And r3(4) «The auxiliary equation of y"" - 4y" - 4y' + 16y = 0 isr³ − 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 equationcan be determined by solving f(r) = 0.Then, applying four iterations of the Bisection Method to f(r), usingthe initial approximations r₁= -0.75 and r2 = 3.75, gives the followingapproximations for the root in [-0.75, 3.75]:(¹1) = [(2)131333)=(64)

Answered: The auxiliary equation of y"" - 4y" -…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

And r3(4)

«
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]:
(¹1) = [
(2)
13
13
33)
=
(64)

Expert Solution

Step 1

The solution of an equation $f (r) = 0$ is defined as the values of r which satisfies the differential equation. In this problem, the auxiliary equation is $r^{3} - 4 r^{2} - 4 r + 16 = 0$ . One of its roots lies on the interval $(- 0.75, 3.75)$ . We have to find this root by four iterations. The bisection method is a step-by-step method to find the approximate solution of the transcendental equation in the form $f (r) = 0$ .