= Solution
For the normalized second-order equation, an <ordinary point> is one where both coefficient functions $p,q$ are analytic. A <singular point of a second-order linear ODE> is a point that is not ordinary. It is a <regular singular point> if $(x-x_0)p(x)$ and $(x-x_0)^2q(x)$ extend analytically to $x_0$; a singular point failing that condition is irregular.
Here $p=0$, $q=1/x$. Thus zero is singular but regular, because $xp=0$ and $x^2q=x$ are analytic. Write the analytic solution as $y=\sum_{n\geq0}a_nx^n$. Its constant equation gives $a_0=0$, and for $n\geq1$,
$$
n(n+1)a_{n+1}+a_n=0.
$$
Using $a_1=1$ gives
$$
\boxed{y(x)=\sum_{n=1}^\infty\frac{(-1)^{n-1}x^n}{(n-1)!n!}
=x-\frac{x^2}{2}+\frac{x^3}{12}-\frac{x^4}{144}+\cdots}.
$$
The factorial denominators show convergence for every finite $x$.
For the other branch, integrating the first correction equation twice and imposing its two conditions gives $y_1'=-\log x$ and
$$
\boxed{y_1=x-x\log x}.
$$
Then $y_2''=\log x-1$, so $y_2'=x\log x-2x+2$ and
$$
\boxed{y_2=\frac{x^2}{2}\log x-\frac{5x^2}{4}+2x}.
$$
Each correction tends to zero at the origin and its <derivative> vanishes at one. Although the second solution has a finite limiting value one, it has a <logarithmically divergent endpoint derivative>. Indeed the differential equation gives $y''\sim-1/x$, and integration gives
$$
\boxed{y'(x)\sim-\log x\longrightarrow+\infty\quad(x\downarrow0)}.
$$
A finite initial <derivative> cannot be prescribed for this second branch. The <logarithm> is consistent with the integer separation of the two exponents in the <Frobenius method>; this is not the repeated-exponent case.
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