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 nonzero terms in a power series expansion of the solution to the given initial value problem.
$y' - e^{x} y = 0; y (0) = 1$

The first four nonzero terms in the power series expansion of the given initial value problem $y' - e^{x} y = 0$ is $y (x) = 1 + x + x^{2} + \frac{5}{6} x^{3} + \dots$ .

See the step by step solution

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 expression after expansion.

Given,

$y' - e^{x} y = 0; y (0) = 1$

From the above equation put $p (x) = e^{x}$ which is analytic over the entire number line.

Use the formula,

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

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

$y' (x) = \sum_{n = 1}^{\infty} n \cdot a_{n} {(x)}^{n - 1} \sum_{n = 1}^{\infty} n \cdot a_{n} {(x)}^{n - 1} - e^{x} \sum_{n = 1}^{\infty} n \cdot a_{n} {(x)}^{n} = 0$

The series expansion for the function is

$e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!} + \dots$

By expanding the series we get,

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

Hence the expression after the expansion is:

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

Step 3: Find the first four nonzero terms.

Expand the expression given in the previous step.

$(a_{1} + 2 a_{2} x + 3 a_{3} x^{2} + 4 a_{4} x^{3} + 5 a_{5} x^{4} + \dots) + (- a_{0} - a_{1} x - a_{2} x^{2} - a_{3} x^{3} - a_{4} x^{4} + \dots) + (- a_{0} x - a_{1} x^{2} - a_{2} x^{3} - a_{3} x^{4} - a_{4} x^{5} + \dots) + (- a_{0} \frac{x^{2}}{2} - a_{1} \frac{x^{3}}{2} - a_{2} \frac{x^{4}}{2} - a_{3} \frac{x^{5}}{2} - \dots)$

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

By equating the coefficients,

$a_{1} - a_{0} = 0 \to a_{1} = a_{0} 2 a_{2} - a_{1} - a_{0} = 0 \to a_{2} = a_{0} 3 a_{3} - a_{2} - a_{1} - \frac{a_{0}}{2} = 0 \to a_{3} = \frac{5}{6} a_{0}$

The general solution was

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

Apply the initial condition and substitute the coefficient.

$y (x) = 1 + x + x^{2} + \frac{5}{6} x^{3} + \dots$

Hence, the first four nonzero terms are $y (x) = 1 + x + x^{2} + \frac{5}{6} x^{3} + \dots$ .

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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 nonzero terms in a power series expansion of the solution to the given initial value problem.
$y' - e^{x} y = 0; y (0) = 1$

Step by Step Solution

Step 1: Define power series expansion.

Step 2: Find the expression after expansion.

Step 3: 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 nonzero terms in a power series expansion of the solution to the given initial value problem.y'-exy=0; y(0)=1

Step by Step Solution

Step 1: Define power series expansion.

Step 2: Find the expression after expansion.

Step 3: Find the first four nonzero terms.

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In Problems 13-19, find at least the first four nonzero terms in a power series expansion of the solution to the given initial value problem.
$y' - e^{x} y = 0; y (0) = 1$