Power Series Solution for Differential Equation Near x0=1

[pinkbabe02]

differential equation help!

Homework Statement

(1+2x-x^2)y''-6xy'-6y=0 find power series solution of the equation near x0=1
(a)show the recurrence relation for an,
(b)derive a formula for an in terms of a0 and a1, and
(c)show the solution in the form y=a0y1(x)+a1y2(x)

Homework Equations

The Attempt at a Solution

I`ll attach the word file I typed it up in if that is easier to read...
(1+2x-x^2 ) y^''-6xy^'-6y=0, x_0=1
Let…
y=∑_(n=0)^∞▒〖a_n x^n 〗
y^'=∑_(n=1)^∞▒〖〖na〗_n x^(n-1) 〗
y^''=∑_(n=2)^∞▒〖n(n-1)a_n x^(n-2) 〗
(1+2x-x^2 ) ∑_(n=2)^∞▒〖n(n-1)a_n x^(n-2) 〗-6x∑_(n=1)^∞▒〖〖na〗_n x^(n-1) 〗-6∑_(n=0)^∞▒〖a_n x^n 〗=0
∑_(n=2)^∞▒〖n(n-1)a_n x^(n-2) 〗+∑_(n=2)^∞▒〖2n(n-1) a_n x^(n-1)-∑_(n=2)^∞▒〖n(n-1)a_n x^n 〗〗-∑_(n=1)^∞▒〖〖6na〗_n x^n 〗-∑_(n=0)^∞▒〖〖6a〗_n x^n 〗=0
∑_(n=0)^∞▒〖(n+2)(n+1)a_(n+2) x^n 〗+∑_(n=1)^∞▒〖2(n+1)(n) a_(n+1) x^n-∑_(n=2)^∞▒〖n(n-1)a_n x^n 〗〗-∑_(n=1)^∞▒〖〖6na〗_n x^n 〗-∑_(n=0)^∞▒〖〖6a〗_n x^n 〗=0
∑_(n=0)^∞▒〖[(n+2)(n+1) a_(n+2)-〖6a〗_n ] x^n 〗+∑_(n=1)^∞▒〖[2(n+1)(n) a_(n+1)-〖6na〗_n ] x^n-∑_(n=2)^∞▒〖n(n-1)a_n x^n 〗〗=0
I can turn the second term into n=0 since when n=0, the value equals 0 anyhow…the same goes for the third term (when n=0, 1 the value is 0), so we have…
∑_(n=0)^∞▒[(n+2)(n+1) a_(n+2)-〖6a〗_n ┤ +2(n+1)(n) a_(n+1)-〖6na〗_n-n(n-1)a_n 〖]x〗^n=0
∑_(n=0)^∞▒〖[(n+2)(n+1) a_(n+2)+(2n(n+1)) a_(n+1)+(-6-6n-n(n-1))a_n ] x^n=0〗
This is where I am stuck. I don’t know how to find the recurrence relation since I have a_n,a_(n+1),and a_(n+2). If someone could give me a hint that would be extremely helpful.
[STRIKE][STRIKE][/STRIKE][/STRIKE]

 

[vela]

Are you sure you copied the original differential equation down correctly?

 

[pinkbabe02]

Yes I just double checked it again and that`s what the problem states.

 

[vela]

Your work looks good so far. If you simplify the last term in the coefficient, you get
$$(n+2)(n+1) a_{n+2}+2n(n+1) a_{n+1}-(n+2)(n+3)a_n = 0$$
This is a perfectly fine recurrence relation. Are you familiar with any techniques on how to solve one like this?

You might try writing out the first few terms and see if you can spot a pattern again.

 

[pinkbabe02]

That`s what I had originally also but when I redid it, I saw that in this case x0=1 so the summation has (x-1) instead of x. then for my final recurrence relation, i have...(i attached the file)