Q28E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution to y’’’ - 3y’ - y = 0 by using Newton’s method or some other numerical procedure to approximate the roots of the auxiliary equation.

Step-by-Step Solution

Answer

_{The general solution is $y\left(x\right)=c_1{e}^{-1.53209}+c_2{e}^{-0.34729}+c_3{e}^{1.87939}$}

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.C

Step 2: Use of Newton’s Approximation method

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

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

Hence, the final answer is :

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