Non Homogeneous Recurrence Relation

[Ethers0n]

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.

 

[epenguin]

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.)

 

[Ethers0n]

well, if I sub in b_m = (n+1)a_n into the original equation of
(n+2)a_n+1 = 2(n+1)a_n+2ⁿ
I get
b_m+1=2b_m+2^m
(1) b_m+1-2b_m=2^m
(2) b_m-2b_m-1=2^m-1
(3) 2b_m-4b_m-1=2^m
(1)-(3)
(4) b_m+1-4b_m+4b_m-1=0
(5) b_m-4b_m-1+4b_m-2=0
(6) r²-4r+4=0
(7) (r-2)(r-2)=0
(8) b_m= c₁2^m+c₂m2^m

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 b_m?
I'm getting the feeling that I'm dong something wrong...