Singular points help

Question

For the differential equation:

2(1-x)y''-3y'+(y/x)=0

What are the singular points and determine whether they are regular or irregular? Then find the roots of the indicial equation

Thank you to whoever can help me :)

Answer 1

First, divide through by 2(1-x) to get:

 

y" -     3     y' +      1        y = 0     (**)

       2(1-x)        2x(1-x)

 

Setting the denominators equal to 0, we get x = 0 and 1 are singular points.

 

Let P(x) = the coefficient of y' and Q(x) = coefficient of y of equation (**).  

 

Since (x-0)P(x) and (x-0)²Q(x) are both analytic at x = 0, 0 is a regular singular point.

 

Since (x-1)P(x) and (x-1)²Q(x) are both analytic at x = 1, 1 is also a regular  singular point.

 

The indicial equation for x = 0 is r(r-1) + p₀r + q₀ = 0, where

         p₀ = limit as x → 0 [(x-0)P(x)] = 0

         q₀ = limit as x→ 0 [(x-0)²Q(x)] = 0

         Indicial equation for x = 0 is:  r² - r = 0

 

The indicial equation for x = 1 is r(r-1) +p₀r+ q₀ = 0, where

        p₀ = limit as x→1 [(x-1)P(x)] = 1.5

        q₀ = limit as x→1[(x-1)²Q(x)] = 0

        Indicial equation for x = 1 is:  r(r-1) +1.5r = 0

                                                            r² + 0.5r = 0