Series Solutions

A series solution is simply an answer written as an endless polynomial, built one term at a time, for equations where no nice formula exists. ("Series" just means an infinite sum, and it's a power series because each term is a power of $t$.) All we're really saying here is: if you can't guess the answer's formula, let the equation tell you its terms. The series $y = \sum a_nt^n$ looks intimidating, but it's just $a_0 + a_1t + a_2t^2 + \cdots$ (a polynomial that never stops), with each $a_n$ a plain number to find, and $y'$ and $y''$ are the same polynomial differentiated term by term. The equation then works like a machine that produces each number from the ones before it. It looks like a lot of sums, but it's just bookkeeping; bear with it.

Every second order equation we've solved so far had constant coefficients, and the guess $y = e^{rt}$ cracked them because each derivative of $e^{rt}$ is just a multiple of $e^{rt}$. Once a coefficient depends on $t$, as in $y'' - ty = 0$, that breaks: $te^{rt}$ isn't a multiple of $e^{rt}$, so no choice of $r$ makes the terms cancel.

So what can we do now? Assume the solution has a power series, $y = \sum a_nt^n$, and let the equation tell us the coefficients. Notice that $a_0 = y(0)$ and $a_1 = y'(0)$, so those two are free, the same two pieces of starting information any second order equation needs.

Steps:

1. Substitute the series for $y$, $y'$, $y''$.
2. Shift indices so every sum has $t^n$.
3. Set the coefficient of each power to zero: that's the recurrence.
4. Generate coefficients from $a_0$ and $a_1$, and recognize the pattern if you can.

Think about why step 3 works. After substituting, the whole left side is one power series that has to equal $0$ for every $t$. A power series is the zero function only when every one of its coefficients is zero, so each power of $t$ gives its own equation.

A common mistake. It's tempting to compare coefficients right after substituting, because each sum seems to come with its own $a_n$. It isn't safe: in the series for $y''$, the term with $a_n$ multiplies $t^{n-2}$, not $t^n$. Shift the indices first so every sum is written in $t^n$, then compare.

Example: $y'' + y = 0$. The recurrence is $a_{n+2} = -\dfrac{a_n}{(n + 2)(n + 1)}$. Starting from $a_0 = 1, a_1 = 0$ gives $1 - \frac{t^2}{2!} + \frac{t^4}{4!} - \cdots = \cos t$; from $a_0 = 0, a_1 = 1$ gives $\sin t$. That's a nice check: the series method lands on exactly the answer the characteristic equation gave us.

Airy's equation $y'' - ty = 0$ has no elementary solutions, but the series works perfectly: $a_{n+2} = \dfrac{a_{n-1}}{(n + 2)(n + 1)}$ with $a_2 = 0$. Here the series isn't a way to rediscover a known function. It's the only way to write the solution down at all.

How far out can you trust the series? It converges at least out to the nearest point where the leading coefficient vanishes (a singular point). The graph further down shows the partial sums for $\cos t$ reaching further as you add terms.

The special functions of physics (Bessel functions for drums and cylinders, Legendre polynomials for spheres, Hermite functions in quantum mechanics) are defined by series solutions.