= Solution
After division by $z(z+2)$, the coefficients are
$$
p(z)=\frac{z+1}{z(z+2)},\qquad q(z)=-\frac{\alpha^2}{z(z+2)}.
$$
Their only finite singularities are \b[$z=0$ and $z=-2$]. At either point $z_0$, the products $(z-z_0)p$ and $(z-z_0)^2q$ are analytic, so both points are regular singular points. All other finite points are ordinary.
The lowest power after substituting the <Frobenius series> gives $\sigma(2\sigma-1)=0$. The coefficient of $z^{n+\sigma}$ then gives
$$
\boxed{\sigma=0\text{ or }\frac12,\qquad a_{n+1}=\frac{\alpha^2-(n+\sigma)^2}{(n+\sigma+1)(2n+2\sigma+1)}a_n.}
$$
For both exponents the denominator is nonzero for every $n\geq0$, so each arbitrary nonzero $a_0$ determines a series. Unless it terminates, its coefficient ratio tends to $-1/2$, proving convergence for $|z|<2$. For the half-integer exponent choose a square-root branch. Their leading powers $1$ and $z^{1/2}$ show that the two local solutions are <linearly independent>.
If $\alpha=m$ is a <positive integer>, choose $\sigma=0$. The recurrence numerator is nonzero for $0\leq n<m$ and vanishes at $n=m$. Thus $a_m\ne0$, $a_{m+1}=0$, and all later coefficients vanish: \b[a <polynomial> solution of exact degree $m$ exists]. It is proportional to the <Chebyshev polynomial> $T_m(z+1)$, since the substitution $t=z+1$ transforms the equation into $(t^2-1)y''+ty'-m^2y=0$.
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