 Solution

After division by $z(z+2)$, the coefficients areTheir only finite singularities are $z=0$ and $z=-2$. At either point $z_0$, the products $(z-z_0)p$ and $(z-z_0{)}^{2}q$ 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 givesFor both exponents the denominator is nonzero for every $n\ge 0$, 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 $0\le n<m$ and vanishes at $n=m$. Thus $a_m\ne 0$, $a_{m+1}=0$, and all later coefficients vanish: 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 $(t^2-1)y^{\prime \prime }+t{y}^{\prime }-m^{2}y=0$.