Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

In the study of the vacuum tube, the following equation is encountered:
$y^{"} + (0.1) (y^{2} - 1) y^{'} + y = 0 .$
Find the Taylor polynomial of degree 4 approximating the solution with the initial values $y (0) = 1$ , $y^{'} (0) = 0$ .

The required polynomial is, $p_{3} (x) = 1 - \frac{x^{2}}{6}$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1:To Find the Taylor polynomial of degree

The formula for the Taylor polynomial of degree n centered at $x_{0}$ , approximating a function $f (x)$ possessing n derivatives at , $x_{0}$ is given by

$p_{n} (x) = f (x_{0}) + f^{'} (x_{0}) \times (x - x_{0}) + f^{''} (x_{0}) \times \frac{{(x - x_{0})}^{2}}{2!} + \dots + f^{n} (x_{0}) \times \frac{{(x - x_{0})}^{n}}{n!}$

It is given that for the function , $y (x)$

$y (0) = 1 and y^{'} (0) = 0$

The Taylor's polynomial cantered around $x_{0} = 0$ is given by

$p_{n} (x) = y (0) + y^{'} (0) \times (x - 0) + y^{''} (0) \times \frac{{(x - 0)}^{2}}{2!} + \dots + y^{n} (0) \times \frac{{(x - 0)}^{n}}{n!}$

We need the value of $y (0), y^{'} (0), y^{''} (0)$ and $y^{'''} (0)$ etc for finding the value of the four non zero terms. The first two are provided by the initial conditions.

The value of $y^{''} (0)$ can be deduced from the differential equation itself and the values of the lower derivatives.

$\begin{matrix} y^{''} = - y - (0.1) (y^{2} - 1) y^{'} \\ y^{''} (0) = - y (0) - (0.1) (y^{2} (0) - 1) y^{'} (0) \\ = - 1 \end{matrix}$

Now since $y^{''} = - y - (0.1) (y^{2} - 1) y^{'}$ holds for some interval around $x_{0} = 0$ , we can differentiate both sides to derive,

$\begin{matrix} y^{'''} = - y^{'} - (0.1) [(y^{2} - 1) (y^{''}) + 2 y {(y^{'})}^{2}] \\ y^{'''} (0) = - y^{'} (0) - (0.1) [({(y (0))}^{2} - 1) (y^{''} (0)) + 2 y (0) {(y^{'} (0))}^{2}] \\ = 0 \end{matrix}$

Similarly,

$\begin{matrix} y^{4} = - y^{''} - (0.1) [(y^{2} - 1) (y^{'''}) + 2 y (y^{'}) (y^{''}) + 2 {(y^{'})}^{3} + 2 y (2 y^{'}) (y^{''})] \\ y^{4} (0) = - y^{''} (0) - (0.1) [(y {(0)}^{2} - 1) (y^{'''} (0)) + 2 y (0) (y^{'} (0)) (y^{''} (0)) + 2 {(y^{'} (0))}^{3} + 2 y (0) (2 y^{'} (0)) (y^{''} (0))] \\ = 1 \end{matrix}$

Step 2:Final proof

Hence by substituting the Taylor polynomial of degree four for the solution is given by. $p_{n} (x) = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!}$

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

In the study of the vacuum tube, the following equation is encountered:
$y^{"} + (0.1) (y^{2} - 1) y^{'} + y = 0 .$
Find the Taylor polynomial of degree 4 approximating the solution with the initial values $y (0) = 1$ , $y^{'} (0) = 0$ .

Step by Step Solution

Step 1:To Find the Taylor polynomial of degree

Step 2:Final proof

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

In the study of the vacuum tube, the following equation is encountered:y"+(0.1)(y2-1)y'+y=0.Find the Taylor polynomial of degree 4 approximating the solution with the initial valuesy(0)=1 , y'(0)=0.

Step by Step Solution

Step 1:To Find the Taylor polynomial of degree

Step 2:Final proof

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In the study of the vacuum tube, the following equation is encountered:
$y^{"} + (0.1) (y^{2} - 1) y^{'} + y = 0 .$
Find the Taylor polynomial of degree 4 approximating the solution with the initial values $y (0) = 1$ , $y^{'} (0) = 0$ .