This document provides an overview of lesson 30 on first order difference equations. It includes: 1) Announcements about upcoming exams and homework due dates. 2) Instructions to make cobweb diagrams of various linear difference equations and a link to an online applet for doing so. 3) Facts and solutions about linear and nonlinear difference equations, including determining equilibria, stability conditions, and general solutions. 4) Practice problems involving guessing solutions, finding equilibria, and making conjectures about stability.

1. Lesson 30
First Order Difference Equations
Math 20
April 27, 2007
Announcements
PS 12 due Wednesday, May 2
MT III Friday, May 4 in SC Hall A
Final Exam (tentative): Friday, May 25 at 9:15am

2. Problem 1
Make cobweb diagrams of lots of different linear difference
equations. Here are a few to get you started:
yk+1 = −1/2yk + 1 yk +1 = 3/2yk + 1
yk+1 = −yk + 1
yk+1 = 1/2yk + 1
yk+1 = −3/2yk + 1 yk+1 = yk + 1

3. Fun with applets
http://math.bu.edu/DYSYS/applets/linear-web.html

4. Problem 2
Make a conjecture: “The difference equation yk+1 = ayk + b has
a stable equilibrium when a and b satisfy the conditions that
...”

5. Problem 2
Make a conjecture: “The difference equation yk+1 = ayk + b has
a stable equilibrium when a and b satisfy the conditions that
...”
Fact
The difference equation yk+1 = ayk + b has a unique
equilibrium when a = 1. The equilibrium is stable when |a| < 1.

6. Problem 3
Guess and check solutions to the following difference
equations:
(ii) yk+1 = −1/2yk ,
(i) yk+1 = 2yk , (iii) yk+1 = ayk ,
y0 = 1/3
y0 = 1 y0 = y0

7. Problem 3
Guess and check solutions to the following difference
equations:
(ii) yk+1 = −1/2yk ,
(i) yk+1 = 2yk , (iii) yk+1 = ayk ,
y0 = 1/3
y0 = 1 y0 = y0
Fact
The general solution to the linear homoegeneous difference
equation is yk+1 = ayk , y0 = y0 is
yk = a k y0

8. Problem 4
(a) Find the equilibrium solution to yk+1 = −1/2yk + 1.
(b) Find the general solution to yk+1 = −1/2yk , y0 = c.
(c) Add the two together and choose c to solve the equation
with initial conditions
yk+1 = −1/2yk + 1, y0 = 0

9. Problem 4
(a) Find the equilibrium solution to yk+1 = −1/2yk + 1.
(b) Find the general solution to yk+1 = −1/2yk , y0 = c.
(c) Add the two together and choose c to solve the equation
with initial conditions
yk+1 = −1/2yk + 1, y0 = 0
Solution
We get
yk = (−1/2)k (−2/3) + 2/3

10. Solving the inhomogenous equation
Fact
The linear first-order difference equation
yk+1 = ayk + b
has solutions given by
ak y − b b

if a = 1
+
0
1−a 1−a
yk =
y0 + kb if a = 1


11. Solving the inhomogenous equation
Fact
The linear first-order difference equation
yk+1 = ayk + b
has solutions given by
ak y − b b

if a = 1
+
0
1−a 1−a
yk =
y0 + kb if a = 1

This establishes our first conjecture about stability, too.

12. Problem 5
Do cobweb diagrams for several members of the family of
difference equations
yk+1 = ryk (1 − yk )
(i) r = 1/2 (iii) r = 2 (v) r = 3
(iv) r = 5/2
(ii) r = 1 (vi) r = 3.1

13. More fun with applets
http://math.bu.edu/DYSYS/applets/nonlinear-web.html

14. Problem 6
Given the equation yk+1 = ryk (1 − yk ), find the equilibria in
terms of r .

15. Problem 6
Given the equation yk+1 = ryk (1 − yk ), find the equilibria in
terms of r .
Solution
Solving
y∗ = 4y∗ (1 − y∗ )
gives
y∗ = 0 and y∗ = 1 − 1/r

16. Problem 7
Make a conjecture about the stability of a fixed point to a
nonlinear systems yk+1 = g(yk ) that involves g (hint: make a
linear approximation near the equilibrium)

17. Problem 7
Make a conjecture about the stability of a fixed point to a
nonlinear systems yk+1 = g(yk ) that involves g (hint: make a
linear approximation near the equilibrium)
Fact
An equilibrium solution y∗ to the equation yk+1 = g(yk ) is stable
if
g (y∗ ) < 1
and unstable if
g (y∗ ) > 1

18. Problem 8
Can you make a conjecture about when the nonzero
equilibrium point of the difference equation yk+1 = ryk (1 − yk ) is
stable? Your answer should involve r .

19. Problem 8
Can you make a conjecture about when the nonzero
equilibrium point of the difference equation yk+1 = ryk (1 − yk ) is
stable? Your answer should involve r .
We have y∗ = 1 − 1 , and g (y∗ ) = 2 − r . So the nonzero
r
equilbrium point is stable if 1 < r < 3 and unstable if r > 3.