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Found in: Page 332

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# 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.

The general solution is $y\left(x\right)={c}_{1}{e}^{-1.53209}+{c}_{2}{e}^{-0.34729}+{c}_{3}{e}^{1.87939}$

See the step by step solution

## 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 :

$\begin{array}{l}{r}^{3}-3r-1=0\\ g\left(x\right)={x}^{3}-3x-1\\ g\text{'}\left(x\right)=3{x}^{2}-3\\ \\ g\left(-2\right)={\left(-2\right)}^{3}-3\left(-2\right)-1=-3\left(-2\right)-1=-3\\ g\left(-1\right)={\left(-1\right)}^{3}-3\left(-1\right)-1=1\\ g\left(0\right)={0}^{3}-3.0-1=-1\\ g\left(1\right)={1}^{3}-3.1-1=-3\\ g\left(2\right)={2}^{3}-3.2-1=1\\ \\ {x}_{n+1}={x}_{n}-\frac{g\left({x}_{n}\right)}{g\text{'}\left({x}_{n}\right)},n=1,2,...\\ {x}_{n+1}={x}_{n}-\frac{{{x}_{n}}^{3}-3{x}_{n}-1}{3{{x}_{n}}^{2}-3},n=1,2,....\\ {x}_{2}=-1.53675\\ {x}_{3}=-1.53211\\ {x}_{4}=-1.53209\\ {x}_{5}=-1.53209\\ {r}_{1}=-1.53209\\ \\ {x}_{n+1}={x}_{n}-\frac{{{x}_{n}}^{3}-3{x}_{n}-1}{3{{x}_{n}}^{2}-3},n=1,2,....\\ {x}_{2}=-0.33333\\ {x}_{3}=-0.34722\\ {x}_{4}=-0.34729\\ {x}_{5}=-0.34729\\ {r}_{2}=-0.34729\\ \\ {x}_{n+1}={x}_{n}-\frac{{{x}_{n}}^{3}-3{x}_{n}-1}{3{{x}_{n}}^{2}-3},n=1,2,....\\ {x}_{2}=1.90935\\ {x}_{3}=1.88003\\ {x}_{4}=1.87939\\ {x}_{5}=1.87939\\ {r}_{3}=1.87939\\ \\ y\left(x\right)={c}_{1}{e}^{-1.53209}+{c}_{2}{e}^{-0.34729}+{c}_{3}{e}^{1.87939}\end{array}$

Hence, the final answer is :

$y\left(x\right)={c}_{1}{e}^{-1.53209}+{c}_{2}{e}^{-0.34729}+{c}_{3}{e}^{1.87939}$