1. ## Regular Singular Points

For those that are interested, the problem has been taken from the Elementary Differential Equations and Boundary Value Problems, 8th Edition, by Boyce. It is problem 1 from Section 5.4.

The problem states to find all singular points of the given equation and determine whether each one is regular or irregular.

$
x y'' + (1 - x) y' + x y = 0
$

$
P(x) = x = 0
$

The point x = 0 is a singular point.

So all other points are ordinary points.

$
\lim_{x \to 0} (x - 0) \frac {(1 - x)} {(x)} = 1
$

$
\lim_{x \to 0} (x - 0)^2 \frac {(x)} {(x)} = 0
$

So as my conclusion the two limits are finite numbers and we can say that the point x = 0 is a regular singular point

2. Originally Posted by fudawala
...
$
P(x) = x = 0
$

...
yes, correct

the above statement is awkward though

you should have said something like:

$xy'' + (1 - x)y' + xy = 0$

$\Rightarrow y'' + \frac {1 - x}{x} y' + y = 0$

we have that the function $\frac {1 - x}{x}$ is discontinuous at the point $x = 0$

thus $x = 0$ is a singular point

...and continue from there