Explanation Given: The given equation is y " + 2 x 2 y ' + x y = 2 cos x Procedure used: The general solution is expressed as y ( x ) = ∑ n = 0 ∞ a n x n ......... ( 2 ) Find y ' ( x ) and y " ( x ) by differentiating the equation (2) Now, in the first summation, replace n by k + 2 , for second summation replace n by k − 1 and for fourth summation replace n by k 2 Separate the terms according to k = 0 , k = 1 Calculation: Let the given equation be y " + 2 x 2 y ' + x y − 2 cos x = 0........ ( 1 ) The general solution is expressed as y ( x ) = ∑ n = 0 ∞ a n x n ......... ( 2 ) Find y ' ( x ) and y " ( x ) by differentiating the equation (2) y ' ( x ) = ∑ n = 1 ∞ n a n x n − 1 y " ( x ) = ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 The value of cos x is, cos x = ∑ n = 0 ∞ ( − 1 ) n 2 n ! x 2 n Substitute the value of y ' ( x ) , y " ( x ) and y ( x ) in (1) ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 + 2 x 2 ∑ n = 1 ∞ n a n x n − 1 + x ∑ n = 0 ∞ a n x n − 2 ∑ n = 0 ∞ ( − 1 ) n 2 n ! x 2 n = 0 ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 + 2 ∑ n = 1 ∞ n a n x 2 ⋅ x n − 1 + ∑ n = 0 ∞ a n x ⋅ x n − 2 ∑ n = 0 ∞ ( − 1 ) n 2 n ! x 2 n = 0 ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 + 2 ∑ n = 1 ∞ n a n x n + 1 + ∑ n = 0 ∞ a n x n + 1 − 2 ∑ n = 0 ∞ ( − 1 ) n 2 n ! x 2 n = 0 Now, in the first summation, replace n by k + 2 , for second summation replace n by k − 1 and for fourth summation replace n by k 2 ∑ k = 0 ∞ ( k + 2 ) ( k + 1 ) a k + 2 x k + 2 ∑ k = 1 ∞ ( k − 1 ) a k − 1 x k + ∑ k = 0 ∞ a k − 1 x k − 1 − 2 ∑ k = 0 ∞ ( − 1 ) k 2 k ! x k = 0 Separate the terms according to k = 0 , k = 1 2 a 2 − 2 + ( 6 a 3 + a 0 − 2 ) x + ∑ k = 2 ∞ [ ( k + 2 ) ( k + 1 ) a k + 2

Differential Equations And Linear Algebra, Books A La Carte Edition (4th Edition)

4th Edition

ISBN: 9780321985811

Author: Stephen W. Goode, Scott A. Annin

Publisher: Pearson (edition 4)

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Chapter 11.2, Problem 20P

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Chapter 11.2, Problem 19P

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Chapter 11 Solutions

Differential Equations And Linear Algebra, Books A La Carte Edition (4th Edition)

Ch. 11.1 - True-False Review For Questions a-j, decide if the...Ch. 11.1 - Prob. 2TFR Ch. 11.1 - Prob. 3TFR Ch. 11.1 - Prob. 4TFR Ch. 11.1 - Prob. 5TFR Ch. 11.1 - True-False Review For Questions a-j, decide if the...Ch. 11.1 - Prob. 7TFR Ch. 11.1 - Prob. 8TFR Ch. 11.1 - Prob. 9TFR Ch. 11.1 - Prob. 10TFR

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The series solution to the initial-value problem.

Solution Summary: The author determines the series solution to the initial-value problem. The differential equation is y(x)=a_0left.

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Chapter 11 Solutions