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3.1.3 Example C By direct substitution, we can show that the equation Yk+2 – 5yk+1 + 6yk (3.27) has the solutions (2) - 2" , (1) - 3*. (3.28) Therefore, for arbitrary constants C1 and С2, Yk = C12k + c23k (3.29) is a solution to equation (3.27). Likewise, we can show by direct substitution that Yk = 1/2(k? + 3k + 5) (3.30) is a solution to the inhomogeneous equation Yk+2 5yk + 6yk = k². (3.31) Therefore, from Theorem 3.3, we conclude that c2* + c23k + 1/2(k² + 3k + 5) (3.32) Yk = is a solution to equation (3.31).

3.1.3 Example C By direct substitution, we can show that the equation Yk+2 – 5yk+1 + 6yk (3.27) has the solutions (2) - 2" , (1) - 3*. (3.28) Therefore, for arbitrary constants C1 and С2, Yk = C12k + c23k (3.29) is a solution to equation (3.27). Likewise, we can show by direct substitution that Yk = 1/2(k? + 3k + 5) (3.30) is a solution to the inhomogeneous equation Yk+2 5yk + 6yk = k². (3.31) Therefore, from Theorem 3.3, we conclude that c2* + c23k + 1/2(k² + 3k + 5) (3.32) Yk = is a solution to equation (3.31).

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section5.6: Exponential And Logarithmic Equations
Problem 64E
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3.1.3 Example C By direct substitution, we can show that the equation Yk+2 – 5yk+1 + 6yk (3.27) has the solutions Y = 2k, (1) (2) – 3k. (3.28) Therefore, for arbitrary constants c1 and c2, Yk C12* + c23k (3.29) is a solution to equation (3.27). Likewise, we can show by direct substitution that (Y = 1/2(k? + 3k + 5) (3.30) is a solution to the inhomogeneous equation Yk+2 – 5yk + 6yk = k². (3.31) Therefore, from Theorem 3.3, we conclude that c12* + c23* + 1/2(k² + 3k + 5) (3.32) Yk = C is a solution to equation (3.31).
Transcribed Image Text:3.1.3 Example C By direct substitution, we can show that the equation Yk+2 – 5yk+1 + 6yk (3.27) has the solutions Y = 2k, (1) (2) – 3k. (3.28) Therefore, for arbitrary constants c1 and c2, Yk C12* + c23k (3.29) is a solution to equation (3.27). Likewise, we can show by direct substitution that (Y = 1/2(k? + 3k + 5) (3.30) is a solution to the inhomogeneous equation Yk+2 – 5yk + 6yk = k². (3.31) Therefore, from Theorem 3.3, we conclude that c12* + c23* + 1/2(k² + 3k + 5) (3.32) Yk = C is a solution to equation (3.31).
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