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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?

Thanks

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?

Thanks

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. ...

en.wikipedia.org/wiki/Recu

Difference Equations

Purpose: To apply linear algebra concepts to study the properties of sequences defined by difference equations. Prerequisites: The concepts of linear ...

www.math.duke.edu/education/ccp/materials/linalg/diffeqs/index.html - 5k - Cached - Similar pages

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

?

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

into

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

:-\

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

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