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

Find at least the first four non-zero terms in a power series expansion about x₀ for a general solution to the given differential equation with the value for x₀.
$(x^{2} - 2 x) y'' + 2 y = 0; x_{0} = 1$

The first four nonzero terms in a power series expansion about x₀ for a general solution:

$y = a_{0} (1 + {(x + 1)}^{2} + \frac{1}{3} {(x - 1)}^{4} + . . .) + a_{1} ((x - 1) + \frac{1}{3} {(x - 1)}^{3} + . . .)$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Power series expansion

A power series expansion of can be obtained simply by expanding the exponential and integrating term-by-term. This series converges for all, but the convergence becomes extremely slow if significantly exceeds unity.

Step 2: To determine the first four nonzero terms in a power series expansion about x0 for a general solution.

$\begin{array}{l} (x^{2} - 2 x) y'' + 2 y = 0 \\ y'' + \frac{2}{x^{2} - 2 x} y = 0 \end{array}$

We get the value of $q (x) = \frac{2}{x^{2} - 2 x}$ .

$\begin{array}{l} y (x) = \sum_{n = 0}^{\infty} a_{n} {(x - 1)}^{n} \\ y'' (x) = \sum_{n = 2}^{\infty} n (n - 1) a_{n} {(x - 1)}^{n - 2} + 2 \sum_{n = 0}^{\infty} a_{n} {(x - 1)}^{n} = 0 \end{array}$

Let, $x - 1 = t$ :

role="math" localid="1664089919710" $\begin{array}{l} (t + 1) (t - 1) \sum_{n = 2}^{\infty} n (n - 1) a_{n} (t)^{n - 2} + 2 \sum_{n = 0}^{\infty} a_{n} {(t)}^{n} = 0 \\ \sum_{n = 2}^{\infty} n (n - 1) a_{n} (t)^{n} - \sum_{n = 2}^{\infty} n (n - 1) a_{n} (t)^{n - 2} + 2 \sum_{n = 0}^{\infty} a_{n} {(t)}^{n} = 0 \\ 2 a_{2} t^{2} + 6 a_{3} t^{3} + 12 a_{4} t^{4} + 20 a_{5} t^{5} + . . . \\ 2 a_{0} + 2 a_{1} t + 2 a_{2} t^{2} + 2 a_{3} t^{3} + 2 a_{4} t^{4} + . . . = 0 \\ (2 a_{0} - 2 a_{2}) + (2 a_{1} - 6 a_{3}) (x - 1) + (2 a_{2} - 12 a_{4} + 2 a_{2}) (x - 1)^{2} + (2 a_{3} - 20 a_{5}) (x - 1)^{3} + . . . . = 0 \\ 2 a_{0} - 2 a_{2} = 0 \Rightarrow a_{2} = a_{0} \\ 2 a_{1} - 6 a_{3} = 0 \Rightarrow a_{3} = \frac{1}{3} a_{1} \\ 2 a_{2} - 12 a_{4} + 2 a_{2} = 0 \Rightarrow a_{4} = \frac{1}{3} a_{0} \end{array}$

In the end,

$\begin{array}{l} y = \sum_{n = 0}^{\infty} a_{n} {(x - 1)}^{n} = a_{0} + a_{1} (x - 1) + a_{2} (x - 1)^{2} + a_{3} (x - 1)_{3} + . . . \\ y = a_{0} (1 + {(x + 1)}^{2} + \frac{1}{3} {(x - 1)}^{4} + . . .) + a_{1} ((x - 1) + \frac{1}{3} {(x - 1)}^{3} + . . .) \end{array}$

Hence, the final answer is $y = a_{0} (1 + {(x + 1)}^{2} + \frac{1}{3} {(x - 1)}^{4} + . . .) + a_{1} ((x - 1) + \frac{1}{3} {(x - 1)}^{3} + . . .)$

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Find at least the first four non-zero terms in a power series expansion about x₀ for a general solution to the given differential equation with the value for x₀.
$(x^{2} - 2 x) y'' + 2 y = 0; x_{0} = 1$

Step by Step Solution

Step 1: Power series expansion

Step 2: To determine the first four nonzero terms in a power series expansion about x0 for a general solution.

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Find at least the first four non-zero terms in a power series expansion about x0 for a general solution to the given differential equation with the value for x0.(x2-2x)y''+2y=0; x0=1

Step by Step Solution

Step 1: Power series expansion

Step 2: To determine the first four nonzero terms in a power series expansion about x0 for a general solution.

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Find at least the first four non-zero terms in a power series expansion about x₀ for a general solution to the given differential equation with the value for x₀.
$(x^{2} - 2 x) y'' + 2 y = 0; x_{0} = 1$