91Ó°ÊÓ

Newton's Method, also known as Newton Method, is important because it's an iterative process that can approximate solutions to an equation with incredible accuracy. And it's a method to approximate numerical solutions (i.e., x-intercepts, zeros, or roots) to equations that are too hard for us to solve by hand.C

We are going to find the roots of auxiliary equation by using Newton’s Approximation method :

$\begin{array}{l}{r}^3−3r−1=0\\ g\left(x\right)=x^3−3x−1\\ g\text{'}\left(x\right)=3x^2−3\\ \\ g(−2)={\left(-2\right)}^3-3\left(-2\right)-1=−3(−2)−1=−3\\ g(−1)={\left(-1\right)}^3−3(−1)−1=1\\ g\left(0\right)=0^3−3.0−1=−1\\ g\left(1\right)=1^3−3.1−1=−3\\ g\left(2\right)=2^3−3.2−1=1\\ \\ x_{n+1}=x_n−\frac{g\left(x_n\right)}{g\text{'}\left(x_n\right)},n=1,2,\mathrm{...}\\ x_{n+1}=x_n−\frac{{x_n}^3−3x_n−1}{3{x_n}^2−3},n=1,2,\mathrm{....}\\ x_2=−1.53675\\ x_3=−1.53211\\ x_4=−1.53209\\ x_5=−1.53209\\ r_1=−1.53209\\ \\ x_{n+1}=x_n−\frac{{x_n}^3−3x_n−1}{3{x_n}^2−3},n=1,2,\mathrm{....}\\ x_2=−0.33333\\ x_3=−0.34722\\ x_4=−0.34729\\ x_5=−0.34729\\ r_2=−0.34729\\ \\ x_{n+1}=x_n−\frac{{x_n}^3−3x_n−1}{3{x_n}^2−3},n=1,2,\mathrm{....}\\ x_2=1.90935\\ x_3=1.88003\\ x_4=1.87939\\ x_5=1.87939\\ r_3=1.87939\\ \\ y\left(x\right)=c_1{e}^{−1.53209}+c_2{e}^{−0.34729}+c_3{e}^{1.87939}\end{array}$

Hence, the final answer is :

$y\left(x\right)=c_1{e}^{−1.53209}+c_2{e}^{−0.34729}+c_3{e}^{1.87939}$