Finding Regular and Irregular Singular Points of the Differential Equation

zan81289

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P(x)y''+Q(x)y'+R(x)y=0? I have to find all singular points of three differential equations and classify them as regular or irregular. The first one worked out perfectly but I can't seem to get the correct answer for the other two. Can you help me with this?

2) x^2(x-2)y''+(5x-7)y'+2(3+5x^2)y=0
answer from book: 0,regular 2,regular

I found the singular points to be 0 and 2 by setting x^2(x-2)=0

However when I try to classify the 0 as regular or irregular I find that (x-x_0)*Q(x)/P(x) is undefined at 0 therefore making it an irregular singular point.

(x-0)*(5x-7)/(x^2(x-2))=(5x-7)/(x^2-2x)
(5*0-7)/(0^2-2*0)=-7/0=undefined

2 worked out fine and i got it as a regular singular point but no matter what I try I can't seem to get 0 as a regular singular point.

3) [(x-2)^(-1)y']'+x^(-5/2)y=0
answer from the book: 0,irregular 2,regular

I believe that this differential equation becomes: (x-2)^(-1)y''-(x-2)^(-2)y'+x^(-5/2)=0

I can then find the singular points to be 0 and 2 from where Q(x)/P(x) and R(x)/P(x) are not analytic.

However I get 2 to be irregular and 0 to be regular, when I do (x-x_0)*Q(x)/P(x) and (x-x_0)^2*R(x)/P(x)

Can anyone help me to get the same answer as the book.
 
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