Explanation Given information: The given differential equation is ( x 2 + 4 ) y " + y = 0 . Concept Used: Power series is defined as an infinite series of the form, ∑ n = 0 ∞ a n x n = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + ......... + a n x n + .... , where x is a variable. Calculation: Assume y = ∑ n = 0 ∞ a n x n is a solution of the differential equation. Then, differentiating the assumed solution with respect to x , y ' = ∑ n = 1 ∞ n a n x n − 1 y " = ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 Now, the given differential equation is, ( x 2 + 4 ) y " + y = 0 Inserting the value of y and y " in above equation, ( x 2 + 4 ) y " + y = 0 ( x 2 + 4 ) ⋅ ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 + ∑ n = 0 ∞ a n x n = 0 ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 ⋅ x 2 + ∑ n = 2 ∞ 4 n ( n − 1 ) a n x n − 2 = − ∑ n = 0 ∞ a n x n ∑ n = 2 ∞ n ( n − 1 ) a n x n + ∑ n = 2 ∞ 4 n ( n − 1 ) a n x n − 2 = − ∑ n = 0 ∞ a n x n To obtain equal power of x , adjust the summation indices by replacing n by n +2 in the left hand side sum, to obtain, ∑ n = 2 ∞ 4 n ( n − 1 ) a n x n − 2 = − ∑ n = 0 ∞ a n x n − ∑ n = 2 ∞ n ( n − 1 ) a n x n ∑ n = 0 ∞ 4 ⋅ ( n + 2 ) ( n + 1 ) a n + 2 x n = − ∑ n = 0 ∞ a n x n ( 1 + n ( n − 1 ) ) Equating the coefficients of like terms, one can obtain the recursion formula

11th Edition

ISBN: 9781337275378

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Chapter 16.4, Problem 7E

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To calculate: The solution of the differential equation (x2+4)y"+y=0, using a power series.

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Chapter 16.4, Problem 6E

Chapter 16.4, Problem 8E

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The solution of the differential equation ( x 2 + 4 ) y " + y = 0 , using a power series.

Solution Summary: The author calculates the solution of the differential equation (x2+4)y"+y=0 using a power series.

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To calculate: The solution of the differential equation (x2+4)y"+y=0, using a power series.

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Chapter 16 Solutions