Regular and irregular points In the following 2 examples, I am trying to find all the singular points of the given equations and determine whether each one is regular or irregular 
I can determine the singular points of the equations but I am having some trouble determining if they are regular or irregular. 
1)  $$x^2 (1-x^2)y'' + \frac {2}{x}y'+4y =0$$
The singular points of the equation are $x=\pm 1, 0$
The answer is that $0$ is irregular and $\pm 1$ regular but I am not sure why 
2) $$xy'' + (1-x)y' +xy =0$$
The singular point is $0$ but why is this considered regular? 
Do I have to consider the $p (x)$ term in each equation?
 A: The definition of wikipedia :
For the ordinary differential equation: $$f''(x)+p_{1}(x)f'(x)+p_{0}(x)f(x)=0.\,$$

*

*Point $a$ is an ordinary point when functions $p_1(x)$ and $p_0(x)$ are analytic at $x = a$.

*Point $a$ is a regular singular point if $p_1(x)$ has a pole up to order $1$ at $x = a$ and $p_0$ has a pole of order up to $2$ at $x = a$.

*Otherwise, point a is an irregular singular point.

For your example 1)
$$x^2 (1-x^2)y'' + \frac {2}{x}y'+4y =0$$
we divide by the coefficient function of $y''$ to obtain:
$$y'' + \frac {2}{x^3 (1-x^2)}y'+\dfrac{4}{x^2 (1-x^2)}y =0.$$
The singular points are $x=0$ and $x=\pm 1$.  From the definition, we see that $x=0$ is an irregular singular point because it is a third order pole. $x=\pm 1$ are order $1$ poles of the coefficient function of $y'$ and also order $1$ poles of the coefficient function of $y$. Hence, we know that these are regular singular points.
Can you do the next example?
A: Consider the first of your equations:

$$x^2 (1-x^2)y'' + \frac {2}{x}y'+4y =0$$

Reducing it to standard form, the coefficients of  $y',y$ are $p(x)=\dfrac{2}{x^3(1-x^2)},q(x)=\dfrac 4{x^2(1-x^3)}$ respectively.
The functions $p,q$ each have poles on $x=0,\pm 1$, so they are singularities, or singular points.

Definition: Now, we call a singular point $a$ to be regular if either $p$ or $q$ diverges as $x\to a$ but $\lim\limits_{x\to a}(x-a)p(x)$ and $\lim\limits_{x\to a}(x-a)^2q(x)$ remain finite. Otherwise, we call it irregular.

This definition is equivalent to saying that $p$ has a pole of order at most $1$ at $x=a$ and $q$ has a pole of order at most $2$ at $x=a$ for $a$ to be regular, otherwise it's irregular
Now, notice that at $x=\pm 1$, $p$ has a pole of order $1\leq 1$ and $q$ has a pole of order $1\leq 2$, so $x=\pm 1$ is regular.
But at $x=0$, $p$ has a pole of order $3\gt 1$, so $x=0$ is irregular.

Can you follow the above and get the answer for your next example?
Here's a head start for you:
The coefficients of $y',y$ after reduction to standard form are:

$$p(x)=\frac{1-x}x^2\quad\textrm{and}\quad q(x)=\frac 1x$$

Clearly, the only pole is at $x=0$ for both $p,q$. What is the order of the pole in $p,q$ ? Compute and compare with the definition.
