= Solution
Here the <indicial equation> has the repeated root $0$. The first <Frobenius method> solution is $y_1=e^{-x}$. Since $p=1+1/x$, the <Abel identity> gives $W=Ce^{-x}/x$. <Reduction of order> with $C=1$ produces
$$
y_2=e^{-x}\int\frac{e^x}{x}\,dx=e^{-x}\left(\log x+\sum_{j=1}^{\infty}\frac{x^j}{j\,j!}\right),\qquad x>0,
$$
where the additive constant is absorbed into $y_1$. In particular the logarithmic coefficient is nonzero: an additional independent <real analytic> solution does not exist.
\b[The general local form is]
$$
\boxed{y(x)=C_1A(x)+C_2\bigl(A(x)\log x+B(x)\bigr),\quad A(x)=e^{-x},\quad B\text{ analytic at }0.}
$$
This is the repeated-root <Logarithmic Frobenius solution>. On the negative real side one may use $\log|x|$, or use a fixed logarithm branch for a complex solution.
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