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.
[tex]y_{n+1}[/tex]-2[tex]y_{n}[/tex]=2
I got as far as saying for the homogenous equation y=[tex]c_{1}[/tex][tex](-1)^{n}[/tex].
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.

 

[blue_raver22]

I understand that learning new material can be challenging, especially when it involves mathematical concepts like undetermined coefficients of difference equations. It seems like you have made some progress in understanding the concept by identifying the homogeneous equation and making a guess for the particular solution. I would suggest looking at some examples to help solidify your understanding of the process. For example, if the equation was y_{n+1}-2y_{n}=2^n, your guess for the particular solution would be y_{p}=A2^n, where A is a constant coefficient to be determined. You can then substitute this guess into the original equation and solve for A by comparing coefficients on both sides. Similarly, if the equation was y_{n+1}-2y_{n}=\cos(n), your guess for the particular solution would be y_{p}=A\cos(n)+B\sin(n), where A and B are constant coefficients to be determined. Again, substitute this guess into the original equation and solve for A and B by comparing coefficients. I hope this helps and I encourage you to keep practicing and seeking help if needed. Understanding new concepts takes time and effort, but it will be worth it in the end.