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Solving and Analyzing Difference Equations: Examples and Techniques

This chapter covers the concepts of solving and analyzing difference equations. It provides step-by-step solutions for two exercises, demonstrating how to find the values of yn and determine their long-term behavior. The examples involve equations with different values for a and b, showcasing various scenarios. By inspecting the obtained values of yn, the pattern and trends in their behavior are identified.

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Solving and Analyzing Difference Equations: Examples and Techniques

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  1. Chapter 11 Section 2 Introduction to Difference Equations II

  2. Exercise 13 (page 538) Part I • Given: yn = 0.4 yn – 1 + 3 , y0 = 7 • Solve the difference equation: a = 0.4 , b = 3 , b/(1 – a ) = 5 • yn = [ b/(1 – a) ] + ( y0 – [ b/(1 – a) ] ) ·an yn = 5 + ( 7 – 5 ) · 0.4n yn = 5 + 2 · 0.4n

  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 yn ?) y0 = 5 + 2 · (0.4)0 = 7 y1 = 5 + 2 · (0.4)1 = 5.8 y2 = 5 + 2 · (0.4)2 = 5.32 y3 = 5 + 2 · (0.4)3 = 5.128 y30 = 5 + 2 · (0.4)30 = 5.000000000002305 By inspection, the yn values APPROACH 5

  4. Exercise 15 (page 538) Part I • Given: yn = – 5 yn – 1 , y0 = 2 • Solve the difference equation: a = – 5 , b = 0 , b/(1 – a ) = 0 • yn = [ b/(1 – a) ] + ( y0 – [ b/(1 – a) ] ) ·an yn = 0 + ( 2 – 0 ) · (– 5)n yn = 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 yn ?) y0 = 2(– 5)0 = 2 y1 = 2(– 5)1 = – 10 y2 = 2(– 5)2 = 50 y3 = 2(– 5)3 = – 250 y4 = 2(– 5)4 = 1250 By inspection, the signs of yn oscillate between positive and negative while size of the values increases.

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