 Solution

For$x_0$ is an ordinary point when $P$ and $Q$ are analytic there, and a singular point otherwise. A singular point is regular singular when $(x-x_0)P(x)$ and $(x-x_0{)}^{2}Q(x)$ are analytic there. These are the ordinary-point and regular-singular criteria.

For Kummer's equation, substituteEquating the coefficient of $x^m$ givesTaking $c_0=1$,where $(a{)}_m=a(a+1)\cdots (a+m-1)$ is the rising factorial. Thus

Putting $y=x^{1-b}u$ and simplifying givesThereforeFor nonintegral $b$, the powers $x^0$ and $x^{1-b}$ at zero are distinct, so these solutions are linearly independent.

When $b\to 1$, both solutions tend to $M(x,a,1)$. Differentiate their difference with respect to $b$. Writing $M_a,M_b$ for derivatives with respect to the second and third arguments,Hence at $b=1$ one may takeas two linearly independent solutions. Their independence follows from the logarithmic term in the second solution.