Verifying that Kummer hypergeometric function is a solution to $xy''+(b-x)y'-ay=0$ The following second order differential equation (see YouTube link)
$$xy''+(b-x)y'-ay=0$$
has two solutions, one of them resmenble Kummer function of the first kind:
$$y=M(a,b,x)=\sum_{n=0}^\infty \frac{a^{(n)}x^n}{b^{(n)}n!},$$
where $a^{(n)}$ are rising factorial of $a$.
While verifying that that function is the solution to the differential equation, I got this:
$$n(n-1)+nb-nx-ax=0$$
I am wondering if there is something wrong. Actually, I thought that all terms will cancel each other.
Thanks.
 A: $\def\d{\mathrm{d}}$If $y(x) = \sum\limits_{n = 0}^∞ \dfrac{a^{(n)} x^n}{b^{(n)} n!} = 1 + \sum\limits_{n = 1}^∞ \dfrac{a^{(n)} x^n}{b^{(n)} n!}$, then$$
ay(x) = a + \sum_{n = 1}^∞ \frac{a · a^{(n)} x^n}{b^{(n)} n!},
$$\begin{gather*}
(b - x) y'(x) = (b - x) \sum_{n = 1}^∞ \frac{a^{(n)} nx^{n - 1}}{b^{(n)} n!} = \sum_{n = 1}^∞ b · \frac{a^{(n)} nx^{n - 1}}{b^{(n)} n!} - \sum_{n = 1}^∞ x · \frac{a^{(n)} nx^{n - 1}}{b^{(n)} n!}\\
= \sum_{n = 1}^∞ \frac{a^{(n)} x^{n - 1}}{(b + 1)^{(n - 1)} (n - 1)!} - \sum_{n = 1}^∞ \frac{a^{(n)} x^n}{b^{(n)} (n - 1)!} = a + \sum_{n = 1}^∞ \frac{a^{(n + 1)} x^n}{(b + 1)^{(n)} n!} - \sum_{n = 1}^∞ \frac{a^{(n)} x^n}{b^{(n)} (n - 1)!},
\end{gather*}\begin{gather*}
xy''(x) = \sum_{n = 2}^∞ x · \frac{a^{(n)} n(n - 1)x^{n - 2}}{b^{(n)} n!} = \sum_{n = 2}^∞ \frac{a^{(n)} x^{n - 1}}{b^{(n)} (n - 2)!} = \sum_{n = 1}^∞ \frac{a^{(n + 1)} x^n}{b^{(n + 1)} (n - 1)!},
\end{gather*}
and\begin{align*}
&\mathrel{\phantom{=}}{} xy''(x) + (b - x)y'(x) - ay(x)\\
&= {\small \sum_{n = 1}^∞ \frac{a^{(n + 1)} x^n}{b^{(n + 1)} (n - 1)!} + \left( a + \sum_{n = 1}^∞ \frac{a^{(n + 1)} x^n}{(b + 1)^{(n)} n!} - \sum_{n = 1}^∞ \frac{a^{(n)} x^n}{b^{(n)} (n - 1)!} \right) - \left( a + \sum_{n = 1}^∞ \frac{a · a^{(n)} x^n}{b^{(n)} n!} \right)}\\
&= \sum_{n = 1}^∞ \left( \frac{a^{(n + 1)}}{b^{(n + 1)} (n - 1)!} + \frac{a^{(n + 1)}}{(b + 1)^{(n)} n!} - \frac{a^{(n)}}{b^{(n)} (n - 1)!} - \frac{a · a^{(n)}}{b^{(n)} n!} \right) x^n\\
&= \sum_{n = 1}^∞ \bigl( (a + n)n + (a + n)b - (b + n)n - a(b + n) \bigr) \frac{a^{(n)}}{b^{(n + 1)} n!} x^n = 0.
\end{align*}
