Given the difference equation y(k + 2) − 2 y(k + 1) + = y(k) = = e(k) where y(0) = y(1) = 0, e(0) = 0, and e(k) = 1, k = 1, 2,.... (a) Solve for y(k) as a function of k, and give the numerical values of y(k), 0≤k≤ 4. (b) Solve the difference equation directly for y(k), 0≤ k ≤ 4, to verify the results of part (a). (c) Repeat parts (a) and (b) for e(k) = 0 for all k, and y(0) = 1, y(1) = -2.
Given the difference equation y(k + 2) − 2 y(k + 1) + = y(k) = = e(k) where y(0) = y(1) = 0, e(0) = 0, and e(k) = 1, k = 1, 2,.... (a) Solve for y(k) as a function of k, and give the numerical values of y(k), 0≤k≤ 4. (b) Solve the difference equation directly for y(k), 0≤ k ≤ 4, to verify the results of part (a). (c) Repeat parts (a) and (b) for e(k) = 0 for all k, and y(0) = 1, y(1) = -2.
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.6: Variation
Problem 2E
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Transcribed Image Text:Given the difference equation
y(k + 2)
3
-y(k + 1) + y(k) = e(k)
where y(0) = y(1) = 0, e(0) = 0, and e(k) = 1, k = 1, 2, ....
(a) Solve for y(k) as a function of k, and give the numerical values of y(k), 0 ≤ k ≤ 4.
(b) Solve the difference equation directly for y(k), 0 ≤ k ≤ 4, to verify the results of part (a).
(c) Repeat parts (a) and (b) for e(k) = 0 for all k, and y(0) = 1, y(1) = −2.
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