Series Solutions of Differential Equations
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 7.4: “Series Solutions of Differential Equations” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/7-4-series-solutions-of-differential-equations |
| Textbook used in class | Stewart, Calculus, Section 17.4: “Series Solutions” (Examples 1, 2) |
Quick Reference
Method: assume $y = \displaystyle\sum_{n=0}^{\infty} a_n x^n$. Substitute into the DE. Collect like powers of $x$ and set each coefficient to zero. This produces a recurrence relation expressing $a_{n+2}$ (or $a_{n+1}$) in terms of earlier coefficients.
Two free parameters $a_0$ and $a_1$ yield two linearly independent series solutions.
Convergence: the series converges at least on the interval $|x| < R$, where $R$ is the distance to the nearest singular point of the DE (a point where the leading coefficient vanishes).
Motivation
The characteristic equation method works when coefficients are constant. For equations like $y'' + xy = 0$ (the Airy equation, which models wave propagation near a turning point in optics and quantum mechanics), the coefficient of $y$ varies with $x$ and there is no characteristic equation. Power series provide an alternative: assume the solution looks like a polynomial with infinitely many terms, substitute it in, and determine the coefficients recursively.
The series usually cannot be summed to a closed form, but the first several terms give excellent approximations near $x = 0$, and the recurrence relation reveals the pattern.
The Procedure
Write $y = \sum_{n=0}^\infty a_n x^n$, $y' = \sum_{n=1}^\infty n a_n x^{n-1}$, $y'' = \sum_{n=2}^\infty n(n-1)a_n x^{n-2}$.
Substitute into the DE and express everything as a series in powers of $x$.
Re-index sums so they all start at the same power (often $x^k$ for the same $k$).
Set the coefficient of each $x^k$ to zero. This gives a recurrence relation.
Solve the recurrence for $a_n$ in terms of $a_0$ and $a_1$.
Write $y = a_0 y_1(x) + a_1 y_2(x)$, where $y_1$ is the even-index series and $y_2$ is the odd-index series.
Worked Example
Find a series solution of $y'' + y = 0$ about $x = 0$ and identify the result.
Step 1. $y'' = \sum_{n=2}^\infty n(n-1)a_n x^{n-2}$. Substitute: $$\sum_{n=2}^\infty n(n-1)a_n x^{n-2} + \sum_{n=0}^\infty a_n x^n = 0.$$
Step 2. Re-index the first sum: let $m = n-2$, so $n = m+2$ and $n(n-1) = (m+2)(m+1)$: $$\sum_{m=0}^\infty (m+2)(m+1)a_{m+2}x^m + \sum_{m=0}^\infty a_m x^m = 0.$$
Step 3. Coefficient of $x^m$: $$(m+2)(m+1)a_{m+2} + a_m = 0 \implies a_{m+2} = \frac{-a_m}{(m+2)(m+1)}.$$
Step 4. Even terms (starting from $a_0$): $$a_2 = \frac{-a_0}{2!}, \quad a_4 = \frac{-a_2}{4\cdot 3} = \frac{a_0}{4!}, \quad a_6 = \frac{-a_4}{6\cdot 5} = \frac{-a_0}{6!}, \;\ldots$$
Odd terms (starting from $a_1$): $$a_3 = \frac{-a_1}{3!}, \quad a_5 = \frac{a_1}{5!}, \quad a_7 = \frac{-a_1}{7!}, \;\ldots$$
Step 5. General solution: $$y = a_0\underbrace{\left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots\right)}_{\cos x} + a_1\underbrace{\left(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\right)}_{\sin x}.$$
The series method recovers exactly $y = a_0\cos x + a_1\sin x$, confirming it agrees with the characteristic equation result.
the series method only works when the DE has no other solution method. The series method works for any DE whose coefficients are analytic at $x_0$ (meaning they have convergent Taylor series there). Applied to $y''+y=0$, it reproduces $\cos x$ and $\sin x$ exactly. The method is most valuable when no closed form exists, but it is not restricted to those cases. In fact, applying it to a solvable DE and recognizing the resulting series as $e^x$, $\cos x$, $\sin x$, etc. is a useful verification technique.
Leveled Practice
Problem 1. Use the power series method to find the recurrence relation for $y'' - y = 0$ and write the first four nonzero terms of each of the two linearly independent solutions.
Show answer
Substituting: $\sum (m+2)(m+1)a_{m+2}x^m - \sum a_m x^m = 0$.
Recurrence: $a_{m+2} = a_m / [(m+2)(m+1)]$.
Even series ($a_0 = 1$, $a_1 = 0$): $1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \cdots = \cosh x$.
Odd series ($a_0 = 0$, $a_1 = 1$): $x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots = \sinh x$.
The solutions are $\cosh x$ and $\sinh x$ (or equivalently $e^x$ and $e^{-x}$, since both pairs span the same space).