Quote:
Originally Posted by jinydu
What has made this problem difficult for me is the repeated root.

Repeated roots in difference equations are similar to repeated roots in differential equations. In this case the general solutions is
A*(4)
^{n} + B*n*(4)
^{n}
You pick A and B to match the first two terms.
One way to derive this is to take the two root solution with alpha = beta + epsilon and figure out what happens in the limit as epsilon goes to zero. You have
[(beta+epsilon)
^{n}  beta
^{n}]/(beta+epsilonbeta)
=[beta
^{n} + n*epsilon* beta
^{n1} + epsilon
^{2}*??  beta
^{n}]/epsilon
= n*beta
^{n1} + epsilon*??
Absobing a factor of beta into the constant, in the limit this is proportional to
n * beta
^{n}
William