1. Chapter 11 Section 2 Introduction to Difference Equations II

2. Exercise 13 (page 538) Part I Given: y n = 0.4 y n – 1 + 3, y 0 = 7 Solve the difference equation: a = 0.4, b = 3, b/(1 – a ) = 5 y n = [ b/(1 – a) ] + ( y 0 – [ b/(1 – a) ] ) · a n y n = 5 + ( 7 – 5 ) · 0.4 n y n = 5 + 2 · 0.4 n

3. Exercise 13 Part II By inspection, determine the long-run behavior of the terms. ( i.e. As n gets larger, what is happening to the values of y n ?) y 0 = 5 + 2 · (0.4) 0 = 7 y 1 = 5 + 2 · (0.4) 1 = 5.8 y 2 = 5 + 2 · (0.4) 2 = 5.32 y 3 = 5 + 2 · (0.4) 3 = 5.128 y 30 = 5 + 2 · (0.4) 30 = 5.000000000002305 By inspection, the y n values APPROACH 5

4. Exercise 15 (page 538) Part I Given: y n = – 5 y n – 1, y 0 = 2 Solve the difference equation: a = – 5, b = 0, b/(1 – a ) = 0 y n = [ b/(1 – a) ] + ( y 0 – [ b/(1 – a) ] ) · a n y n = 0 + ( 2 – 0 ) · (– 5) n y n = 2(– 5) n

5. Exercise 15 Part II By inspection, determine the long-run behavior of the terms. ( i.e. As n gets larger, what is happening to the values of y n ?) y 0 = 2(– 5) 0 = 2 y 1 = 2(– 5) 1 = – 10 y 2 = 2(– 5) 2 = 50 y 3 = 2(– 5) 3 = – 250 y 4 = 2(– 5) 4 = 1250 By inspection, the signs of y n oscillate between positive and negative while size of the values increases.