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For $a>0$, $b\ne 0$, and a positive integer $\beta$, consider $x^2{y}^{\prime \prime }+x{y}^{\prime }+(-a^2+b{x}^{\beta })y=0$ near $x=0$. Its indicial roots are $a,-a$. The first Frobenius solution, normalized to leading coefficient one, is $y_{+}=x^a{\sum }_{k\ge 0}{c}_{k\beta }{x}^{k\beta }$, with $c_{k\beta }=-b{c}_{(k-1)\beta }/[k\beta (k\beta +2a)]$. If $N=2a$ is a positive integer, a second pure Frobenius solution with leading coefficient one would requireBefore $n=N$, induction makes $d_n=0$ for indices not divisible by $\beta$. If $\beta$ divides $N$, the coefficient $d_{N-\beta }$ is nonzero, and the exceptional equation $0=-b{d}_{N-\beta }$ is impossible. If $\beta$ does not divide $N$, the right-hand side is zero and $d_N$ is free. Thus, in the Frobenius case classification, a positive integer gap gives Case IIIa exactly when $\beta$ divides $2a$, and Case IIIb otherwise.

The coefficient of the necessary logarithm can also be calculated. Let $L=x^2{d}^2/d{x}^2+xd/dx-a^2+b{x}^{\beta }$. The product rule gives $L(y_{+}\log x)=2x{y}_{+}^{\prime }$. Therefore a second solution $x^{-a}\sum d_n{x}^n+K{y}_{+}\log x$, with $d_0=1$, hasFor $N=m\beta$ the preceding recurrence relation gives $d_{(m-1)\beta }=b^{m-1}/[{\beta }^{2m-2}((m-1)!{)}^2]$, soThis proves that the logarithm cannot disappear for a special nonzero value of $b$. For $a=0$, reduction of order instead produces the repeated-root Logarithmic Frobenius solution directly.

For a putative second Frobenius solution with the smaller indicial root $-a$, set $y_{-}=x^{-a}{\sum }_{n\ge 0}{d}_n{x}^n$, with $d_0=1$ and $d_j=0$ for $j<0$. Its recurrence relation isThe 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 Case I: no denominator vanishes. There are two linearly independent pure Frobenius solutions, withTheir different leading powers establish linear independence.

If $a=0$, this is 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 givesHere $y_{+}(t)=1+O(t^{\beta })$, so the integrand is $t^{-1}+O(t^{\beta -1})$. ConsequentlyThe 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 readsIf $N=m\beta$ for some positive integer $m$, all coefficients $d_0,d_{\beta },\dots ,d_{(m-1)\beta }$ are nonzero because $b\ne 0$. The exceptional equation is inconsistent. This is Case IIIa, and the independent second solution has the formThe 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 $2x{y}_{+}^{\prime }$. Cancellation of the coefficient of $x^a$ requires $K=-b{d}_{N-\beta }/(2a)\ne 0$.

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 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 thereforeIn particular, for $\beta =1$ every positive integer gap gives Case IIIa. For $\beta =2$, positive integer $a$ gives Case IIIa, while $a=1/2,3/2,5/2,\dots$ gives Case IIIb; the remaining positive $a$ give Case I. The condition $b\ne 0$ matters: at $b=0$ the two Cauchy-Euler equation powers already give independent solutions whenever $a>0$.