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Find the solution to the following linear, homogeneous recurrence with constant coefficients: an = +10an-1-22an-2 for n ≥ 2 with initial conditions ap 10, a₁ = 53. The solution is of the form: an = (a + B√s) (r + √5)² + (a − B√s) (r - √s)" for suitable real constants a, p, r, s. Note that the variable in this problem doesn't represent a characteristic value. Find these constants. T = S= α = B =

Find the solution to the following linear, homogeneous recurrence with constant coefficients: an = +10an-1-22an-2 for n ≥ 2 with initial conditions ap 10, a₁ = 53. The solution is of the form: an = (a + B√s) (r + √5)² + (a − B√s) (r - √s)" for suitable real constants a, p, r, s. Note that the variable in this problem doesn't represent a characteristic value. Find these constants. T = S= α = B =

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Find the solution to the following linear, homogeneous recurrence with constant coefficients: 53. The solution is of the form: an = +10an-1-22an-2 for n ≥ 2 with initial conditions ag = 10, a₁ = an = : (a + B√√s) (r + √5)² + (a − B√s) (r - √s)" for suitable real constants a, p, r, s. Note that the variable r in this problem doesn't represent a characteristic value. Find these constants. T = S= R || B ||
Transcribed Image Text:Find the solution to the following linear, homogeneous recurrence with constant coefficients: 53. The solution is of the form: an = +10an-1-22an-2 for n ≥ 2 with initial conditions ag = 10, a₁ = an = : (a + B√√s) (r + √5)² + (a − B√s) (r - √s)" for suitable real constants a, p, r, s. Note that the variable r in this problem doesn't represent a characteristic value. Find these constants. T = S= R || B ||
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