Power Series Solutions near Ordinary Points

Equations like $y'' - xy' - y = 0$ have variable coefficients, so the characteristic equation from constant-coefficient equations does not apply. But near a point where the coefficients are well behaved, the solution is a power series, and we can find its coefficients one at a time. The tool is the fact that you can differentiate a power series term by term.

Idea clave

At an ordinary point $x = 0$ (the coefficients of $y'' + p(x)y' + q(x)y = 0$ have power series there), every solution is $y = \sum_{n=0}^{\infty} a_nx^{n}$, with $a_0 = y(0)$ and $a_1 = y'(0)$ free. Substitute, shift every sum to the same power $x^{n}$, and set each coefficient to zero; this gives a recurrence for $a_{n+2}$. The series converges at least out to the nearest singular point (complex ones count).

Por qué funciona

Term by term, $y' = \sum_{n\ge 1} na_nx^{n-1}$ and $y'' = \sum_{n\ge 2} n(n - 1)a_nx^{n-2}$. Replacing $n$ by $n + 2$ in the second sum gives $\sum_{n\ge 0} (n + 2)(n + 1)a_{n+2}x^{n}$: the same terms, renamed so that the power is $x^{n}$. Once every sum is written with $x^{n}$, the equation reads $\sum c_nx^{n} = 0$, and a power series is zero on an interval only if every $c_n = 0$ (its coefficients are its derivatives at 0 divided by $n!$). Each $c_n = 0$ ties $a_{n+2}$ to earlier coefficients, so $a_0$ and $a_1$ determine everything, just as $y(0)$ and $y'(0)$ should.

Ejemplo resuelto: A familiar answer

Solve $y'' + y = 0$ with $y(0) = 1$, $y'(0) = 0$ by power series.

$\sum (n + 2)(n + 1)a_{n+2}x^{n}$ $+ \sum a_nx^{n}$ $= 0$, so $a_{n+2} = -\frac{a_n}{(n + 2)(n + 1)}$.
$a_0 = 1$, $a_1 = 0$: the odd coefficients vanish; $a_2 = -\frac{1}{2}$, $a_4 = \frac{1}{24}$, $a_6 = -\frac{1}{720}$.
$y = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \dots$

Respuesta: $y = \cos x$, as it should be

Ejemplo resuelto: A recurrence with $x$ in it

Solve $y'' - xy' - y = 0$ with $y(0) = 1$, $y'(0) = 0$.

$xy' = \sum na_nx^{n}$ already has power $x^{n}$; only $y''$ needs the shift.
Coefficient of $x^{n}$: $(n + 2)(n + 1)a_{n+2} - na_n - a_n$ $= 0$, so $a_{n+2} = \frac{a_n}{n + 2}$.
$a_2 = \frac{1}{2}$, $a_4 = \frac{1}{8}$, $a_6 = \frac{1}{48}$, and in general $a_{2k} = \frac{1}{2^{k}k!}$.

Respuesta: $y = \sum \frac{(x^{2}/2)^{k}}{k!} = e^{x^{2}/2}$ (check: $y'' - xy' - y$ $= (1 + x^{2} - x^{2} - 1)e^{x^{2}/2} = 0$)

Ejemplo resuelto: Airy's equation

Find the first terms of the solution of $y'' = xy$ with $y(0) = 1$, $y'(0) = 0$.

$xy = \sum a_nx^{n+1}$; shift it so the power is $x^{n}$: $\sum_{n\ge 1} a_{n-1}x^{n}$.
Coefficient of $x^{0}$: $2a_2 = 0$. For $n \ge 1$: $(n + 2)(n + 1)a_{n+2} = a_{n-1}$.
$a_3 = \frac{a_0}{6} = \frac{1}{6}$, $a_6 = \frac{a_3}{30} = \frac{1}{180}$; $a_1 = a_2 = 0$ kill the other chains.

Respuesta: $y = 1 + \frac{x^{3}}{6} + \frac{x^{6}}{180} + \dots$, a function with no elementary formula

Error común

Shifting the index on one sum but not the others. In the Airy example, writing $\sum (n + 2)(n + 1)a_{n+2}x^{n}$ $= \sum a_nx^{n+1}$ and then equating $a_{n+2}$ with $a_n$ compares different powers of $x$. Before matching, every sum must show the same power $x^{n}$ and start at the same $n$; peel off any leftover first terms (here $2a_2 = 0$) separately.

How far does the series reach? For $(1 + x^{2})y'' + y = 0$ the leading coefficient vanishes at $x = \pm i$, distance 1 from 0, so the series is guaranteed to converge for $|x| < 1$, even though nothing goes wrong on the real line.

Partial sums of the series solution from the second example, $1 + \frac{x^{2}}{2} + \frac{x^{4}}{8} + \dots$, closing in on $e^{x^{2}/2}$. Raise the degree and watch the interval of good agreement widen.

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