Find a value of alpha such that all solutions of the differential equation exist I was recently challenged to solve a question that is as follows:
Determine the two values of the constant $\alpha$ for which all solutions of
$xy'' + (x-1)y' - \alpha y = 0$
can be written as
$y(x) = x^s\sum_{n=0}^{\infty}[A_kx^k]$
To solve this problem, I tried using the Frobenius method given that $x_0 = 0$ is a regular singular point and found that the indicial equation was
$s(s-2) = 0$
And obtained that 
$A_n = \frac{(\alpha + 1 - n - s)A_{n-1}}{(n+s)(n+s-2)}$
From here on out, I do not know how to find the two values of $\alpha$ that satisfy the problem conditions. Any help would be appreciated!
 A: The recurrence relation (RR) for $s_2 = 2$ is $n (n + 2) A_n = (\alpha - n - 1) A_{n - 1}$, which gives one power series solution $y_2$.
The RR for $s_1 = 0$ is $n (n - 2) A_n = (\alpha - n + 1) A_{n - 1}$,
which for $n = 1$ and $n = 2$ becomes
$$-A_1 = \alpha A_0, \\
0 = (\alpha - 1) A_1.$$
If the last equation doesn't hold, we don't get a power series solution. Therefore we need either $\alpha = 1$ or $A_1 = 0$. The latter implies $\alpha = 0$.
In both cases, $A_2$ is arbitrary. If we take $A_2 = 0$, we get $y_1 = A_0$ for $\alpha = 0$ and $y_1 = A_0 - A_0 x$ for $\alpha = 1$. $y_1$ is linearly independent of $y_2$ because $y_2 = O(x^2)$.
If we take $A_0 = A_1 = 0$ and $A_2 \neq 0$, we get a solution linearly dependent on $y_2$ from the RR for $s_1$. Sometimes it's inaccurately stated that one of the solutions is not a Frobenius series because the RR for $s_1$ is not solvable.  Rather, the logarithmic case arises when the RR's for $s_1$ and $s_2$ give the same solution (because we have to set the first coefficients to zero to satisfy the RR for $s_1$).
