StudySmarter AI is coming soon!

- :00Days
- :00Hours
- :00Mins
- 00Seconds

A new era for learning is coming soonSign up for free

Suggested languages for you:

Americas

Europe

Q28E

Expert-verifiedFound in: Page 332

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

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

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

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

Hence, the final answer is :

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

94% of StudySmarter users get better grades.

Sign up for free