Solve Recurrence Relation for DE x(x-2)y''+(1-x)y'+xy=0@x=2

[esuess]

Im trying to determine the recurrence relation of the following differential equation : x(x-2)y'' + (1-x)y' + xy =0 about the regular singular point x =2.

I've tried rewriting the DE as (x-2+2)(x-2)y'' -(x-2+1)y' +(x-2+2)y =0, but it doesn't seem to work. any ideas?

 

[HallsofIvy]

Why doesn't it work? You have (x-2)²y"- 2(x-2)y"- (x-2)y'- y'+ (x-2)y+ 2y= 0
Let
$$y= \Sigma_{n=0}^\infty a_n (x-2)^{n+c}$$
(You need the c precisely because this is a "regular singular point")
$$y'= \Sigma_{n=0}^\infty (n+c)a_n(x-2)^{n+c-1}$$
$$y"= \Sigma_{n=0}^\infty (n+c)(n+c-1)a_n(x-2)^{n+c-2)$$
The equation becomes
$$\Sigma( n+c)(n+c-1)a_n(x-2)^{n+c}+\Sigma 2(n+c)(n+c-1)a_n(x-2)^{n+c-2}- \Sigma (n+c)a_n(x-2)^{n+c}- \Sigma (n+c)a_n(x-2)^{n+c-1}+ \Sigma a_n(x-2)^{n+c}+ \Sigma 2a_n(x-2)^{n+c}= 0$$
Now we need to determine c. Get the indicial equation by looking at the lowest possible exponent in each sum. Those will be when n= 0. The lowest power of x is, then, (x-2)^c-2 and has coefficient c(c-1)a₀. We want to make sure that a_0 is not 0 so the indicial equation is c(c-1)= 0. either c= 0 or c= 1.
Put those in for c and find the recurrance relations.