# Non Homogeneous Recurrence Relation

1. solve the following recurrence relation for an

2. (n+2)an+1= 2(n+1)an+2$$^{n}$$, n>=0, a0=1
I shifted the index, multiplied through by the 2$$^{n}$$ term and then subtracted the resulting equation from the original equation to get rid of the 2$$^{n}$$ term...

3. I have gotten to this point
(n+1)an-4(n)an-14(n-1)an-2=0

I'm not really sure how to handle the (n+1), n, or (n-1) terms when looking for the particular/ homogeneous solution parts.

## The Attempt at a Solution

Related Calculus and Beyond Homework Help News on Phys.org
epenguin
Homework Helper
Gold Member
can you see a way to change your variable a(n) that would do it? There is something consistemt between the successive terms.

(You have missed a + out of your formula BTW.)

well, if I sub in bm = (n+1)an into the original equation of
(n+2)an+1 = 2(n+1)an+2n
I get
bm+1=2bm+2m
(1) bm+1-2bm=2m
(2) bm-2bm-1=2m-1
(3) 2bm-4bm-1=2m
(1)-(3)
(4) bm+1-4bm+4bm-1=0
(5) bm-4bm-1+4bm-2=0
(6) r2-4r+4=0
(7) (r-2)(r-2)=0
(8) bm= c12m+c2m2m

but I don't really know where to go from there?
do I sub back in, or is there a way to use a0=1 with bm?
I'm getting the feeling that I'm dong something wrong....