# Stuck on Taylor series DE

1. Oct 30, 2013

### vermin

So, I have this DE which is 2nd order, w/ variable coefficients, it goes;
xy''+(x-5)y'+(x^2-4)y=0 revolving around x_0=4.

I know there's a singular point at 0 and I assume to use a summation y(x)=[∞,Ʃ,n=0] a_n(x-x_0)^n
pardon me I don't know how to type the summation symbol, but that's supposed to be starting at n=0 going to infinity, I'm using x_0 for 'x sub 0', etc.

Well I take the 1st and 2nd derivatives of that sum and plug them into the ODE, having divided the y' and y coefficients by x so that y'' is alone. Then I'm left with the following;

http://files.royw.airpost.net/summm.jpg [Broken]

my problem is, I was told on other problems that any x-terms as coefficients outside (such as the two in blue, to the left of the summation symbols) should be re-written to include the form of the x-term inside the sums. So, I need them to involve (x-4), so I can factor them out later. But maybe my algebra is weak, because I don't know what to do here. I've already stumped two math majors with this situation so I'm thinking maybe I took a wrong turn to lead me into this.

Any help would be appreciated! thanks.

Last edited by a moderator: May 6, 2017
2. Oct 31, 2013

### HallsofIvy

Staff Emeritus
The first thing you need to do is get rid of that "x" in the denominator by multiplying through by x:
$$\sum_{n=0}^\infty n(n-1)x(x- 4)^{n-2}+ (x- 5)\sum_{n=0}^\infty a_nnx(x- 4)^{n-1}+ (x^2- 4)\sum_{n=0}^\infty a_nn(x- 4)^n= 0$$
Now you will have to convert the "x" in the first sum, the "x- 5" in the second, and both "x- 2" and "x+ 2" (factors of $x^2- 4$) to "x- 4" so it can be included in the powers.

x= x- 4+ 4 so "$x(x-4)^{n-2}= (x- 4+ 4)(x- 4)^{n-2}= (x- 4)^{n-1}+ 4(x- 4)^{n-2}$".

x- 5= x- 4- 1 so $(x- 5)(x- 4)^{n-1}= (x-4- 1)(x- 4)^{n-1}= (x-4)^n- (x- 4)^{n-1}$.

$x^2- 4= (x- 2)(x+ 2)= (x- 4+ 2)(x- 4+ 6)= (x- 4)^2+ 8(x- 4)+ 12$ so $(x^2- 4)(x- 4)^n= (x- 4)^{n+2}+ 8(x- 4)^{n+1}+ 12(x- 4)^n$.

3. Oct 31, 2013

### vermin

ok, thanks! I didn't realize I could move the x inside first.

4. Oct 31, 2013

### vermin

wait, on the 3rd term you didn't include an X in the summation argument. I'm assuming that was a typo...?
also, it looks like I included an extra 'n' that shouldn't have been there on the 3rd summation between a_n and (x-4), sorry.

Last edited: Oct 31, 2013