Power Series Solutions of ODEs

When power series solutions are the right tool

Many ODEs — Legendre, Bessel, Hermite, Airy — have no closed form in elementary functions, but they do have power-series solutions around a regular point. The strategy is always the same:

1. Guess $y(x) = \sum_{n = 0}^{\infty} a_n x^{n}$.
2. Compute derivatives term by term.
3. Plug into the ODE and collect coefficients of like powers of $x$.
4. Set each coefficient to zero → get a recursion relating $a_n$ to earlier $a_k$.
5. Solve the recursion given enough initial data (usually $a_0$ from $y(0)$ and $a_1$ from $y'(0)$).
6. Recognise the resulting series if you can.

This problem applies the method to the simplest nontrivial ODE in existence — $y' = y$ — so every step is visible and the final series is something you already know ($e^{x}$).

The given ODE

$y' = y, \qquad y(0) = 1 \text{ (for definiteness)}.$

Step-by-step

Step 1 — Ansatz.

$y(x) = \sum_{n = 0}^{\infty} a_n \, x^{n}.$

Step 2 — Differentiate term by term.

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

To match the index with $y$'s, reindex by $m = n - 1$ (so $n = m + 1$): $y'(x) = \sum_{m = 0}^{\infty} (m + 1) \, a_{m + 1} \, x^{m}.$

Rename $m \to n$ — still the same series: $y'(x) = \sum_{n = 0}^{\infty} (n + 1) \, a_{n + 1} \, x^{n}.$

Step 3 — Substitute into the ODE.

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

Step 4 — Match coefficients of $x^{n}$.

$(n + 1) \, a_{n + 1} = a_n \qquad \text{for every } n \ge 0.$

Step 5 — Solve the recursion.

$a_{n + 1} = \frac{a_n}{n + 1}.$

Start from $a_0$ and roll forward: $a_1 = \frac{a_0}{1}, \quad a_2 = \frac{a_1}{2} = \frac{a_0}{2!}, \quad a_3 = \frac{a_2}{3} = \frac{a_0}{3!}, \ldots$ $a_n = \frac{a_0}{n!}.$

Step 6 — Assemble the series.

$y(x) = a_0 \sum_{n = 0}^{\infty} \frac{x^{n}}{n!}.$

With $y(0) = a_0 = 1$: $\boxed{\, y(x) = \sum_{n = 0}^{\infty} \frac{x^{n}}{n!} = e^{x} \,}$

The series is the Taylor series of $e^{x}$, converging for all $x \in \mathbb{R}$ (radius of convergence $R = \infty$).

Verification

Differentiate the series term by term: $y'(x) = \sum_{n = 1}^{\infty} \frac{n \, x^{n - 1}}{n!} = \sum_{n = 1}^{\infty} \frac{x^{n - 1}}{(n - 1)!} = \sum_{m = 0}^{\infty} \frac{x^{m}}{m!} = y(x). \; \checkmark$

This is the classic reason $e^{x}$ is its own derivative — the factorial in each term exactly cancels the power that falls out when you differentiate.

Radius of convergence

The recursion gives $a_{n+1}/a_n = 1/(n + 1) \to 0$. Ratio test for the series: $\lim_{n \to \infty} \left|\frac{a_{n+1} x^{n+1}}{a_n x^{n}}\right| = |x| \cdot \lim \frac{1}{n+1} = 0 < 1 \;\text{for all } x.$

So the series converges everywhere — the solution is entire (analytic on the whole real line). This is a general feature of linear ODEs with analytic coefficients: the series always converges at least as far as the nearest singularity of the coefficient functions.

When the ODE has non-constant coefficients — regular vs singular points

For the ODE $y'' + p(x) y' + q(x) y = 0$:

• $x_0$ is an ordinary point if $p$ and $q$ are analytic at $x_0$. Power series centred at $x_0$ converge in a disk around $x_0$ at least up to the nearest singularity — Frobenius isn't needed, the usual ansatz works.
• $x_0$ is a regular singular point if $p$ or $q$ fails to be analytic but $(x - x_0) p(x)$ and $(x - x_0)^{2} q(x)$ are analytic. Use the Frobenius method — ansatz $y = \sum a_n (x - x_0)^{n + r}$ with an unknown exponent $r$.
• Otherwise, it's an irregular singular point — power series may diverge; asymptotic methods are needed instead.

Bessel's equation has a regular singular point at $x = 0$; Hermite and Legendre have ordinary points. Each generates its characteristic family via this machinery.

A second worked example — Airy's equation

$y'' = x y.$

Ansatz $y = \sum a_n x^{n}$. Then $y'' = \sum n(n-1) a_n x^{n-2} = \sum (n+2)(n+1) a_{n+2} x^{n}$ (reindexed). Substituting: $(n + 2)(n + 1) a_{n + 2} = a_{n - 1} \quad \text{for } n \ge 1,$ with $a_2 = 0$ (from the $x^0$ matching, since $x y$ has no constant term). The recursion steps by $3$, producing two independent series: one from $a_0$ (even powers shifted by $0$) and one from $a_1$ (powers shifted by $1$). Both converge on all of $\mathbb{R}$. These are the famous Airy functions.

Common mistakes

• Reindexing errors. When substituting, line up powers of $x$ carefully by shifting indices. Off-by-one slips are the most common failure mode.
• Forgetting that coefficients vanish independently. If two series agree, their $x^{n}$ coefficients must match for every $n$. One recursion per power; that's how you extract all the info.
• Not supplying enough initial data. For a first-order ODE you need $a_0$. For second-order you need $a_0$ and $a_1$. The rest is determined by the recursion.
• Assuming the series converges everywhere without checking. It usually does for the textbook examples, but singular points of $p, q$ limit the radius.

Try it in the visualization

Watch the partial-sum series $y_N(x) = \sum_{n=0}^{N} a_n x^{n}$ grow toward $e^{x}$ as $N$ increases. Slide $N$ and see each new term "fix" the approximation over a wider interval — classic Taylor-series visualisation of how more terms buy more accuracy.