 Solution

Move the expansion point to the origin with $z=x+1$. The transformed ordinary differential equation isAfter division by $z$, the functions $z(1-1/z)=z-1$ and $z^2(c/z)=cz$ are analytic functions at zero, so this is a regular singular point. For a Frobenius solution with leading term $a_0{z}^{\rho }$, $a_0\ne 0$, the lowest power is $z^{\rho -1}$ and its coefficient isTherefore the indicial equation and its indicial exponents areThe exponent $\rho$ denotes the quantity called $\alpha$ in the paper, and avoids confusion with other uses of that letter.