At a singular point, where the leading coefficient vanishes, a plain power series can fail. Bessel's equation $x^{2}y'' + xy' + (x^{2} - \nu^{2})y = 0$ is the standard example. If the singularity is mild, a series times a power $x^{r}$ still works. The method combines power series solutions with the $x^{r}$ guess of Cauchy–Euler equations.
🔑 Key idea
Write the equation as $y'' + p(x)y' + q(x)y = 0$. The point $x = 0$ is a regular singular point if $p$ or $q$ fails to be analytic there but $xp(x)$ and $x^{2}q(x)$ are analytic; otherwise it is irregular. At a regular singular point, try $y = \sum_{n=0}^{\infty} a_nx^{n+r}$ with $a_0 \ne 0$. The lowest power gives the indicial equation $$r(r - 1) + p_0r + q_0 = 0,$$ $$p_0 = \lim_{x\to 0} xp(x),\ q_0 = \lim_{x\to 0} x^{2}q(x).$$ If the roots do not differ by an integer, each gives a Frobenius solution.
💡 Why it works
Multiply by $x^{2}$: $x^{2}y'' + x\left(xp\right)y' + \left(x^{2}q\right)y$ $= 0$. Near 0, $xp \approx p_0$ and $x^{2}q \approx q_0$, so the equation looks like the Cauchy–Euler equation $x^{2}y'' + p_0xy' + q_0y = 0$, whose solutions are $x^{r}$. The Frobenius series corrects that leading behavior with a power series. Substituting, the lowest power $x^{r}$ appears only from $a_0$, with coefficient $a_0\left(r(r - 1) + p_0r + q_0\right)$; since $a_0 \ne 0$, the bracket must vanish. Higher powers give a recurrence. When the roots differ by an integer $N$, the recurrence for the smaller root can break down at $n = N$, and the second solution may need a $\ln x$ term.
✏️ Worked example: Classify the point
Classify $x = 0$ for $x^{2}y'' + xy' + \left(x^{2} - \frac{1}{4}\right)y = 0$ and find the indicial roots.
- $p = \frac{1}{x}$ and $q = \frac{x^{2} - 1/4}{x^{2}}$ blow up at 0, but $xp = 1$ and $x^{2}q = x^{2} - \frac{1}{4}$ are polynomials: regular singular.
- $p_0 = 1$, $q_0 = -\frac{1}{4}$: $r(r - 1) + r - \frac{1}{4} = r^{2} - \frac{1}{4} = 0$.
Answer: $r = \pm\frac{1}{2}$ (this is Bessel's equation of order $\frac{1}{2}$; its solutions are $\frac{\sin x}{\sqrt{x}}$ and $\frac{\cos x}{\sqrt{x}}$)
✏️ Worked example: A recurrence
For $2x^{2}y'' - xy' + (1 + x)y = 0$, find the indicial roots and the first terms of the solution for the larger root.
- Substitute $y = \sum a_nx^{n+r}$; the term $xy$ contributes $a_{n-1}$ to the power $x^{n+r}$.
- Coefficient of $x^{n+r}$: $\left(2(n + r)(n + r - 1) - (n + r) + 1\right)a_n$ $+ a_{n-1}$ $= 0$. With $F(s) = 2s^{2} - 3s + 1$ $= (2s - 1)(s - 1)$, that is $F(n + r)a_n = -a_{n-1}$.
- At $n = 0$: $F(r) = 0$, so $r = 1$ or $r = \frac{1}{2}$.
- For $r = 1$: $F(n + 1) = n(2n + 1)$, so $a_n = -\frac{a_{n-1}}{n(2n + 1)}$: $a_1 = -\frac{1}{3}$, $a_2 = \frac{1}{30}$.
Answer: $y_1 = x\left(1 - \frac{x}{3} + \frac{x^{2}}{30} - \dots\right)$; the root $\frac{1}{2}$ gives $y_2 = x^{1/2}\left(1 - x + \frac{x^{2}}{6} - \dots\right)$
✏️ Worked example: An irregular point
Classify $x = 0$ for $x^{3}y'' + y = 0$.
- $q = \frac{1}{x^{3}}$, so $x^{2}q = \frac{1}{x}$, which is not analytic at 0.
Answer: irregular singular point: the Frobenius method does not apply
⚠️ Common mistake
Testing $p$ and $q$ before dividing by the leading coefficient, or checking $xq$ instead of $x^{2}q$. Put the equation in the form $y'' + py' + qy = 0$ first; then $p$ may have at most a simple pole and $q$ at most a double pole at a regular singular point.
When the indicial roots are equal, or differ by an integer, one Frobenius solution always exists (for the larger root). The second one may contain $y_1\ln x$; for Bessel's equation of order 0 it does, and that solution is unbounded at $x = 0$.