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.

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

$\begin{array}{l}{r}^4+2r^3−3r^2−r+\frac{1}{2}=0\\ g\left(x\right)=x^4+2x^3−3x^2−x+\frac{1}{2}\\ g\text{'}\left(x\right)=4x^3+6x^2−6x−1\\ \\ g(−3)=3,5\\ g(−2)=−9,5\\ g(−1)=−2,5\\ g\left(0\right)=0,5\\ g\left(1\right)=−0,5\\ g\left(2\right)=18,5\\ \\ x_{n+1}=x_n−\frac{g\left(x_n\right)}{g\text{'}\left(x_n\right)},n=1,2,3,\mathrm{...}\\ x_{n+1}=x_n−\frac{{x_n}^4+2x_n^3−3x_n^2−x_n+0,5}{4x_n^3+6x_n^2−6x_n−1},n=1,2,3,\mathrm{...}\\ \\ x_2=−3.0381\\ x_3=−2.91276\\ x_4=−2.89634\\ x_5=−2.89607\\ x_6=−2.89607\\ r_1=−2.89607\\ \\ x_{n+1}=x_n−\frac{{x_n}^4+2x_n^3−3x_n^2−x_n+0,5}{4x_n^3+6x_n^2−6x_n−1},n=1,2,3,\mathrm{...}\\ x_2=−0.55095\\ x_3=−0.52145\\ x_4=−0.52022\\ x_5=−0.52022\\ r_2=−0.52022\\ \\ x_{n+1}=x_n−\frac{{x_n}^4+2x_n^3−3x_n^2−x_n+0,5}{4x_n^3+6x_n^2−6x_n−1},n=1,2,3,\mathrm{...}\\ x_2=0.28125\\ x_3=0.29641\\ x_4=0.29632\\ x_5=0.29632\\ r_3=0.29632\\ \\ x_{n+1}=x_n−\frac{{x_n}^4+2x_n^3−3x_n^2−x_n+0,5}{4x_n^3+6x_n^2−6x_n−1},n=1,2,3,\mathrm{...}\\ x_2=1.12085\\ x_3=1.11997\\ x_4=1.11997\\ r_4=1.11997\\ \\ y\left(x\right)=c_1{e}^{−2.89607}+c_2{e}^{−0.52022x}+c_3{e}^{0.29632}+c_4{e}^{1.11997}\end{array}$

Hence, the final answer is :

$y\left(x\right)=c_1{e}^{−2.89607}+c_2{e}^{−0.52022x}+c_3{e}^{0.29632}+c_4{e}^{1.11997}$