Nonhomogeneous second order difference equation

x(n+1) - 4x(n) + 3x(n-1) = 36n^2

What form will the particular solution take?

I tried x(n)=an^2+bn+c, but when I plugged it in, I got a 0 coefficient for the n^2 terms on the LHS.
What is an alternative to try?

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ozoConnect With a Mentor Commented:
a*b^n + c*n^3 + d*n^2 + e*n + f
It has been a while since I have worked with difference equations but the following two reference might be of help.

Recurrence relation - Wikipedia, the free encyclopedia
A difference equation is a specific type of recurrence relation. ... Certain difference equations can be solved using z-transforms. ... - 50k - Cached - Similar pages

Difference Equations
Purpose: To apply linear algebra concepts to study the properties of sequences defined by difference equations. Prerequisites: The concepts of linear ... - 5k - Cached - Similar pages

Beta07Author Commented:
Interesting. The next question is:

x(n+1) - 4x(n) + 3x(n-1) = 3^n

Which I'm having the same problem with (getting 0 coefficients).

Would the particular solution take the form:

a*3^n + c*n^3 + d*n^2 + e*n + f

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Beta07Author Commented:
> a*b^n + c*n^3 + d*n^2 + e*n + f

Isn't that the form of the general solution?
And is there a reason for having a cubic rather than a quadratic? ("Because it works" will suffice)
x(n+1) - x(n)  = 36n^2 would be cubic
x(n+1) - 4x(n) = 0 would be exponential
Beta07Author Commented:
Ahhh, very clever!

I'll give it a shot, thanks
Beta07Author Commented:
Hmm, when I plug

x(n) = a*b^n + c*n^3 + d*n^2 + e*n + f


x(n+1) - 4x(n) + 3x(n-1) = 36n^2

And equate coefficients, the only information I get out of it is:

d = 3c
c+d+e = 0

Beta07Author Commented:
I think I may have figured it (this method has worked for 3^n instead of 36n²), I'm just stuck on this one step;

The Shift Operator, E, is defined as:  E x(n) = x(n+1)
So obviously, E^k x(n) = x(n+k)

I need to define an annihilator A(E), such that:

A(E) n^2 = 0

An example for n is:

(E-1)^2 n = 0
Beta07Author Commented:
Oh, lol

(E-1)^4  n^2 = 0


Which makes sense from more than one perspective ..
Beta07Author Commented:
Ah, done it! :)

I used a slightly varied method to my previous attempts (obviously), but it practically paralleled ozo's solution; and in doing so, I realised where I went wrong with my first attempt with ozo's suggestion..

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