Find a general solution to

$y^{\left(4\right)}+2y\text{'}\text{'}\text{'}-3y\text{'}\text{'}-y\text{'}+\frac{1}{2}y=0$

by using Newton’s method (Appendix B) or some othernumerical procedure to approximate the roots of the auxiliaryequation.

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 :

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