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Power-Series Solutions

When an equation has no closed-form answer, guess the answer is a power series and let the equation tell you its coefficients — one recurrence at a time. This is the doorway through which Legendre, Hermite, and Bessel will all arrive.

When no formula will answer

By now you can solve a wide menu of ordinary differential equations with formulas: separable ones, linear first-order ones with an integrating factor, constant-coefficient ones with a characteristic equation. But these are the lucky cases. Write down something as innocent-looking as y'' - x y = 0 (the Airy equation) and every closed-form trick fails — there is no combination of powers, exponentials, logs, and sines that satisfies it. The honest response is not to give up but to change what we mean by 'an answer'. We will build the solution as an infinite series.

The idea rests squarely on something you already met in Volume I. A power series is an infinite polynomial, y = sum from n = 0 to infinity of a_n x^n = a_0 + a_1 x + a_2 x^2 + ..., and you learned that within its radius of convergence it behaves like a polynomial you can trust: you can differentiate it term by term and add and multiply such series freely. A Taylor series is the special case where the coefficients come from derivatives of a known function. Here we flip the logic. We do NOT know the function; we treat the coefficients a_n as unknowns and let the differential equation hunt them down for us.

Ordinary points: where the method is allowed to work

Before guessing a series we must ask where we are allowed to. Put a second-order linear equation into the standard form y'' + P(x) y' + Q(x) y = 0. A point x_0 is an ordinary point if both P and Q are well-behaved there — precisely, if each is analytic (itself equal to a convergent power series) at x_0. If either coefficient blows up, x_0 is instead a singular point, and a plain power series may not suffice. This distinction between ordinary and singular points is not bureaucratic fine print: it is exactly the line between 'a tidy power series will work' and 'you need a heavier tool'.

At an ordinary point a guarantee kicks in, and it is worth stating plainly because it is what makes the whole method legitimate rather than wishful: if x_0 is an ordinary point, the equation HAS two independent solutions, each given by a power series in (x - x_0) that converges in a real disk around x_0. So we are not hoping a series might exist — a theorem promises it does, and even tells us, as we'll see, how far it converges. The two free constants a_0 and a_1 you leave undetermined are exactly the two initial conditions y(x_0) and y'(x_0); they are the room a second-order equation always leaves for a general solution.

The machine: substitute, align, and match

The whole procedure is a small piece of bookkeeping done with care. You write y as an undetermined series, differentiate it term by term to get y' and y'' (which is allowed inside the radius of convergence — that licence comes straight from the Volume I theory of power series), substitute all three into the equation, and then demand that the result be identically zero. The key insight is what 'identically zero' forces: a power series is zero for ALL x in an interval only if every single coefficient is zero. So matching the coefficient of each power of x turns one differential equation into infinitely many ordinary algebraic equations, one per power.

  1. Write the guess: y = sum a_n x^n (centered at the ordinary point, here x_0 = 0). Differentiate term by term: y' = sum n a_n x^{n-1}, y'' = sum n(n-1) a_n x^{n-2}.
  2. Substitute all the pieces into the equation, so every term is now some constant times a power of x.
  3. Shift indices so that every sum runs over the SAME power x^k. This re-labelling is the only fiddly step — it lines the sums up so you can read off one coefficient bundle per power.
  4. Set the coefficient of each x^k to zero. Solving that for the highest-index a expresses it in terms of earlier ones — that is your recurrence relation.

Step three deserves a word, because it trips everyone the first time. After substituting, y'' starts at the power x^0 but its index n starts at 2; meanwhile a term like x times y starts at x^1. To compare them you re-label the summation index so all the sums speak about the same power x^k — for instance writing y'' = sum from k = 0 of (k+2)(k+1) a_{k+2} x^k. It feels like a sleight of hand the first time, but it is just renaming the variable you sum over, exactly like renaming a dummy variable in a definite integral. Once the powers align, the coefficient-matching reads off cleanly.

A worked picture: building the cosine from scratch

Let us run the machine on an equation whose answer we already know, so we can watch it work and trust the output: the harmonic oscillator y'' + y = 0, expanded around the ordinary point x_0 = 0. Substitute y = sum a_n x^n. The y'' term contributes (k+2)(k+1) a_{k+2} x^k after the index shift, and the y term contributes a_k x^k directly. Adding and demanding the coefficient of x^k vanish gives (k+2)(k+1) a_{k+2} + a_k = 0 for every k. Solve for the newest coefficient: a_{k+2} = -a_k / [(k+2)(k+1)]. That single line is the recurrence relation.

y'' + y = 0,   y = sum_{n>=0} a_n x^n

recurrence:  a_{k+2} = - a_k / [ (k+2)(k+1) ]

leave a_0, a_1 free  (= y(0), y'(0))

even chain (from a_0):           odd chain (from a_1):
 a_2 = -a_0/2!                    a_3 = -a_1/3!
 a_4 = +a_0/4!                    a_5 = +a_1/5!
 a_6 = -a_0/6!                    a_7 = -a_1/7!

y = a_0 (1 - x^2/2! + x^4/4! - ...)  +  a_1 (x - x^3/3! + x^5/5! - ...)
  = a_0 cos(x)  +  a_1 sin(x)
The recurrence splits into an even chain fed by a_0 and an odd chain fed by a_1; summing them recovers cos(x) and sin(x) — the two solutions the theorem promised.

Notice how the recurrence relation organizes everything. It jumps by two, so the even coefficients form one chain reaching back to a_0 and the odd ones form a separate chain reaching back to a_1. The two starting values never mix — which is exactly why we end up with two independent solutions and two free constants. Following the even chain gives 1 - x^2/2! + x^4/4! - ..., the Taylor series of cos(x); the odd chain gives the Taylor series of sin(x). The method rebuilt the trig functions out of pure algebra, without ever assuming we knew them. When the equation is one whose answer is NOT a familiar function, this same machine still runs, and the series IS the new function.

How far does it reach? The radius of convergence

A series solution is only worth anything where it converges, so the last question is: how big is the disk on which our infinite polynomial actually makes sense? Recall from Volume I that every power series has a radius of convergence R, a distance out to which it converges and beyond which it diverges. For a series built from a recurrence you can often find R directly with the ratio test on the coefficients. For y'' + y = 0 the ratios shrink without bound, so R is infinite — cos and sin really are entire, valid for every x, which matches what we already knew.

But there is a far slicker way to know R in advance, before computing a single coefficient, and it ties the whole story together. The theorem at an ordinary point guarantees that the radius of convergence of the series solution is AT LEAST the distance from the center x_0 to the nearest singular point of the equation — and here is the twist that earns its keep: that nearest singularity may live off the real line, in the complex plane. The equation (1 + x^2) y'' + ... has coefficients that go bad where 1 + x^2 = 0, that is at x = plus or minus i. Though those points are invisible on the real axis, they sit a distance 1 from the origin, and they cap the real radius of convergence at R = 1. The real solution feels a wall it cannot see.

Why this opens the rung

What you have in hand is a universal opener. Whenever an equation is linear, has an ordinary point, and refuses every closed-form trick, the power-series method still produces two solutions, honestly and constructively, as series whose coefficients obey a recurrence and whose reach is fixed by the nearest singularity. It is the calculus equivalent of dividing a hard number by hand: tedious in places, but it always terminates with a digit you can trust.

And it is the gateway the rest of this rung walks through. Feed the machine y'' - 2x y' + 2n y = 0 and the recurrence terminates after finitely many terms whenever n is a whole number — out fall the Hermite polynomials. Feed it (1 - x^2) y'' - 2x y' + l(l+1) y = 0 and the Legendre polynomials appear the same way. Bessel's equation has a singular point at the origin, so it needs the Frobenius upgrade — but that is the very next guide, and it is nothing more than this method given permission to start with a fractional power. Every great named function of mathematical physics is, at bottom, a recurrence relation waiting to be unrolled.