Regular Singular Points

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• Jul 8th 2007, 10:39 AM
fudawala
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.

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

$\displaystyle P(x) = x = 0$

The point x = 0 is a singular point.

So all other points are ordinary points.

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

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

So as my conclusion the two limits are finite numbers and we can say that the point x = 0 is a regular singular point
• Jul 8th 2007, 10:57 AM
Jhevon
Quote:

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

yes, correct

the above statement is awkward though

you should have said something like:

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

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

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

thus $\displaystyle x = 0$ is a singular point

...and continue from there