How Do I Transform Coefficients in a Taylor Series Differential Equation?

[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

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.

 

[HallsofIvy]

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

 

[vermin]

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

 

[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.

 

[blue_raver22]

As a scientist, my first suggestion would be to double check your work and make sure all calculations and algebraic manipulations are correct. This is a complex problem and it's easy to make a mistake along the way.

Additionally, it may be helpful to seek guidance from a professor or someone with more expertise in this specific type of problem. They may be able to offer insights or point you in the right direction for solving it.

In terms of the x-terms as coefficients outside the summation symbols, it may be helpful to consider using a change of variables to rewrite them in terms of (x-4). This can help simplify the problem and make it easier to factor out later.

Lastly, don't be discouraged if it takes some time to solve this problem. It's important to approach it with patience and determination, and to seek help when needed. Good luck!