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

To derive the general solutions given by equations (17)- (20) for the non-homogeneous equation (16), complete the following steps.
(a) Substitute $y (x) = \sum_{n = 0}^{\infty} a_{n} x^{n}$ and the Maclaurin series into equation (16) to obtain
$(2 a_{2} - a_{0}) + \sum_{k = 1}^{\infty} [(k + 2) (k + 1) a_{k + 2} - (k + 1) a_{k}] x^{k} = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1)!} x^{2 n + 1}$
(b) Equate the coefficients of like powers on both sides of the equation in part (a) and thereby deduce the equations
$a_{2} = \frac{a_{0}}{2}, a_{3} = \frac{1}{6} + \frac{a_{1}}{3}, a_{4} = \frac{a_{0}}{8}, a_{5} = \frac{1}{40} + \frac{a_{1}}{15}, a_{6} = \frac{a_{0}}{48}, a_{7} = \frac{19}{5040} + \frac{a_{1}}{105}$
(c) Show that the relations in part (b) yield the general solution to (16) given in equations (17)-(20).

(a) After substituting $y (x) = \sum_{n = 0}^{\infty} a_{n} x^{n}$ and Maclaurin series, we get the solution $(a_{2} - a_{0}) + \sum_{k = 1}^{\infty} [(k + 1) (k + 2) a_{k + 2} - (k + 1) a_{k}] x^{k} = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \dots$ .

(b) After equating the coefficients, the deduced equations are

$2 a_{2} - a_{0} = 0 \to a_{2} = \frac{a_{0}}{2} 6 a_{3} - 2 a_{1} = 1 \to a_{3} = \frac{1}{6} + \frac{a_{1}}{3} 12 a_{4} - 3 a_{2} = 0 \to a_{4} = \frac{a_{0}}{8} 20 a_{5} - 4 a_{3} = - \frac{1}{6} \to a_{5} = \frac{1}{40} + \frac{a_{1}}{15} 30 a_{6} - 5 a_{4} = 0 \to a_{6} = \frac{a_{0}}{48} 42 a_{7} - 6 a_{5} = \frac{1}{120} \to a_{7} = \frac{19}{5040} + \frac{a_{1}}{105}$

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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.

The equation given in (16) is:

$y'' - xy' - y = sinx$

Use the formula

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

Take derivative

$y' (x) = \sum_{n = 1}^{\infty} {na}_{n} x^{n - 1} y'' (x) = \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2}$

The Maclaurin series is $sinx = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \dots$

Substitute in the above equation.

role="math" localid="1664105464082" $\sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2} - x \cdot \sum_{n = 1}^{\infty} {na}_{n} x^{n - 1} - \sum_{n = 0}^{\infty} a_{n} x^{n} = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \dots \sum_{k = 0}^{\infty} (k + 1) (k + 2) a_{k + 2} x^{k} - \sum_{k = 1}^{\infty} {ka}_{k} x^{k} - \sum_{k = 0}^{\infty} a_{k} x^{k} = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \dots a_{2} + \sum_{k = 1}^{\infty} (k + 1) (k + 2) a_{k + 2} x^{k} - \sum_{k = 1}^{\infty} {ka}_{k} x^{k} - a_{0} - \sum_{k = 1}^{\infty} a_{k} x^{k} = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \dots (a_{2} - a_{0}) + \sum_{k = 1}^{\infty} [(k + 1) (k + 2) a_{k + 2} - (k + 1) a_{k}] x^{k} = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \dots$

Hence the relation is $(a_{2} - a_{0}) + \sum_{k = 1}^{\infty} [(k + 1) (k + 2) a_{k + 2} - (k + 1) a_{k}] x^{k} = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \dots$ .

Step 3: Find the equations of coefficients.

Expand the expression.

$(2 a_{2} - a_{0}) + (6 a_{3} - 2 a_{1}) x + (12 a_{4} - 3 a_{3}) x^{2} + + (20 a_{5} - 4 a_{3}) x^{3} + (30 a_{6} - 5 a_{4}) x^{4} + (42 a_{7} - 6 a_{5}) x^{5} + (56 a_{8} - 6 a_{6}) x^{6} + \dots = x - \frac{x^{3}}{6} + \frac{x^{5}}{120} - \frac{x^{7}}{5040} + \dots$

Equate the coefficients of like powers on both sides of the equation.

Step 4: Find the expression after expansion.

Substitute the above coefficients in the equation:

$y (x) = a_{0} + a_{1} x + \frac{a_{0}}{2} x^{2} + (\frac{1}{6} + \frac{a_{1}}{3}) x^{3} + \frac{a_{0}}{8} x^{4} + (\frac{1}{40} + \frac{a_{1}}{15}) x^{5} + \frac{a_{0}}{48} x^{6} + (\frac{19}{5040} + \frac{a_{1}}{105}) x^{7} + \dots y (x) = a_{0} (1 + \frac{x^{2}}{2} + \frac{x^{4}}{8} + \frac{x^{6}}{48} + \dots) + a_{1} (x + \frac{x^{3}}{3} + \frac{x^{5}}{15} + \frac{x^{7}}{105} + \dots) + \frac{x^{3}}{6} + \frac{x^{5}}{40} + \frac{19 x^{7}}{5040} + \dots$

Where we can write:

$y_{1} = 1 + \frac{x^{2}}{2} + \frac{x^{4}}{8} + \frac{x^{6}}{48} + \dots y_{2} = x + \frac{x^{3}}{3} + \frac{x^{5}}{15} + \frac{x^{7}}{105} + \dots y_{p} = \frac{x^{3}}{6} + \frac{x^{5}}{40} + \frac{19 x^{7}}{5040} + \dots$

Hence, we showed that the results are similar to equations (17)-(20).

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Step by Step Solution

Step 1: Define power series expansion.

Step 2: Find the relation.

Step 3: Find the equations of coefficients.

Step 4: Find the expression after expansion.

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Step by Step Solution

Step 1: Define power series expansion.

Step 2: Find the relation.

Step 3: Find the equations of coefficients.

Step 4: Find the expression after expansion.

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