Difference equations – exercise sheet
This sheet is intended, first, to give you some practice, and then to guide you through
as you explore methods of solution for various types of difference equation.
1. Use the methods described in class to find the solutions of
(a) xn = 5xn −1 − 6 xn − 2 for n ≥ 2 with x0 = 0 and x1 = 1 ;
(b) xn = 4 xn − 2 for n ≥ 2 with x0 = 1 and x1 = 2 .
2. The formula for the solution of the difference equation that describes the Fibonacci
sequence:
xn = xn −1 + xn − 2 for n ≥ 2 with x0 = 0 and x1 = 1 ,
is
LMd
N
i d
i OPQ
n
n
1 1
1+ 5 − 1− 5 .
xn = n
2 5
Find x3 (and so confirm that it is indeed an integer).
3. Solve the difference equation xn = nxn −1 , for n ≥ 1 with x0 given, by direct
construction i.e. substitute for xn−1 and then for xn−2 , and so on (but only in
principle, of course !) until you get to x0 .
4. Use the same method as in Q.1 to solve the difference equation
xn = 4 xn −1 − 4 xn − 2 for n ≥ 2 with x0 and x1 given.
You should find that your solution contains only one arbitrary constant (although
you may have noted a complication in this calculation). It is therefore impossible to
incorporate both given values. Now look for a solution in the form
xn = ( A + Bn)2 n ,
where A and B are constants; what do you find ?
5. Use the method of direct construction (as in Q.3) to find the solution of
xn = xn −1 + 1 for n ≥ 1 and x0 given.
Now seek a solution of the equation in the form xn = a + bn , where a and b are
constants to be determined. Use your values of a and b (and what do you notice
about a ?) to write xn = X n + a + bn and find the equation for X n .
[You should have found that you may set a = 0 .]
xn = xn −1 + n
in the form xn = a + bn + cn 2 , where a, b and c are constants to be determined (but
note what we did in Q.5).
6. Find a solution of the difference equation
Word Doc\Difference Equations\Year13MasterclassDifferenceTutorial