[Math] Regular and irregular points

Question

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?

Answer 1

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?