The auxiliary equation of of y'''-4y''-4y'+16y=0 is r^(3)-4r^(2)-4r+16=0 . The auxiliary equation has a root in the interval [3,6]. Let f(r)=r^(3) -4r^(2)-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=3 and r2=6 , gives the following approximations for the root in [3,6]: r3^(1)= r3^(2)= r3^(3)= r3^(4)=
The auxiliary equation of of y'''-4y''-4y'+16y=0 is r^(3)-4r^(2)-4r+16=0 . The auxiliary equation has a root in the interval [3,6]. Let f(r)=r^(3) -4r^(2)-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=3 and r2=6 , gives the following approximations for the root in [3,6]: r3^(1)= r3^(2)= r3^(3)= r3^(4)=
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.4: Complex And Rational Zeros Of Polynomials
Problem 14E
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The auxiliary equation of of y'''-4y''-4y'+16y=0 is r^(3)-4r^(2)-4r+16=0
. The auxiliary equation has a root in the interval [3,6].
Let f(r)=r^(3) -4r^(2)-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=3 and r2=6 , gives the following approximations for the root in [3,6]:
r3^(1)=
r3^(2)=
r3^(3)=
r3^(4)=
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