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 Problems 13-19, find at least the first four non-zero terms in a power series expansion of the solution to the given initial value problem.
$y'' + ty' + e^{t} y = 0; y (0) = 0, y' (0) = - 1$

The first four nonzero terms in the power series expansion of the given initial value problem $y'' + ty' + e^{t} y = 0$ is $y (t) = - t + \frac{t^{3}}{3} + \frac{t^{4}}{12} - \frac{t^{5}}{24}$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Define power series expansion.

The power series approach is used in mathematics to find a power series solution to certain differential equations. In general, such a solution starts with an unknown power series and then plugs that solution into the differential equation to obtain a coefficient recurrence relation.

A differential equation's power series solution is a function with an infinite number of terms, each holding a different power of the dependent variable. It is generally given by the formula,

$y (x) = \sum_{n = 0}^{\infty} a_{n} x^{n}$

Step 2: Find the relation.

Given,

$y'' + ty' + e^{t} y = 0; y (0) = 0, y' (0) = - 1$

Use the formula,

$y (x) = \sum_{n = 0}^{\infty} a_{n} x^{n}$

Taking derivative and substituting in the equation, we get the relation,

$y' (x) = \sum_{n = 1}^{\infty} n \cdot a_{n} {(t)}^{n - 1} y'' (x) = \sum_{n = 2}^{\infty} n (n - 1) \cdot a_{n} {(t)}^{n - 2} \sum_{n = 2}^{\infty} n (n - 1) \cdot a_{n} {(t)}^{n - 2} + t \cdot \sum_{n = 1}^{\infty} n \cdot a_{n} {(t)}^{n - 1} + e^{t} \cdot \sum_{n = 0}^{\infty} a_{n} {(t)}^{n} = 0$

Hence we get the relation $\sum_{n = 2}^{\infty} n (n - 1) \cdot a_{n} {(t)}^{n - 2} + t \cdot \sum_{n = 1}^{\infty} n \cdot a_{n} {(t)}^{n - 1} + e^{t} \cdot \sum_{n = 0}^{\infty} a_{n} {(t)}^{n} = 0$ .

Step 3: Find the expression after expansion.

The series expansion for the function is

$(2 a_{2} + 6 a_{3} t + 12 a_{4} t^{2} + 20 a_{5} t^{3} + 30 a_{6} t^{4} + \dots) + t \cdot (a_{1} + 2 a_{2} t + 3 a_{3} t^{2} + 4 a_{4} t^{4} + 5 a_{5} t^{4} + \dots) + (1 + t + \frac{t^{2}}{2} + \frac{t^{3}}{6} + \dots) (a_{0} + a_{1} t + a_{2} t^{2} + a_{3} t^{3} + \dots) = 0$

By expanding the series we get,

role="math" localid="1664095204500" $(2 a_{2} + 6 a_{3} t + 12 a_{4} t^{2} + 20 a_{5} t^{3} + 30 a_{6} t^{4} + \dots) + (a_{1} t + 2 a_{2} t^{2} + 3 a_{3} t^{3} + 4 a_{4} t^{5} + 5 a_{5} t^{6} + \dots) + (a_{0} + a_{1} t + a_{2} t^{2} + a_{3} t^{3} + \dots) + (a_{0} t + a_{1} t^{2} + a_{2} t^{3} + a_{3} t^{4} + \dots) + (a_{0} \frac{t^{2}}{2} + a_{1} \frac{t^{3}}{2} + a_{2} \frac{t^{4}}{2} + a_{3} \frac{t^{5}}{2} + \dots) + (a_{0} \frac{t^{3}}{6} + a_{1} \frac{t^{4}}{6} + a_{2} \frac{t^{5}}{6} + a_{3} \frac{t^{6}}{6} + \dots) + \dots = 0$

Simplify the expression.

$(2 a_{2} + a_{0}) + (6 a_{3} + a_{1} + a_{1} + a_{0}) t + (12 a_{4} + a_{2} + a_{2} + a_{1} + \frac{a_{0}}{2}) t^{2} + (20 a_{5} + 3 a_{3} + a_{3} + a_{2} + \frac{a_{1}}{2} + \frac{a_{0}}{6} + \dots) t^{3} + \dots = 0$

Hence, the expression after the expansion is:

Step 4: Find the first four nonzero terms.

By equating the coefficients, we get,

$6 a_{3} + a_{1} + a_{1} + a_{0} = 0 \to a_{3} = - \frac{a_{1}}{3} = \frac{1}{3} 12 a_{4} + a_{2} + a_{2} + a_{1} + \frac{a_{0}}{2} = 0 \to a_{4} = \frac{- a_{0}}{12} = \frac{1}{12} 20 a_{5} + 3 a_{3} + a_{3} + a_{2} + \frac{a_{1}}{2} + \frac{a_{0}}{6} \to a_{5} = \frac{- 4 a_{3} - \frac{a_{1}}{2}}{20} = - \frac{1}{24}$

The general solution was

$y (t) = \sum_{n = 0}^{\infty} a_{n} t^{n} = a_{0} + a_{1} t + a_{2} t^{2} + a_{3} t^{3} + \dots$

Apply the initial condition and substitute the coefficient.

$y (t) = - t + \frac{t^{3}}{3} + \frac{t^{4}}{12} - \frac{t^{5}}{24}$

Hence, the first four nonzero terms are $y (t) = - t + \frac{t^{3}}{3} + \frac{t^{4}}{12} - \frac{t^{5}}{24}$ .

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

In Problems 13-19, find at least the first four non-zero terms in a power series expansion of the solution to the given initial value problem.
$y'' + ty' + e^{t} y = 0; y (0) = 0, y' (0) = - 1$

Step by Step Solution

Step 1: Define power series expansion.

Step 2: Find the relation.

Step 3: Find the expression after expansion.

Step 4: Find the first four nonzero terms.

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

In Problems 13-19, find at least the first four non-zero terms in a power series expansion of the solution to the given initial value problem.y''+ty'+ety=0; y(0)=0, y'(0)=-1

Step by Step Solution

Step 1: Define power series expansion.

Step 2: Find the relation.

Step 3: Find the expression after expansion.

Step 4: Find the first four nonzero terms.

Most popular questions for Math Textbooks

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In Problems 13-19, find at least the first four non-zero terms in a power series expansion of the solution to the given initial value problem.
$y'' + ty' + e^{t} y = 0; y (0) = 0, y' (0) = - 1$