1. When series methods apply
The method targets linear ODEs whose coefficients are either polynomials or, more generally, analytic functions — functions that have their own power-series expansions. The typical setup is a second-order equation
$$ P(x)\,y'' + Q(x)\,y' + R(x)\,y = 0, $$or equivalently, after dividing through by $P(x)$ where it's nonzero,
$$ y'' + p(x)\,y' + q(x)\,y = 0. $$If $p(x)$ and $q(x)$ are well-behaved (analytic) at a point $x_0$, you can hope for a solution of the form
$$ y(x) = \sum_{n=0}^{\infty} a_n\,(x - x_0)^n, $$and find the coefficients $a_n$ by plugging the series into the ODE and matching powers of $x - x_0$.
Equations like $y'' + xy = 0$ (Airy) or $x^2 y'' + xy' + (x^2 - \nu^2)y = 0$ (Bessel) show up constantly in physics — quantum mechanics, optics, vibrating membranes — and their solutions are their power series. You don't get a "nicer" formula by working harder; the series is the formula.
2. Ordinary points vs singular points
A point $x_0$ is an ordinary point of $y'' + p(x)y' + q(x)y = 0$ if both $p(x)$ and $q(x)$ are analytic at $x_0$ — meaning they have convergent Taylor series around it. Otherwise $x_0$ is a singular point.
Singular points get a finer classification:
A singular point $x_0$ is regular if both $(x - x_0)\,p(x)$ and $(x - x_0)^2\,q(x)$ are analytic at $x_0$ — i.e., the singularities are mild enough that multiplying by a small power of $(x - x_0)$ cleans them up. Otherwise $x_0$ is irregular, and series methods don't directly apply.
The classification determines which method to use:
| Point type | Method | Solution shape |
|---|---|---|
| Ordinary | Plain power series | $\sum a_n (x - x_0)^n$ |
| Regular singular | Frobenius's method | $(x - x_0)^r \sum a_n (x - x_0)^n$ |
| Irregular singular | Beyond elementary methods | Asymptotic expansions, transforms, special techniques |
For example, in Bessel's equation $x^2 y'' + xy' + (x^2 - \nu^2)y = 0$, dividing through by $x^2$ gives $p(x) = 1/x$ and $q(x) = 1 - \nu^2/x^2$. Both are singular at $x_0 = 0$, but $x \cdot (1/x) = 1$ and $x^2 \cdot (1 - \nu^2/x^2) = x^2 - \nu^2$ are both analytic — so $x = 0$ is a regular singular point, and Frobenius's method applies.
3. The algorithm at an ordinary point
Assume the solution as a power series, plug it in, force the equation to hold term by term. Here's the recipe in five steps, with $x_0 = 0$ for clarity.
- Posit the series. Write $y(x) = \sum_{n=0}^{\infty} a_n x^n$.
- Differentiate term by term. $y'(x) = \sum_{n=1}^{\infty} n\,a_n\,x^{n-1}$ and $y''(x) = \sum_{n=2}^{\infty} n(n-1)\,a_n\,x^{n-2}$.
- Substitute into the ODE. Replace $y$, $y'$, $y''$ by their series.
- Shift indices so every series uses the same power of $x$. This is the bookkeeping step where most mistakes happen — see the pitfalls section.
- Collect coefficients and set each coefficient of $x^n$ to zero. Solve the resulting recurrence relation for $a_n$ in terms of earlier coefficients.
The first two coefficients, $a_0$ and $a_1$, remain free — they're the two constants of integration that a second-order ODE must have. Everything else is determined.
4. The recurrence relation
The recurrence is what makes the method practical: instead of solving for infinitely many $a_n$ at once, you solve for each one in terms of those before it. A typical recurrence has the form
$$ a_{n+2} = f(n)\,a_n \quad \text{or} \quad a_{n+2} = f(n)\,a_n + g(n)\,a_{n+1}, $$and you generate the coefficients one at a time:
$$ a_2 = f(0)\,a_0, \quad a_3 = f(1)\,a_1, \quad a_4 = f(2)\,a_2 = f(2)\,f(0)\,a_0, \quad \ldots $$The even-indexed coefficients trace back to $a_0$; the odd-indexed ones trace back to $a_1$. The two free constants split the solution into two independent families:
$$ y(x) = a_0\,\underbrace{(1 + \cdots)}_{y_1(x)} + a_1\,\underbrace{(x + \cdots)}_{y_2(x)}. $$Those two pieces $y_1$ and $y_2$ are linearly independent and span the two-dimensional solution space. The free constants $a_0$ and $a_1$ are then fixed by initial conditions $y(0)$ and $y'(0)$.
5. Convergence: how far does the series carry you?
A series solution is only useful where it converges. The key theorem is reassuring:
If $x_0$ is an ordinary point of $y'' + p(x)y' + q(x)y = 0$ and the Taylor series of $p$ and $q$ around $x_0$ both have radius of convergence at least $R$, then any series solution has radius of convergence at least $R$ as well.
So $R$ is determined by the distance from $x_0$ to the nearest singularity of the coefficients in the complex plane. For polynomial coefficients with no singularities, $R = \infty$ — the series converges everywhere.
To check convergence of a specific series, the ratio test on the recurrence usually does the job. If
$$ \left|\frac{a_{n+1}}{a_n}\right| \to L \quad \text{as } n \to \infty, $$then the radius of convergence is $1/L$ (Cauchy-Hadamard). For most well-behaved ODEs, $L = 0$ and the series converges on the whole real line.
Even when the radius of convergence is infinite, the rate of convergence often isn't great for large $|x|$. To get accurate values far from $x_0$ you may need hundreds of terms. Series solutions are usually combined with asymptotic methods that work where the series struggles.
6. Frobenius's method at a regular singular point
At a regular singular point, a plain power series usually fails — but a tweaked one succeeds. Frobenius's idea: try a series with an extra power of $x$ out front:
$$ y(x) = x^r \sum_{n=0}^{\infty} a_n x^n = \sum_{n=0}^{\infty} a_n x^{n+r}, \qquad a_0 \neq 0, $$where $r$ is an unknown exponent to be determined. Substitute into the ODE, set the lowest-order coefficient to zero, and you get the indicial equation — a polynomial in $r$ whose roots are the allowed exponents.
The indicial equation
For the standard form $y'' + p(x)y' + q(x)y = 0$ at a regular singular point $x_0 = 0$, write
$$ p(x) = \frac{p_0 + p_1 x + \cdots}{x}, \qquad q(x) = \frac{q_0 + q_1 x + \cdots}{x^2}. $$The indicial equation reads
$$ r(r - 1) + p_0\,r + q_0 = 0. $$It has two roots $r_1 \geq r_2$ (real or complex). The structure of the two independent solutions depends on how the roots relate:
| Case | Form of the second solution |
|---|---|
| $r_1 - r_2$ not an integer | Two independent Frobenius series: $x^{r_1}\sum a_n x^n$ and $x^{r_2}\sum b_n x^n$. |
| $r_1 = r_2$ (repeated root) | Second solution contains a $\log x$ term: $y_2 = y_1 \ln x + x^{r_1}\sum c_n x^n$. |
| $r_1 - r_2$ is a nonzero integer | Second solution may need a $\log x$ term — sometimes it doesn't, depending on the recurrence. |
7. Famous examples: Airy and Bessel
Airy's equation
The equation $y'' - xy = 0$ governs quantum particles in a linear potential and the diffraction pattern near a caustic. Series solution around $x_0 = 0$ (an ordinary point):
$$ y = a_0\!\left(1 + \tfrac{x^3}{6} + \tfrac{x^6}{180} + \cdots\right) + a_1\!\left(x + \tfrac{x^4}{12} + \tfrac{x^7}{504} + \cdots\right). $$The two independent solutions are the Airy functions $\text{Ai}(x)$ and $\text{Bi}(x)$ — distinguished from each other by their behavior at infinity, not at the origin.
Bessel's equation of order zero
The equation $x^2 y'' + xy' + x^2 y = 0$ has a regular singular point at $0$ with indicial roots $r = 0, 0$ (a repeated root). The Frobenius series with $r = 0$ produces
$$ J_0(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(n!)^2}\!\left(\tfrac{x}{2}\right)^{2n} = 1 - \tfrac{x^2}{4} + \tfrac{x^4}{64} - \tfrac{x^6}{2304} + \cdots, $$the Bessel function of the first kind, order zero. The second independent solution $Y_0(x)$ contains a logarithm (because of the repeated root) and blows up at $x = 0$.
These functions show up so often in physics that mathematicians simply gave them names and tabulated their values. There's no "simpler" form to reduce them to — they're as elementary as $\sin$ and $\cos$ once you accept that an "elementary function" is any function whose properties you've internalized.
8. Common pitfalls
When you differentiate $\sum_{n=0}^\infty a_n x^n$ you get $\sum_{n=1}^\infty n a_n x^{n-1}$. To combine it with the original series under a common $x^n$, you have to re-index: let $m = n - 1$, then the sum is $\sum_{m=0}^\infty (m+1) a_{m+1} x^m$. Skip this step and your recurrence will be wrong by an offset.
$x^2 y'' + y = 0$ looks like a polynomial-coefficient ODE, but dividing through by $x^2$ gives $y'' + (1/x^2) y = 0$ — singular at $x = 0$. Always normalize to leading coefficient $1$ before classifying the point.
"I plugged in the series, got a recurrence, therefore I have a solution" — only partly true. You have a formal series. Whether that series converges (and where) is a separate question. The theorem in section 5 gives you a guarantee at ordinary points; at singular points you have to check explicitly.
If you try $y = \sum a_n x^n$ at a regular singular point, you'll usually get a recurrence that forces $a_0 = 0$ — and then $a_1 = 0$, and then everything is zero. The whole-zero "solution" is the universe's way of telling you to switch to the Frobenius form $y = x^r \sum a_n x^n$ instead.
9. Worked examples
Example 1 · Solve $y'' + y = 0$ by power series (recover sine and cosine)
Step 1. Posit $y = \sum_{n=0}^{\infty} a_n x^n$. Then
$$ y'' = \sum_{n=2}^{\infty} n(n-1)\,a_n\,x^{n-2}. $$Step 2. Shift: let $m = n - 2$, so $y'' = \sum_{m=0}^{\infty} (m+2)(m+1)\,a_{m+2}\,x^m$. Rename $m \to n$ for clarity.
Step 3. Substitute into $y'' + y = 0$ and collect:
$$ \sum_{n=0}^{\infty} \!\Big[(n+2)(n+1)\,a_{n+2} + a_n\Big]\,x^n = 0. $$Each coefficient must vanish, giving the recurrence:
$$ a_{n+2} = -\frac{a_n}{(n+1)(n+2)}. $$Step 4. Iterate. Even-indexed: $a_2 = -a_0/2!$, $a_4 = a_0/4!$, $a_6 = -a_0/6!$. Odd-indexed: $a_3 = -a_1/3!$, $a_5 = a_1/5!$, $a_7 = -a_1/7!$.
Step 5. Assemble:
$$ y = a_0\!\left(1 - \tfrac{x^2}{2!} + \tfrac{x^4}{4!} - \cdots\right) + a_1\!\left(x - \tfrac{x^3}{3!} + \tfrac{x^5}{5!} - \cdots\right) = a_0 \cos x + a_1 \sin x. $$The series method recovers cosine and sine — as you'd hope, since they really are defined by being the solutions to this equation.
Example 2 · First few terms of the Airy function from $y'' = xy$
Step 1. Posit $y = \sum_{n=0}^\infty a_n x^n$. Then $y'' = \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n$ (after shifting). And $xy = \sum_{n=0}^\infty a_n x^{n+1} = \sum_{n=1}^\infty a_{n-1} x^n$.
Step 2. Subtract: $y'' - xy = 0$ becomes
$$ 2 a_2 + \sum_{n=1}^\infty \Big[(n+2)(n+1) a_{n+2} - a_{n-1}\Big] x^n = 0. $$Step 3. Setting each coefficient to zero: $a_2 = 0$, and
$$ a_{n+2} = \frac{a_{n-1}}{(n+1)(n+2)} \quad \text{for } n \geq 1. $$Step 4. Iterate. $a_3 = a_0 / 6$, $a_4 = a_1 / 12$, $a_5 = 0$, $a_6 = a_3 / 30 = a_0 / 180$, $a_7 = a_4 / 42 = a_1 / 504$. The series is:
$$ y = a_0\!\left(1 + \tfrac{x^3}{6} + \tfrac{x^6}{180} + \cdots\right) + a_1\!\left(x + \tfrac{x^4}{12} + \tfrac{x^7}{504} + \cdots\right). $$Every third coefficient is zero, and the two independent pieces are the standard Airy functions up to normalization.
Example 3 · Indicial roots of a regular singular point
Consider $2x^2 y'' + 3xy' - (1 + x) y = 0$. Verify $x = 0$ is a regular singular point and find the indicial roots.
Step 1. Standard form: divide by $2x^2$ to get $y'' + \tfrac{3}{2x} y' - \tfrac{1 + x}{2x^2} y = 0$. So $p(x) = 3/(2x)$ and $q(x) = -(1+x)/(2x^2)$.
Step 2. Check: $x p(x) = 3/2$ (analytic) and $x^2 q(x) = -(1+x)/2$ (analytic). Both are analytic at $0$, so $x = 0$ is a regular singular point. Read off $p_0 = 3/2$ and $q_0 = -1/2$.
Step 3. Indicial equation: $r(r - 1) + \tfrac{3}{2} r - \tfrac{1}{2} = 0$, i.e. $r^2 + \tfrac{1}{2} r - \tfrac{1}{2} = 0$. Multiply by 2: $2r^2 + r - 1 = 0$, giving $r = \tfrac{1}{2}$ or $r = -1$.
Since the roots differ by $\tfrac{3}{2}$ (not an integer), there are two independent Frobenius series — one starting with $x^{1/2}$ and one with $x^{-1}$.
Example 4 · Convergence radius of the Legendre equation series
Legendre's equation is $(1 - x^2) y'' - 2x y' + n(n+1) y = 0$, with $n$ a constant. Dividing through by $1 - x^2$ gives $p(x) = -2x/(1 - x^2)$ and $q(x) = n(n+1)/(1 - x^2)$. Both are analytic at $x = 0$, so it's an ordinary point.
The nearest singularities of $p$ and $q$ are at $x = \pm 1$ (where $1 - x^2 = 0$). The convergence theorem then guarantees: any power-series solution around $x_0 = 0$ converges at least for $|x| < 1$.
For integer values of $n$, one of the two series actually terminates into a polynomial of degree $n$ — these are the Legendre polynomials $P_n(x)$, and they're (trivially) entire. The other solution still has only a unit disk of convergence and blows up at $x = \pm 1$.