The auxiliary equation of y"" - 4y" - 4y' + 16y = 0 is r³-4r² - 4r + 16 = 0. The auxiliary equation has a root in the interval [-0.75,3.75]. Let f(r) = r³ - 4r² - 4r + 16, so that the roots of the auxiliary equation can be determined by solving f(r) = 0. Then, applying four iterations of the Bisection Method to f(r), using the initial approximations r₁= -0.75 and r2 = 3.75, gives the following approximations for the root in [-0.75,3.75]: r₂(²) = 13

The auxiliary equation of y"" - 4y" - 4y' + 16y = 0 is r³-4r² - 4r + 16 = 0. The auxiliary equation has a root in the interval [-0.75,3.75]. Let f(r) = r³ - 4r² - 4r + 16, so that the roots of the auxiliary equation can be determined by solving f(r) = 0. Then, applying four iterations of the Bisection Method to f(r), using the initial approximations r₁= -0.75 and r2 = 3.75, gives the following approximations for the root in [-0.75,3.75]: r₂(²) = 13

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter2: Equations And Inequalities

Section2.1: Equations

Problem 54E

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