Math 4791/5791 - Solution Set 8
Fall 1998
We look for solutions of the form 
, where 
is
to be determined. The resulting characteristic polynomial is
which has roots 
and 
. Thus the general
solution is
The initial conditions imply that 
and 
. The
solution is 
which makes the solution to the initial value
problem
Check your results! Notice the both the difference equation and the
solution give the sequence 
.
which has roots 
(one of which is
the golden mean). So the general solution is
The initial conditions imply that 
and 
. A little algebra gives the constants as
The solution of the initial value problem is
Unlikely as it seems, this solution generates the sequence of integers
.
Let y denote the limit of the ratio of successive terms in the
sequence: 
; this is the
quantity that we want to determine. We can form ratios of terms by
dividing the difference equation by 
as follows.
Now take the limit as 
across the difference
equation:
In the limit, the difference equation becomes
Solving this quadratic equation, we see that the positive root of
interest is 
.
which has the double root 
. This gives us only one linearly
independent solution 
. As with the ODE analog of this case,
the second solution is 
. This means the general solution is
The initial conditions give us that 
and 
. The
solution to this system is 
, producing the solution to the
initial value problem
Check that both the difference equation and the solution give the
sequence 
.
which has the complex roots
As in the ODE case, we need to come up with two linearly independent, real-valued solutions. Reasoning by analogy, we see that the general solution is
Evaluating the constants gives us 
and 
, so the solution
to the initial value problem is
Check that both the difference equation and the solution give the
sequence 
.