• Apr 10th 2009, 04:36 PM
mrtwigx
I've got a question I'm really not sure how to do. here it is:

Consider the following difference equation:
q(n) = 2*q(n-1) + q(n-2) - 2*q(n-3)

Let q(n) = ar^n. Show that r must satisfy (r-1)(r+1)(r-2) = 0

It doesnt look to hard, im just not quiet sure exactly how to go about it...
• Apr 10th 2009, 05:11 PM
mr fantastic
Quote:

Originally Posted by mrtwigx
I've got a question I'm really not sure how to do. here it is:

Consider the following difference equation:
q(n) = 2*q(n-1) + q(n-2) - 2*q(n-3)

Let q(n) = ar^n. Show that r must satisfy (r-1)(r+1)(r-2) = 0

It doesnt look to hard, im just not quiet sure exactly how to go about it...

Substitute the trial solution into the difference equation:

$\displaystyle ar^n = 2 a r^{n-1} + a r^{n-2} - 2a r^{n-3}$.

Simplify and re-arrange:

$\displaystyle \Rightarrow r^{n-3} \left( r^3 - 2r^2 - r + 2\right) = 0$.

Since $\displaystyle r \neq 0$ you're left with $\displaystyle r^3 - 2r^2 - r + 2 = 0$.
• Apr 12th 2009, 07:50 PM
mrtwigx
Thanks for your reply, it realy was helpfull. I feel a little stupid that I couldnt get it myself...