Short Answer

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 $y (x) = c_{1} e^{- 1.53209} + c_{2} e^{- 0.34729} + c_{3} e^{1.87939}$}

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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 :

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

Hence, the final answer is :

$y (x) = c_{1} e^{- 1.53209} + c_{2} e^{- 0.34729} + c_{3} e^{1.87939}$

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

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

Step 1: Newton’s Approximation method

Step 2: Use of Newton’s Approximation method

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

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

Step 1: Newton’s Approximation method

Step 2: Use of Newton’s Approximation method

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