a regular singular point
Among singular points there is a forgiving kind. At a regular singular point the coefficients do blow up — so it is genuinely singular — but they blow up only mildly, slowly enough that a slightly modified series still works. This is the precise dividing line that decides whether the method of Frobenius will rescue you (it does at a regular singular point) or whether you are in deeper trouble (an irregular singular point).
Here is the test, and it is mechanical. Put the equation in standard form y'' + P(x) y' + Q(x) y = 0 and look at a singular point x0. Form the two quantities (x - x0) P(x) and (x - x0)^2 Q(x). If both of these are analytic at x0 — that is, the factor of (x - x0) is enough to cancel the blow-up in P, and (x - x0)^2 is enough for Q — then x0 is a regular singular point. The intuition: P is allowed to blow up no worse than 1/(x - x0), and Q no worse than 1/(x - x0)^2. Bessel's equation passes this test at x = 0: x times (1/x) = 1 and x^2 times (x^2 - nu^2)/x^2 = x^2 - nu^2, both perfectly analytic.
Why the fuss? Because at a regular singular point the solution behaves like (x - x0)^r times an ordinary power series, where r is a possibly fractional or even negative exponent fixed by the indicial equation. The mild singularity shows up exactly as that extra power (x - x0)^r out front. Recognising a regular singular point is the green light to launch a Frobenius series rather than a plain Taylor series.
For 2x^2 y'' + 3x y' - (x + 1) y = 0, standard form gives P = 3/(2x), Q = -(x+1)/(2x^2), both singular at x = 0. Test: x P = 3/2 (analytic) and x^2 Q = -(x+1)/2 (analytic). Both pass, so x = 0 is a regular singular point and Frobenius applies.
The factors (x - x0) and (x - x0)^2 must fully cancel the blow-ups in P and Q respectively; if they do, the point is regular singular.
Both conditions must hold. If (x - x0) P is fine but (x - x0)^2 Q still blows up (or vice versa), the point is irregular singular, and Frobenius is not guaranteed to work.