= Solution
Insert the <Frobenius solution> and its differentiated series into the equation. Collecting the terms before shifting an index gives
$$
\sum_{n=0}^{\infty}(n+\sigma)(n+\sigma+\gamma-1)a_nx^{n+\sigma-1}
-\sum_{n=0}^{\infty}(n+\sigma+\alpha)(n+\sigma+\beta)a_nx^{n+\sigma}=0.
$$
The coefficient of the lowest power $x^{\sigma-1}$ is $a_0\sigma(\sigma+\gamma-1)$. Because $a_0\ne0$, the <indicial equation> gives
$$
\boxed{\sigma=0\quad\text{or}\quad\sigma=1-\gamma.}
$$
For the coefficient of $x^{n+\sigma-1}$ with $n\geq1$, shift the index in the second sum. The <hypergeometric Frobenius recurrence> is
$$
\boxed{(n+\sigma)(n+\sigma+\gamma-1)a_n
=(n+\sigma+\alpha-1)(n+\sigma+\beta-1)a_{n-1}.}
$$
Whenever the left coefficient is nonzero, division yields
$$
\boxed{a_n=\frac{(n+\sigma+\alpha-1)(n+\sigma+\beta-1)}{(n+\sigma)(n+\sigma+\gamma-1)}a_{n-1}.}
$$
For noninteger $\gamma$, the two independent local solutions can therefore be identified as
$$
y_1={}_2F_1(\alpha,\beta;\gamma;x),\qquad
y_2=x^{1-\gamma}{}_2F_1(\alpha-\gamma+1,\beta-\gamma+1;2-\gamma;x),
$$
with a consistent local branch for the fractional power. The <hypergeometric function> coefficients follow by iterating the recurrence using <rising factorials>.
There is a parameter qualification to the PDF's statement for all real constants: a zero denominator cannot be divided through. At an <undetermined coefficient at Frobenius resonance>, use the undivided coefficient equation to test compatibility. For example, $\gamma=0,\sigma=0,n=1$ requires $0=\alpha\beta a_0$, so a pure exponent-zero series with $a_0\ne0$ is impossible if $\alpha\beta\ne0$. When $\gamma=1$ the two <indicial roots> coincide, and a second independent solution generally needs a logarithmic term. The <indicial roots> and undivided relation remain the valid coefficient conclusions in these exceptional cases.
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