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.
$y'' - (cosx) y' - y = 0 y (π / 2) = 1, y' (π / 2) = 1$

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

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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,

$y'' - (cosx) y' - y = 0 y (π / 2) = 1, y' (π / 2) = 1$

Apply a substitution and transform the equation,

$y'' - \cos (t + \frac{π}{2}) \cdot y' - y = 0 y'' - sint \cdot y' - y = 0$

Use the formula

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

Substitute it in the above equation we get,

$\sum_{n = 2}^{\infty} n (n - 1) \cdot a_{n} {(t)}^{n - 2} - (t - \frac{t^{3}}{3!} + \frac{t^{5}}{5!} - \frac{t^{7}}{7!} + \dots) \sum_{n = 1}^{\infty} n \cdot a_{n} {(t)}^{n - 1} - \sum_{n = 0}^{\infty} a_{n} {(t)}^{n} = 0$

Hence we get the relation:

Step 3: Find the expression after expansion.

The series expansion for the function is

$(2 a_{2} + 6 a_{3} t + 12 a_{4} t^{2} + 20 a_{5} t^{3} + \dots) - (a_{1} t + 2 a_{2} t^{2} + 3 a_{3} t^{3} + 4 a_{4} t^{4} + \dots) + (a_{1} \frac{t^{3}}{3!} + 2 a_{2} \frac{t^{4}}{3!} + 3 a_{3} \frac{t^{5}}{3!} + 4 a_{4} \frac{t^{6}}{3!} + \dots) - (a_{1} \frac{t^{5}}{5!} + 2 a_{2} \frac{t^{6}}{5!} + 3 a_{3} \frac{t^{7}}{5!} + 4 a_{4} \frac{t^{8}}{5!} + \dots) + \dots - (a_{0} + a_{1} t + a_{2} t^{2} + a_{3} t^{3} + a_{4} t^{4} + \dots) = 0$

Taking coefficients and exponents of the same power.

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

Simplify the expression:

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

Hence, the expression after the expansion is:

Step 4: Find the first four nonzero terms.

By equating the coefficients, we get,

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

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.

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

Hence, the first four nonzero terms are:

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

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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.
$y'' - (cosx) y' - y = 0 y (π / 2) = 1, y' (π / 2) = 1$

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 nonzero terms in a power series expansion of the solution to the given initial value problem.y''-(cosx)y'-y=0y(π/2)=1, y'(π/2)=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 nonzero terms in a power series expansion of the solution to the given initial value problem.
$y'' - (cosx) y' - y = 0 y (π / 2) = 1, y' (π / 2) = 1$