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.

 

[tacman]

First, let's rewrite the DE as (x-2)^2y'' + (1-x)(x-2)y' + x(x-2)y = 0. This will make it easier to find the recurrence relation.

Next, let's substitute x = 2 into the DE to get (2-2)^2y'' + (1-2)(2-2)y' + 2(2-2)y = 0. This simplifies to 0y'' + 0y' + 0y = 0, which is not very helpful.

To find the recurrence relation, we will use the method of Frobenius. We will assume that the solution has the form y = x^r, where r is a constant to be determined.

Substituting this into the DE, we get (x-2)^2r(r-1)x^(r-2) + (1-x)(x-2)rx^(r-1) + x(x-2)x^r = 0.

Simplifying and collecting terms, we get x^r[(x-2)^2r(r-1) + (1-x)(x-2)r + x(x-2)] = 0.

Since x = 2 is a regular singular point, we know that the coefficient of x^(r-2) must be 0. This means that (x-2)^2r(r-1) = 0.

Solving for r, we get r = 0 or r = 1. These are the two roots of the indicial equation.

Now, we can write our solution as y = c1x^0 + c2x^1 = c1 + c2x, where c1 and c2 are arbitrary constants.

To find the recurrence relation, we need to find a relation between the coefficients c1 and c2. To do this, we will use the method of reduction of order.

Let's assume that the second solution has the form y2 = u(x)y1. Substituting this into the DE, we get (x-2)^2u''y1 + (1-x)(x-2)u'y1 + x(x-2)uy1 = 0.

Since y1 = c1 + c2x, we can write this as (x-2)^2u''(c1 + c2x) + (1