Singular Points of the ODE: Identifying and Understanding

[cabellos6]

Homework Statement

For the ODE xy" + (2-x)y' + y = 0

i want to show it has one singular point and identify its nature

Homework Equations

The Attempt at a Solution

I have read the topic and I see that a point Xo is called and ordinary point of the equation if both p(x) and q(x) (once converted to SF) are anlytic at Xo.

I really don't understand the method to work this out though...

Help would be much appreciated

 

[HallsofIvy]

What "method" are you talking about? Just finding p(x) and q(x) and determining whether they are analytic?

Unfortunately, you haven't said what you mean by "SF" nor what p(x) and q(x) are here.

I might guess that by "SF" (standard form?) You mean the equation solved for y". Here that would be y"= ((2-x)/x) y'+ (1/x)y and perhaps you mean p(x)= (2-x)/x and q(x)= 1/x.

 

[cabellos6]

yes i am aware that when we divide by x this ODE is then in standard form. What i don't understand is how to show this equation has only one singular point?

 

[HallsofIvy]

What IS a singular point?

You've already said that an "ordinary point" is a point where the coefficents of y' and y are not analytic. A "singular point" is a point where that is not true. For what values of x and y are (2-x)/x and 1/x not analytic?