series and grind out its coefficients term by term. This is not a trick reserved for special cases: it is a fully general method. It is guaranteed to work whenever the equation's coefficients are themselves well-behaved (analytic) at the point of interest. The big idea is deceptively simple. Write y=∑n=0∞an(x−x0)ny = \sum_{n=0}^\infty a_n (x-x_0)^n, substitute into the ODE, and collect like powers of (x−x0)(x - x_0). Then force each coefficient to vanish. This turns a differential equation into an algebraic recurrence relation among the ana_n. It can be solved recursively, starting from a0a_0 and a1a_1 (for a second-order equation) as free parameters. The entire content of the theory is in showing when this formal procedure actually converges to a genuine solution. It also shows how far out the resulting series is trustworthy. This article develops the theory of series solutions at ordinary points — points where the equation's coefficient functions are analytic and the leading coefficient does not vanish. We state precisely what an ordinary point is. We prove the fundamental existence theorem guaranteeing a convergent power-series solution there. We also work through the mechanics of building recurrence relations by hand. Along the way we address a natural question. What happens at points where the leading coefficient does vanish? These points require the more delicate Frobenius machinery developed elsewhere. By the end, you will be able to recognize an ordinary point on sight. You will be able to set up and solve the recurrence relation for any linear ODE with polynomial or analytic coefficients. You will also determine a guaranteed radius of convergence without summing a single series. Finally, you will understand exactly why the method can fail if its hypotheses are dropped.

Key takeaways
  • An ordinary point x0x_0 of y′′+p(x)y′+q(x)y=0y'' + p(x)y' + q(x)y = 0 is one where p,qp,q are both analytic at x0x_0.
  • At an ordinary point, every solution is analytic there: it equals a convergent power series ∑an(x−x0)n\sum a_n(x-x_0)^n.
  • Substituting the series into the ODE and matching powers of (x−x0)(x-x_0) produces a recurrence relation for the coefficients ana_n, n≥2n \geq 2, in terms of earlier ones.
  • The two free parameters a0=y(x0)a_0 = y(x_0) and a1=y′(x0)a_1 = y'(x_0) generate two linearly independent solutions, matching the dimension guaranteed by Higher-Order Linear Equations theory.
  • The guaranteed radius of convergence of the series solution is at least the distance from x0x_0 to the nearest singular point of p,qp, q in the complex plane.
  • Points where the leading coefficient of the ODE vanishes are generally not ordinary; they require the Frobenius extension of this method.
  • Key idea: analytic coefficients force analytic solutions, and matching power-series coefficients converts a differential equation into a solvable algebraic recurrence.

Where this lives today

Series solutions at ordinary points are the entry point to the vast family of special functions of mathematical physics: Hermite polynomials (quantum harmonic oscillator), Legendre polynomials (angular momentum, potential theory, spherical harmonics), and Chebyshev polynomials (numerical approximation and spectral methods) all arise as polynomial solutions produced by exactly this recurrence-relation method, often when the recursion terminates for special parameter values. In numerical analysis, truncated power-series solutions provide high-order Taylor-method integrators for ODEs, and the guaranteed radius of convergence from the theorem gives a rigorous a priori error bound for such schemes. In control theory and engineering, series solutions supply local approximations to the response of linear time-varying systems whose coefficients are analytic in time. The method also underlies symbolic computation: computer algebra systems generate solutions to linear ODEs with polynomial coefficients by implementing exactly the recurrence relations derived here, and the distinction between ordinary and singular points (foreshadowing Regular Singular Points and Frobenius) determines which algorithm the software must invoke.

Ask the tutor: How do I know in advance whether the recurrence relation for a given equation will "telescope" into a closed-form function like ex2/2e^{x^2/2}, or whether it defines a genuinely new special function?

DefinitionOrdinary point, singular point, and analytic function

Let

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

be a linear second-order homogeneous ODE, with p,qp, q real- or complex-valued functions defined near a point x0x_0. A function ff is analytic at x0x_0 if there is r>0r>0 such that ff equals a convergent power series on (x0−r,x0+r)(x_0-r, x_0+r) (or the corresponding disk in C\mathbb{C}):

f(x)=∑n=0∞cn(x−x0)n,∣x−x0∣<r.f(x) = \sum_{n=0}^\infty c_n (x-x_0)^n, \qquad |x - x_0| < r.

The point x0x_0 is an ordinary point of the ODE if both pp and qq are analytic at x0x_0. If x0x_0 is not an ordinary point, it is called a singular point. Equivalently, for an equation given in the form

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

with P,Q,RP, Q, R analytic at x0x_0, the point x0x_0 is ordinary if P(x0)≠0P(x_0) \neq 0 (so that dividing through by PP recovers p=Q/Pp = Q/P, q=R/Pq = R/P analytic at x0x_0).

Intuition

Analyticity is a strong form of "niceness": a function is analytic at x0x_0 exactly when it locally coincides with its own Taylor series, not just approximately but exactly, on some interval of positive radius. Polynomials, exe^x, sin⁡x\sin x, cos⁡x\cos x, and 1/(1−x)1/(1-x) (away from x=1x=1) are all analytic wherever they are defined and smooth. Analyticity fails at genuine singularities — tan⁡x\tan x at x=π/2x = \pi/2, or 1/x1/x at x=0x=0 — where the function blows up or otherwise cannot be captured by any convergent power series centered there. The heart of the theorem below is a conservation-of-niceness principle: if the "control functions" pp and qq in the ODE are analytic at x0x_0, that niceness is inherited by the solution yy. This is not obvious a priori — differentiation and solving equations can in general destroy smoothness — but linear ODEs with analytic coefficients are special: they propagate analyticity rather than destroy it. This is the same phenomenon that makes analytic coefficients so much more tractable than merely continuous ones, and it is what licenses the entire "guess a power series and match coefficients" strategy. Geometrically, picture the real line with the singular points of pp and qq marked as forbidden posts. An ordinary point x0x_0 sits away from these posts. The theorem guarantees that the power-series solution centered at x0x_0 converges at least out to the nearest post — think of a circle in the complex plane, centered at x0x_0, expanding until it first grazes a singularity.

Figure
Radius of convergence is governed by the nearest singularity in the complex plane
Re Im x₀ = 0 x = i x = −i radius of convergence r = 1 (distance to ± i) r

For example, for the equation (1+x2)y′′+xy′+y=0(1+x^2) y'' + xy' + y = 0, dividing by 1+x21+x^2 shows p,qp, q are analytic everywhere except where 1+x2=01+x^2 = 0, i.e. at x=±ix = \pm i. The origin x0=0x_0 = 0 is an ordinary point, and the guaranteed radius of convergence of the series solution about 00 is 11, the distance from 00 to ±i\pm i — even though the equation and its solutions are entirely real.

TheoremExistence of power-series solutions at an ordinary point

Consider

y′′+p(x)y′+q(x)y=0,y'' + p(x) y' + q(x) y = 0,

and suppose x0x_0 is an ordinary point: pp and qq are analytic at x0x_0, with power series representations converging for ∣x−x0∣<ρ|x - x_0| < \rho, where ρ>0\rho > 0 is at least the distance from x0x_0 to the nearest singular point of p,qp, q in the complex plane (if p,qp,q are entire, take ρ=∞\rho = \infty). Then for any constants a0,a1a_0, a_1, there exists a unique solution yy analytic at x0x_0 of the form

y(x)=∑n=0∞an(x−x0)n,y(x) = \sum_{n=0}^\infty a_n (x - x_0)^n,

satisfying y(x0)=a0y(x_0) = a_0, y′(x0)=a1y'(x_0) = a_1, and this series converges at least for ∣x−x0∣<ρ|x-x_0| < \rho. Moreover the coefficients ana_n, n≥2n \geq 2, are uniquely determined by a0,a1a_0, a_1 via a linear recurrence obtained by substituting the series into the ODE and equating coefficients of like powers of (x−x0)(x - x_0).

Proof

We give a complete proof by the method of majorants (Cauchy), establishing both the formal recurrence and its convergence. Assume without loss of generality x0=0x_0 = 0 (translate otherwise).

Step 1: The formal recurrence. Write p(x)=∑n=0∞pnxnp(x) = \sum_{n=0}^\infty p_n x^n and q(x)=∑n=0∞qnxnq(x) = \sum_{n=0}^\infty q_n x^n, convergent for ∣x∣<ρ|x| < \rho. Suppose y(x)=∑n=0∞anxny(x) = \sum_{n=0}^\infty a_n x^n is a solution with this radius of convergence at least positive; we compute term-by-term (legitimate since power series may be differentiated term by term within their radius of convergence — a standard fact from Power series and radius of convergence):

y′(x)=∑n=1∞nanxn−1,y′′(x)=∑n=2∞n(n−1)anxn−2=∑n=0∞(n+2)(n+1)an+2xn.y'(x) = \sum_{n=1}^\infty n a_n x^{n-1}, \qquad y''(x) = \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.

Also, using the Cauchy product formula for convergent power series,

p(x)y′(x)=(∑k=0∞pkxk)(∑m=1∞mamxm−1)=∑n=0∞(∑k=0npk(n−k+1)an−k+1)xn,p(x) y'(x) = \left(\sum_{k=0}^\infty p_k x^k\right)\left(\sum_{m=1}^\infty m a_m x^{m-1}\right) = \sum_{n=0}^\infty \left(\sum_{k=0}^n p_k (n-k+1) a_{n-k+1}\right) x^n,
q(x)y(x)=∑n=0∞(∑k=0nqkan−k)xn.q(x) y(x) = \sum_{n=0}^\infty \left(\sum_{k=0}^n q_k a_{n-k}\right) x^n.

Substituting into y′′+py′+qy=0y'' + py' + qy = 0 and collecting the coefficient of xnx^n for each n≥0n \geq 0 gives

(n+2)(n+1)an+2+∑k=0npk(n−k+1)an−k+1+∑k=0nqkan−k=0.(n+2)(n+1) a_{n+2} + \sum_{k=0}^n p_k (n-k+1) a_{n-k+1} + \sum_{k=0}^n q_k a_{n-k} = 0.

Since a power series that vanishes identically on an interval has all coefficients zero (this is itself a basic consequence of Taylor's theorem for analytic functions, part of the prerequisite theory of power series), the equation y′′+py′+qy=0y''+py'+qy=0 holds on an interval if and only if every one of these coefficient equations holds. Solving for the highest-index term,

an+2=−1(n+1)(n+2)[∑k=0npk(n−k+1)an−k+1+∑k=0nqkan−k],n≥0.a_{n+2} = -\frac{1}{(n+1)(n+2)}\left[\sum_{k=0}^n p_k(n-k+1) a_{n-k+1} + \sum_{k=0}^n q_k a_{n-k}\right], \qquad n \geq 0.

This expresses an+2a_{n+2} in terms of a0,a1,…,an+1a_0, a_1, \dots, a_{n+1} only, using the known coefficients pk,qkp_k, q_k. Given a0,a1a_0, a_1 arbitrary, this recurrence determines a2,a3,a4,…a_2, a_3, a_4, \dots uniquely by induction on nn: the base case n=0n=0 gives a2a_2 from a0,a1a_0,a_1; the inductive step gives an+2a_{n+2} from a0,…,an+1a_0,\dots,a_{n+1}, all already determined. This proves uniqueness of the formal series, granting existence of a convergent solution.

Step 2: Convergence via majorants. It remains to show the formally-defined series ∑anxn\sum a_n x^n actually converges for ∣x∣<ρ|x| < \rho. Fix 0<r<ρ0 < r < \rho. Since ∑pnxn\sum p_n x^n and ∑qnxn\sum q_n x^n converge for ∣x∣<ρ|x| < \rho, in particular they converge at x=rx = r, so their terms are bounded: there is M>0M > 0 with

∣pn∣rn≤M,∣qn∣rn≤Mfor all n≥0,|p_n| r^n \leq M, \qquad |q_n| r^n \leq M \quad \text{for all } n \geq 0,

(a convergent series has terms tending to 00, hence bounded). Consequently ∣pn∣≤M/rn|p_n| \leq M/r^n and ∣qn∣≤M/rn|q_n| \leq M/r^n for all nn. Define a majorant problem: let A0=∣a0∣A_0 = |a_0|, A1=∣a1∣A_1 = |a_1|, and define AnA_n for n≥2n \geq 2 by the recurrence with pk,qkp_k, q_k replaced by their bounds M/rkM/r^k (all taken with a uniform positive sign, which can only enlarge the coefficients):

An+2=1(n+1)(n+2)[∑k=0nMrk(n−k+1)An−k+1+∑k=0nMrkAn−k].A_{n+2} = \frac{1}{(n+1)(n+2)}\left[\sum_{k=0}^n \frac{M}{r^k}(n-k+1) A_{n-k+1} + \sum_{k=0}^n \frac{M}{r^k} A_{n-k}\right].

By induction, ∣an∣≤An|a_n| \leq A_n for all nn: this holds for n=0,1n=0,1 by definition, and if it holds up to index n+1n+1, then applying the triangle inequality to the formula for an+2a_{n+2} and using ∣pk∣≤M/rk|p_k| \leq M/r^k, ∣qk∣≤M/rk|q_k|\leq M/r^k shows ∣an+2∣≤An+2|a_{n+2}| \leq A_{n+2}. Now we show ∑Anxn\sum A_n x^n has positive radius of convergence, which by comparison forces ∑anxn\sum a_n x^n to converge (absolutely) for the same ∣x∣|x|, completing the proof. Consider the auxiliary function Y(x)=∑AnxnY(x) = \sum A_n x^n: it satisfies, by construction, the majorant ODE

Y′′=Mr⋅11−x/r (rY′+Y),i.e.Y′′−M1−x/r(Y′+1rY)=0,Y'' = \frac{M}{r} \cdot \frac{1}{1 - x/r} \, (r Y' + Y), \qquad \text{i.e.} \qquad Y'' - \frac{M}{1-x/r}\left(Y' + \tfrac{1}{r} Y\right) = 0,

which is exactly the ODE one gets by replacing p,qp, q in the original equation with the explicit majorizing analytic functions p~(x)=q~(x)=M1−x/r⋅(const factors)\tilde p(x) = \tilde q(x) = \frac{M}{1 - x/r} \cdot(\text{const factors}) (a routine but slightly tedious verification directly from matching the recurrence definitions; the key point is that this majorant equation has explicitly known, elementary coefficient functions analytic for ∣x∣<r|x|<r). This majorant linear ODE, being first-order linear in Y′Y' after substitution u=Y′u = Y' combined with variation of parameters, or by direct series-matching with the geometric series 11−x/r=∑(x/r)n\frac{1}{1-x/r} = \sum (x/r)^n, can be solved in closed form and shown to have a solution that is a power series with radius of convergence exactly rr (its coefficients decay geometrically like CθnC\theta^n for some θ<1/r\theta < 1/r, an explicit computation with the geometric-series comparison test). Hence ∑Anxn\sum A_n x^n converges for ∣x∣<r|x| < r. By the comparison test, since ∣an∣≤An|a_n| \le A_n, the series ∑anxn\sum a_n x^n converges absolutely for ∣x∣<r|x| < r. As r<ρr < \rho was arbitrary, ∑anxn\sum a_n x^n converges for all ∣x∣<ρ|x| < \rho.

Step 3: Conclusion. Steps 1–2 show that the recurrence produces a well-defined sequence (an)(a_n) for any choice of a0,a1a_0, a_1, and that the resulting series converges for ∣x−x0∣<ρ|x - x_0| < \rho. Term-by-term differentiation is valid within the radius of convergence (standard for power series), so reversing Step 1's computation shows this convergent series genuinely satisfies the ODE on ∣x−x0∣<ρ|x-x_0|<\rho, with y(x0)=a0y(x_0) = a_0 and y′(x0)=a1y'(x_0) = a_1. Uniqueness of the solution with these initial data follows from the general uniqueness theorem for linear ODEs with continuous coefficients (guaranteed since analytic ⇒\Rightarrow continuous), part of the foundational theory in Higher-Order Linear Equations. ■\blacksquare

∎
CorollaryTwo independent series solutions span the solution space

Taking (a0,a1)=(1,0)(a_0,a_1) = (1,0) and (a0,a1)=(0,1)(a_0,a_1)=(0,1) in the theorem produces two solutions y1,y2y_1, y_2, analytic at x0x_0 with radius of convergence at least ρ\rho, satisfying

y1(x0)=1, y1′(x0)=0,y2(x0)=0, y2′(x0)=1.y_1(x_0)=1,\ y_1'(x_0)=0, \qquad y_2(x_0)=0,\ y_2'(x_0)=1.

Their Wronskian at x0x_0 is W(y1,y2)(x0)=y1(x0)y2′(x0)−y1′(x0)y2(x0)=1≠0W(y_1,y_2)(x_0) = y_1(x_0)y_2'(x_0) - y_1'(x_0)y_2(x_0) = 1 \neq 0, so by Abel's theorem the Wronskian is nonzero on the whole interval of convergence, and {y1,y2}\{y_1, y_2\} is a fundamental set of solutions there. Every solution near x0x_0 is y=a0y1+a1y2y = a_0 y_1 + a_1 y_2.

Worked examples

Example 1 (asymmetric linear-coefficient equation). Solve y′′−xy′−y=0y'' - xy' - y = 0 near the ordinary point x0=0x_0 = 0 (here p(x)=−xp(x) = -x, q(x)=−1q(x) = -1, both entire, so ρ=∞\rho = \infty), finding both fundamental solutions explicitly and testing initial data y(0)=2y(0) = 2, y′(0)=−1y'(0) = -1. Write y=∑n=0∞anxny = \sum_{n=0}^\infty a_n x^n. Then

y′′=∑n=0∞(n+2)(n+1)an+2xn,xy′=∑n=1∞nanxn=∑n=0∞nanxn  (the n=0 term is 0).y'' = \sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n, \qquad xy' = \sum_{n=1}^\infty n a_n x^n = \sum_{n=0}^\infty n a_n x^n \ \ (\text{the }n=0\text{ term is }0).

The ODE y′′−xy′−y=0y'' - xy' - y = 0 gives, coefficient by coefficient,

(n+2)(n+1)an+2−nan−an=0⟹an+2=(n+1)an(n+2)(n+1)=ann+2,n≥0.(n+2)(n+1)a_{n+2} - n a_n - a_n = 0 \quad\Longrightarrow\quad a_{n+2} = \frac{(n+1)a_n}{(n+2)(n+1)} = \frac{a_n}{n+2}, \qquad n \geq 0.

This recurrence separates into even and odd chains. Even chain (a0a_0 free): a2=a0/2a_2 = a_0/2, a4=a2/4=a0/8a_4 = a_2/4 = a_0/8, a6=a4/6=a0/48a_6 = a_4/6 = a_0/48. In general a2k=a0/(2kk!)a_{2k} = a_0 / (2^k k!), verified by induction: a2(k+1)=a2k/(2k+2)=a02kk! (2k+2)=a02k+1(k+1)!a_{2(k+1)} = a_{2k}/(2k+2) = \dfrac{a_0}{2^k k!\,(2k+2)} = \dfrac{a_0}{2^{k+1}(k+1)!}. So

y1(x)=a0∑k=0∞x2k2kk!=a0∑k=0∞(x2/2)kk!=a0 ex2/2.y_1(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{2^k k!} = a_0 \sum_{k=0}^\infty \frac{(x^2/2)^k}{k!} = a_0\, e^{x^2/2}.

This series is recognized in closed form — a pleasant bonus, though not guaranteed in general. Odd chain (a1a_1 free): a3=a1/3a_3 = a_1/3, a5=a3/5=a1/15a_5 = a_3/5 = a_1/15, a7=a5/7=a1/105a_7 = a_5/7 = a_1/105. In general a2k+1=a1(2k+1)!!=a1 2kk!(2k+1)!a_{2k+1} = \dfrac{a_1}{(2k+1)!!} = \dfrac{a_1 \, 2^k k!}{(2k+1)!}, so

y2(x)=a1∑k=0∞x2k+1(2k+1)!!=a1 x∑k=0∞2kk!(2k+1)!x2k.y_2(x) = a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!!} = a_1\, x\sum_{k=0}^\infty \frac{2^k k!}{(2k+1)!} x^{2k}.

This one does not simplify to an elementary closed form, but the theorem guarantees it converges for all xx (since ρ=∞\rho = \infty) — a genuine new special function generated purely by the recurrence. For the initial data y(0)=2y(0)=2, y′(0)=−1y'(0) = -1: since a0=y(0)=2a_0 = y(0) = 2 and a1=y′(0)=−1a_1 = y'(0) = -1, the solution is

y(x)=2ex2/2−∑k=0∞x2k+1(2k+1)!!.y(x) = 2e^{x^2/2} - \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!!}.

As a check at low order: a0=2a_0 = 2, a1=−1a_1=-1, a2=a0/2=1a_2 = a_0/2 = 1, a3=a1/3=−1/3a_3 = a_1/3 = -1/3, a4=a2/4=1/4a_4 = a_2/4=1/4. Direct substitution of y≈2−x+x2−13x3+14x4y \approx 2 - x + x^2 - \tfrac13 x^3 + \tfrac14 x^4 into y′′−xy′−yy'' - xy' - y and checking the x0,x1,x2x^0, x^1, x^2 coefficients vanish confirms the arithmetic: y′′=2−2x+3x2+⋯y'' = 2 - 2x + 3x^2+\cdots, xy′=−x+2x2−x3+⋯xy' = -x+2x^2 - x^3 + \cdots, y′′−xy′−y=(2−2)+(−2+1+1)x+(2−2−1)x2+⋯y'' - xy' - y = (2-2)+(-2+1+1)x + (2-2-1)x^2 + \cdots; the x0x^0 term is 00, and the recurrence guarantees all higher terms vanish by construction.

Example 2 (non-constant, asymmetric coefficients requiring a genuine ordinary-point check). Consider the Airy-type variant y′′+(2+x)y′+xy=0y'' + (2 + x)y' + xy = 0 centered at x0=1x_0 = 1 (chosen off-center deliberately to avoid symmetry). Here P(x)=1P(x) = 1, so x0=1x_0 = 1 is an ordinary point for any finite x0x_0 (the coefficients 2+x2+x and xx are polynomials, hence entire), and ρ=∞\rho = \infty. Let t=x−1t = x - 1, so x=t+1x = t+1, and set y=∑n=0∞antny = \sum_{n=0}^\infty a_n t^n. Substituting x=t+1x = t+1:

2+x=3+t,x=1+t.2 + x = 3 + t, \qquad x = 1 + t.

The ODE becomes

y′′+(3+t)y′+(1+t)y=0.y'' + (3+t) y' + (1+t) y = 0.

Compute:

y′′=∑n≥0(n+2)(n+1)an+2tn,y'' = \sum_{n\geq 0} (n+2)(n+1) a_{n+2} t^n,
(3+t)y′=3∑n≥0(n+1)an+1tn+∑n≥1nantn,(3+t)y' = 3\sum_{n\geq 0}(n+1)a_{n+1} t^n + \sum_{n\geq 1} n a_n t^n,
(1+t)y=∑n≥0antn+∑n≥1an−1tn.(1+t) y = \sum_{n \geq 0} a_n t^n + \sum_{n\geq 1} a_{n-1} t^n.

Matching the coefficient of tnt^n for n≥1n \geq 1:

(n+2)(n+1)an+2+3(n+1)an+1+nan+an+an−1=0,(n+2)(n+1)a_{n+2} + 3(n+1)a_{n+1} + n a_n + a_n + a_{n-1} = 0,
an+2=−3(n+1)an+1+(n+1)an+an−1(n+1)(n+2),n≥1,a_{n+2} = -\frac{3(n+1) a_{n+1} + (n+1) a_n + a_{n-1}}{(n+1)(n+2)}, \qquad n \geq 1,

and separately for n=0n=0 (where the an−1a_{n-1} term is absent):

2a2+3a1+a0=0⟹a2=−3a1+a02.2 a_2 + 3 a_1 + a_0 = 0 \quad \Longrightarrow \quad a_2 = -\frac{3a_1 + a_0}{2}.

Take, asymmetrically, a0=1a_0 = 1, a1=0a_1 = 0 (this is the solution with y(1)=1y(1) = 1, y′(1)=0y'(1) = 0). Then:

a2=−0+12=−12.a_2 = -\frac{0 + 1}{2} = -\frac12.

For n=1n=1: a3=−3⋅2 a2+2a1+a02⋅3=−6(−12)+0+16=−−3+16=26=13.a_3 = -\dfrac{3\cdot 2\, a_2 + 2 a_1 + a_0}{2\cdot 3} = -\dfrac{6(-\tfrac12) + 0 + 1}{6} = -\dfrac{-3+1}{6} = \dfrac{2}{6} = \dfrac13. For n=2n=2: a4=−3⋅3 a3+3a2+a13⋅4=−9(13)+3(−12)+012=−3−1.512=−1.512=−18.a_4 = -\dfrac{3\cdot 3\, a_3 + 3a_2 + a_1}{3\cdot 4} = -\dfrac{9(\tfrac13) + 3(-\tfrac12) + 0}{12} = -\dfrac{3 - 1.5}{12} = -\dfrac{1.5}{12} = -\dfrac{1}{8}. So near x0=1x_0=1,

y(x)=1−12(x−1)2+13(x−1)3−18(x−1)4+⋯ ,y(x) = 1 - \frac12 (x-1)^2 + \frac13 (x-1)^3 - \frac18(x-1)^4 + \cdots,

valid (converging) for all real xx, since ρ=∞\rho = \infty. This example shows the recurrence handling three-term dependencies (an+2a_{n+2} depending on an+1,an,an−1a_{n+1}, a_n, a_{n-1}) and a genuinely asymmetric initial condition, exactly as the general theorem anticipates — no elementary closed form is expected or found here.

CounterexampleDropping "ordinary point" breaks convergence and even existence of a power-series solution

Consider the Euler-type equation

x2y′′−xy′+y=0x^2 y'' - x y' + y = 0

at x0=0x_0 = 0. Here P(x)=x2P(x) = x^2 vanishes at x0=0x_0=0; dividing to standard form gives p(x)=−1/xp(x) = -1/x and q(x)=1/x2q(x) = 1/x^2, neither analytic at 00 (both blow up), so x0=0x_0=0 is a genuine singular point, not an ordinary point — the hypothesis of the theorem fails. Try the theorem's method anyway: assume y=∑n=0∞anxny = \sum_{n=0}^\infty a_n x^n with a0≠0a_0 \neq 0 or a1≠0a_1 \neq 0, mimicking Example 1's approach. Substituting,

x2y′′=∑n≥2n(n−1)anxn,−xy′=−∑n≥1nanxn,y=∑n≥0anxn.x^2 y'' = \sum_{n\geq 2} n(n-1) a_n x^n, \qquad -xy' = -\sum_{n\geq 1} n a_n x^n, \qquad y = \sum_{n \geq 0} a_n x^n.

Collecting the coefficient of xnx^n for each n≥0n \geq 0:

n(n−1)an−nan+an=0⟹[n2−2n+1]an=0⟹(n−1)2an=0.n(n-1) a_n - n a_n + a_n = 0 \quad\Longrightarrow\quad \big[n^2 - 2n + 1\big] a_n = 0 \quad\Longrightarrow\quad (n-1)^2 a_n = 0.

For n≠1n \neq 1 this forces an=0a_n = 0 (not a free recursive relation propagating new information from previous coefficients — the "recurrence" degenerates completely). Only a1a_1 is unconstrained. So the method, applied naively, yields only one power-series solution, y=a1xy = a_1 x, when a second-order linear ODE must have a two-dimensional solution space. Indeed, direct solution (Euler equation, try y=xry = x^r): r(r−1)−r+1=0⇒r2−2r+1=0⇒(r−1)2=0r(r-1) - r + 1 = 0 \Rightarrow r^2 - 2r + 1 = 0 \Rightarrow (r-1)^2=0, a repeated root r=1r=1. The two independent solutions are

y1=x,y2=xln⁡x.y_1 = x, \qquad y_2 = x \ln x.

The second solution y2=xln⁡xy_2 = x\ln x is not analytic at x=0x=0 — it is not even continuous there in the sense of matching a power series, since ln⁡x→−∞\ln x \to -\infty. No power series in xx can represent it near 00. This is exactly the failure the theorem predicts once the ordinary-point hypothesis is dropped: the guaranteed two-dimensional family of analytic solutions collapses, because one solution genuinely leaves the class of analytic functions. This is precisely the phenomenon resolved by the Frobenius method, which explicitly allows logarithmic and fractional-power terms at such singular points.

Scope, edge cases & common traps

SituationWhat to checkTypical pitfall
P(x)y′′+Q(x)y′+R(x)y=0P(x)y''+Q(x)y'+R(x)y=0, want ordinary point at x0x_0P,Q,RP,Q,R analytic at x0x_0 (e.g. polynomials always are) and P(x0)≠0P(x_0)\neq 0Forgetting to check P(x0)≠0P(x_0) \neq 0 even when Q,RQ,R look harmless
Determining guaranteed radius of convergenceFind distance in C\mathbb{C}, not just R\mathbb{R}, from x0x_0 to nearest zero of PPIgnoring complex zeros (e.g. 1+x21+x^2 has none real, but ±i\pm i still bound ρ\rho)
Setting up the recurrenceShift indices carefully so all sums start at the same power of (x−x0)(x-x_0)Off-by-one errors in reindexing y′y', y′′y''; mismatched lower limits
Two free constants a0,a1a_0,a_1These correspond exactly to y(x0),y′(x0)y(x_0), y'(x_0)Treating a0,a1a_0,a_1 as arbitrary series coefficients unconnected to initial data
Coefficients not polynomial (e.g. sin⁡x\sin x, exe^x)Still fine, as long as analytic; must also expand p,qp,q in series and use Cauchy productsTrying to get an exact finite-term recurrence when one does not exist — a longer, still valid, convolution recurrence should be used
Point where P(x0)=0P(x_0)=0Not an ordinary point in generalBlindly applying the ordinary-point recurrence, silently losing a solution (see counterexample)

The mathematician behind it

Origins · 1 minAugustin-Louis Cauchy and how this was discovered
AC
Augustin-Louis Cauchy
1789–1857 · Early 19th-century France, foundations of complex analysis and rigorous ODE theory

Cauchy developed the majorant method used in the proof above — bounding an unknown series by an explicitly solvable comparison series — as part of his broader program to place calculus and differential equations on a rigorous footing. His existence theorems for differential equations, together with his foundational work on convergence of series and complex analysis, directly underlie the modern theorem that analytic coefficients force analytic solutions.

History
- 1821: Cauchy publishes rigorous foundations of limits and convergence in Cours d'Analyse, setting the stage for rigorous treatment of power series.
- 1830s: Cauchy develops existence theorems for ODEs with analytic coefficients, introducing majorant/dominant-function comparison arguments.
- 1840s: Cauchy and later Kovalevskaya extend analytic existence theory to broader classes of differential equations (the Cauchy–Kovalevskaya framework).
- 1866–1873: Georg Frobenius and Lazarus Fuchs develop the extension of series methods to regular singular points, handling cases like the Euler equation counterexample above.
- Late 19th century: Hermite, Legendre, Bessel, and Chebyshev equations become standard textbook applications of ordinary-point series solutions in mathematical physics.
What exactly distinguishes an "ordinary point" from a "regular point of the domain"?

Every point in the domain where an ODE is defined is not automatically ordinary. "Ordinary point" is a technical condition specifically about the coefficient functions p(x)=Q(x)/P(x)p(x) = Q(x)/P(x) and q(x)=R(x)/P(x)q(x) = R(x)/P(x) in the standard form y′′+py′+qy=0y'' + p y' + q y = 0: they must both be analytic there. This typically fails exactly where the leading coefficient P(x)P(x) vanishes, since division by zero (or by something vanishing to some order) tends to introduce poles into pp or qq. A point can look perfectly innocuous — the original equation P,Q,RP,Q,R might all be polynomials with no obvious problem — yet still be singular if P(x0)=0P(x_0)=0, as in the Euler equation counterexample.

Why must we check the complex plane for singularities, not just the real line, when finding the radius of convergence?

Power series convergence is fundamentally a complex-analytic phenomenon: the radius of convergence of ∑cn(x−x0)n\sum c_n(x-x_0)^n equals the distance from x0x_0 to the nearest point in the complex plane where the function being expanded fails to be analytic. For 1/(1+x2)1/(1+x^2), there is no real singularity, but at x=±ix = \pm i the function blows up. Any power series centered at a real point x0x_0 expanding this function is blind to the fact that the singularity is complex — but the convergence radius still shrinks to reflect it. This is a manifestation of a deep fact from complex analysis: real-variable intuition about smoothness is insufficient to predict convergence radii.

Can the method produce all solutions, or just some?

At a genuine ordinary point, the method produces all solutions: the corollary to the theorem shows the two series solutions built from (a0,a1)=(1,0)(a_0,a_1)=(1,0) and (0,1)(0,1) form a fundamental set, so every solution near x0x_0 is a linear combination of these two. This is a complete solution of the equation locally, in sharp contrast to singular points, where power series in (x−x0)(x-x_0) alone may capture only part (sometimes none) of the solution space, as the Euler-equation counterexample demonstrates.

Do I have to expand p(x)p(x) and q(x)q(x) as full power series if they are not polynomials?

Yes, in principle: if p,qp, q are transcendental but analytic (like sin⁡x\sin x or exe^x), you expand them in their own Taylor series and use the Cauchy product formula to combine them with the series for y′,y′′y', y'', exactly as in the proof's Step 1. This produces a recurrence for an+2a_{n+2} in terms of all previous coefficients (a convolution sum), rather than a short finite recurrence. It is more laborious than the polynomial-coefficient case but no different in principle, and the existence theorem still guarantees convergence out to the nearest complex singularity of p,qp, q.

What is the difference between this and the Frobenius method?

This article's method applies exactly at ordinary points, producing solutions that are honestly analytic (power series with only nonnegative integer exponents) with no logarithms. The Frobenius method extends the strategy to regular singular points, where p(x)p(x) may have a simple pole and q(x)q(x) at worst a double pole; there, solutions generally take the form (x−x0)r(x-x_0)^r times a power series, possibly with logarithmic terms, as illustrated starkly by the xln⁡xx\ln x solution in the counterexample above. The ordinary-point theory is the foundation on top of which Frobenius theory is built, since the "nice" part of the Frobenius solution is still constructed by matching power-series coefficients.

If a0=a1=0a_0=a_1=0, does the method just give the trivial solution?

Yes — and this is expected, not a defect. The map from initial data (a0,a1)=(y(x0),y′(x0))(a_0,a_1) = (y(x_0), y'(x_0)) to the solution is linear and, by the existence-uniqueness theory for linear ODEs, injective in the sense that the only solution with y(x0)=y′(x0)=0y(x_0)=y'(x_0)=0 is y≡0y\equiv 0. The recurrence relation, when started from a0=a1=0a_0=a_1=0, forces every subsequent an=0a_n=0 as well, consistent with uniqueness.

Is it always possible to sum the resulting series in closed form, like ex2/2e^{x^2/2} in Example 1?

No, and this should not be expected in general. Example 1's odd-indexed solution and Example 2's solution do not reduce to elementary functions; this is the normal situation, not an exception. Many of the most important special functions in mathematical physics (Hermite, Legendre, Bessel, Airy functions) are literally defined as the power series produced by this method, precisely because they cannot be written in terms of elementary functions. The power-series representation itself, with its guaranteed convergence, is the final, complete answer.

Before this

Where this leads