= Solution
For
$$
y''+P(x)y'+Q(x)y=0,
$$
$x_0$ 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 $(x-x_0)^2Q(x)$ are analytic there. These are the ordinary-point and <regular singular point criterion for a second-order equation>[regular-singular] criteria.
For <Kummer differential equation>[Kummer's equation], substitute
$$
y=\sum_{m=0}^{\infty}c_mx^m.
$$
Equating the coefficient of $x^m$ gives
$$
(m+1)(m+b)c_{m+1}=(m+a)c_m.
$$
Taking $c_0=1$,
$$
\boxed{
c_m(a,b)=\frac{(a)_m}{(b)_m\,m!}},
$$
where $(a)_m=a(a+1)\cdots(a+m-1)$ is the rising factorial. Thus
$$
y_1=M(x,a,b)
=\sum_{m=0}^{\infty}\frac{(a)_m}{(b)_m\,m!}x^m.
$$
Putting $y=x^{1-b}u$ and simplifying gives
$$
xu''+(2-b-x)u'-(a-b+1)u=0.
$$
Therefore
$$
\boxed{
y_2=x^{1-b}M(x,a-b+1,2-b)}.
$$
For nonintegral $b$, the powers $x^0$ and $x^{1-b}$ at zero are distinct, so these solutions are linearly independent.
When $b\to1$, 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,
$$
\begin{aligned}
\lim_{b\to1}\frac{y_2-y_1}{b-1}
&=-M(x,a,1)\log x\\
&\quad-M_a(x,a,1)-2M_b(x,a,1).
\end{aligned}
$$
Hence at $b=1$ one may take
$$
\boxed{
M(x,a,1),\qquad
-M(x,a,1)\log x-M_a(x,a,1)-2M_b(x,a,1)}
$$
as two linearly independent solutions. Their independence follows from the logarithmic term in the second solution.
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