Undetermined Coefficients of Difference Equations

[kathrynag]

I am working on a section on undetermined coefficients of a difference. I'm really trying to understand the material, but am struggling.
$$y_{n+1}$$-2$$y_{n}$$=2
I got as far as saying for the homogenous equation y=$$c_{1}$$$$(-1)^{n}$$.
I'm assuming the guess next would be A, but I'm confused.
Maybe if someone could help me with some examples on how to do this using the 2 or something like 2^n or cos(n). I'm trying to learn something something new on my own and it's hard.

 

[tiny-tim]

Hi kathrynag!

Do you mean y_n+1 - 2y_n = 2 ?

If so, the homogenous equation for this recurrence relation is y_n+1 - 2y_n = 0, and C(-1)ⁿ is not a solution, is it?

For the particular solution, try a polynomial of lowest possible degree (start with zero!)

 

[kathrynag]

Hi kathrynag!

Do you mean y_n+1 - 2y_n = 2 ?

If so, the homogenous equation for this recurrence relation is y_n+1 - 2y_n = 0, and C(-1)ⁿ is not a solution, is it?

For the particular solution, try a polynomial of lowest possible degree (start with zero!)

Yeah that's what i meant. I tried letting the polynomial =A.
Then A(n+1)+A=2
An+2A=2
This is where I get stuck.
Now if I had a number like n^2 on the end polynomial would be An^2+Bn+C
An^2(n+1)+Cn(n+1)+Cn+C+An^2+Bn+C=n^2
An^3+n^2(2A+C)+n(2C+B)+(2C)=n^2
I guess I get stuck at the end each time. Say I had cosn. Then polynomial would be Acosn+Bsinn.
Acos(n+1)+Bsin(n+1)+Acosn+Bsin(n)=cosn

Maybe I'm doing something wrong somewhere.

 

[tiny-tim]

Hi kathrynag!

Yeah that's what i meant. I tried letting the polynomial =A.
Then A(n+1)+A=2
An+2A=2 …

Now you're mystifying me …

where did A(n+1) come from? ​

If the polynomial is A, then y_n+1 - 2y_n is -A …

carry on from there.