= Solution
{c}
For the first <Frobenius solution>, take the larger <indicial root> $r=a$ and normalize the first coefficient to one:
$$
y_+(x)=x^a\sum_{n=0}^\infty c_nx^n,\qquad c_0=1,
\qquad c_j=0\quad(j<0).
$$
Substitution and comparison of the coefficient of $x^{n+a}$ give
$$
\big[(n+a)^2-a^2\big]c_n+bc_{n-\beta}=0,
\qquad\boxed{c_n=-\frac{b\,c_{n-\beta}}{n(n+2a)}\quad(n\geq1).}
$$
Every denominator is nonzero because $a\geq0$.
For $\beta=1$, the <recurrence relation> is $c_n=-bc_{n-1}/[n(n+2a)]$, so
$$
\boxed{y_+(x)=x^a\left[1-\frac{b}{1+2a}x+
\frac{b^2}{2(1+2a)(2+2a)}x^2+\cdots\right],\quad
c_n=\frac{(-b)^n}{\prod_{j=1}^n j(j+2a)}.}
$$
For $\beta=2$, $c_1=0$ and all odd coefficients vanish; the even coefficients satisfy $c_{2k}=-bc_{2k-2}/[2k(2k+2a)]$. Hence
$$
\boxed{y_+(x)=x^a\left[1-\frac{b}{4(1+a)}x^2+
\frac{b^2}{32(1+a)(2+a)}x^4+\cdots\right].}
$$
For any positive integer $\beta$, induction in the <recurrence relation> shows that only indices divisible by $\beta$ survive:
$$
\boxed{y_+(x)=x^a\sum_{k=0}^\infty
\frac{(-b)^k x^{k\beta}}{\prod_{j=1}^k j\beta(j\beta+2a)}.}
$$
The empty product for $k=0$ equals one. Successive terms in the <power series> factor have ratio of magnitudes
$$
\frac{|b|\,|x|^\beta}{(k+1)\beta((k+1)\beta+2a)}\longrightarrow0.
$$
Thus the <ratio test> gives convergence for every finite $x$; a nonintegral $a$ still requires a chosen side of zero or a complex branch for the factor $x^a$.
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