= Solution
{c}
For a putative second <Frobenius solution> with the smaller <indicial root> $-a$, set $y_-=x^{-a}\sum_{n\geq0}d_nx^n$, with $d_0=1$ and $d_j=0$ for $j<0$. Its <recurrence relation> is
$$
\boxed{n(n-2a)d_n=-b\,d_{n-\beta}.}
$$
The <Frobenius case classification> concerns the gap $2a$ between the <indicial roots>. In the course convention, Case IIIa is the obstructed recurrence requiring a <Logarithmic Frobenius solution>; Case IIIb permits two pure <Frobenius series>.
If $a>0$ and $2a\notin\mathbb Z$, this is \b[Case I]: no denominator vanishes. There are two <linearly independent> pure <Frobenius solutions>, with
$$
y_-(x)=x^{-a}\sum_{k=0}^\infty
\frac{(-b)^kx^{k\beta}}{\prod_{j=1}^k j\beta(j\beta-2a)}.
$$
Their different leading powers establish <linear independence>.
If $a=0$, this is \b[Case II], with a repeated <indicial root>. There is only one pure <Frobenius series> with a nonzero leading coefficient. To see the second solution's form, <reduction of order> gives
$$
y_2(x)=y_+(x)\int^x\frac{dt}{t\,y_+(t)^2}.
$$
Here $y_+(t)=1+O(t^\beta)$, so the integrand is $t^{-1}+O(t^{\beta-1})$. Consequently
$$
\boxed{y_2(x)=y_+(x)\log x+\text{an analytic power series}.}
$$
The <logarithm> has a nonzero coefficient and cannot be removed by adding a multiple of $y_+$.
If $N=2a$ is a positive integer, the only potentially obstructed coefficient is $n=N$. Below that index, $d_n=0$ unless $\beta$ divides $n$. At the exceptional index the equation reads
$$
0=-b\,d_{N-\beta}.
$$
If $N=m\beta$ for some positive integer $m$, all coefficients $d_0,d_\beta,\ldots,d_{(m-1)\beta}$ are nonzero because $b\ne0$. The exceptional equation is inconsistent. This is \b[Case IIIa], and the independent second solution has the form
$$
\boxed{y_2=x^{-a}\sum_{n=0}^\infty d_nx^n+K y_+\log x,\qquad K\ne0.}
$$
The <Frobenius resonance for an Euler equation with monomial potential> explains why divisibility, rather than the integer gap alone, decides this outcome. For example, with $d_0=1$, applying the original <differential operator> to $y_+\log x$ gives $2xy_+'$. Cancellation of the coefficient of $x^a$ requires $K=-bd_{N-\beta}/(2a)\ne0$.
If $\beta$ does not divide $N$, then $d_{N-\beta}=0$; this includes $N<\beta$ through the negative-index convention. The exceptional equation is $0=0$, so $d_N$ is free. This is \b[Case IIIb]: set $d_N=0$ to select an independent pure <Frobenius series> starting with $x^{-a}$. Subsequent coefficients again follow the <recurrence relation>. Changing $d_N$ simply adds a multiple of $y_+$, the <undetermined coefficient at Frobenius resonance>.
The complete answer is therefore
$$
\boxed{\begin{array}{c|c}
\text{parameters}&\text{case}\\\hline
a=0&\mathrm{II}\\
a>0,\quad 2a\notin\mathbb Z&\mathrm{I}\\
2a\in\mathbb N_{>0},\quad 2a/\beta\in\mathbb N_{>0}&\mathrm{IIIa}\\
2a\in\mathbb N_{>0},\quad 2a/\beta\notin\mathbb N_{>0}&\mathrm{IIIb}.
\end{array}}
$$
In particular, for $\beta=1$ every positive integer gap gives Case IIIa. For $\beta=2$, positive integer $a$ gives Case IIIa, while $a=1/2,3/2,5/2,\ldots$ gives Case IIIb; the remaining positive $a$ give Case I. The condition $b\ne0$ matters: at $b=0$ the two <Cauchy-Euler equation> powers already give independent solutions whenever $a>0$.
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