In example 1 of Differential Equations 9th edition by Nagle page 436, we are asked to find the singular points for the second-order linear differential equation
$$xy''+\frac{x}{(x+1)}y'+(\sin x)y=0$$
Dividing the equation by $x$, we have the coefficient of the middle term is $p(x) = \frac{x}{x(x+1)}$. Clearly $p(x)$ is not defined at x=0, however, he states that $p(x)$ is indeed analytic at $x=0$ since we can cancel the $x$ in the numerator and denominator and this new function $ \frac{1}{x+1}$ does have a power series expansion around $x=0$. He references the following footnote:
†Such points are called removable singularities. In this chapter we assume in such cases that the function has been defined (or redefined) so that it is analytic at the point.
I'd like to have more explanation on why we can make this assumption. Does it have something to do with... changing one point will not change the solution to a differential equation? Why not?
I'm also not sure why he uses the phrase "In this chapter..." and "...in such cases.." in the footnote. Does the rule not apply in other cases? Maybe not for nonlinear equations? Why not? When is the assumption not valid?
It's possible the answer to these questions is found further along in the textbook, but I'm looking for some explanation now if possible as I'm having trouble wrapping my mind around what is meant by a function being analytic at a point. Thank you in advance for any help.
*If I can learn the answers to the questions above, I should be able to understand why the coefficient of the last term $q(x) = \frac{\sin(x)}{x}$ is analytic at $x=0$ as well. It's the same idea I believe.