Q27E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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.

Step-by-Step Solution

Answer

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

Step 1: Newton’s Approximation method

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.

Step 2: Use of Newton’s Approximation method

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 :

 

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