Explanation Formula used: Power Series Method : The power series method for solving a second-order homogenous differential equation consists of finding the coefficients of a power series y ( x ) = ∑ n = 0 ∞ c n x n = c 0 + c 1 x + c 2 x 2 + … Calculation: The given equation is y ″ + 2 y ′ + y = 0 . Assume that the series solution takes the form of y = ∑ n = 0 ∞ c n x n . Calculate the derivative of y = ∑ n = 0 ∞ c n x n . y ′ = ∑ n = 1 ∞ n c n x n − 1 Now calculate the second derivative of y = ∑ n = 0 ∞ c n x n . y ″ = ∑ n = 2 ∞ n ( n − 1 ) c n x n − 1 − 1 = ∑ n = 2 ∞ n ( n − 1 ) c n x n − 2 Substitute the derivatives of y ″ and y ′ in y ″ + 2 y ′ + y = 0 as follows. ∑ n = 2 ∞ n ( n − 1 ) c n x n − 2 + ∑ n = 1 ∞ 2 n c n x n − 1 + ∑ n = 0 ∞ c n x n = 0 Equate the coefficient of each power of x to zero as shown in the following table. Power of x Coefficient equation x 0 2 ( 1 ) c 2 + 2 ( 1 ) c 1 + c 0 = 0 or c 2 = − c 1 − 1 2 c 0 x 1 3 ( 2 ) c 3 + 2 ( 2 ) c 2 + c 1 = 0 or c 3 = − 2 3 c 2 − 1 6 c 1 = 1 2 c 1 + 1 3 c 0 x 2 4 ( 3 ) c 4 + 2 ( 3 ) c 3 + c 2 = 0 or c 4 = − 1 2 c 3 − 1 12 c 2 = − 1 6 c 1 − 1 8 c 0 x 3 5 ( 4 ) c 5 + 2 ( 4 ) c 4 + c 3 = 0 or c 5 = − 2 5 c 4 − 1 20 c 3 = 1 24 c 1 + 1 30 c 0 x 4 6 ( 5 ) c 6 + 2 ( 5 ) c 5 + c 4 = 0 or c 6 = − 1 3 c 5 − 1 30 c 4 = − 1 120 c 1 − 1 144 c 0 x n ( n + 2 ) ( n + 1 ) c n + 2 + 2 ( n + 1 ) c n + 1 + c n = 0 or c n + 2 = − 2 n + 2 c n + 1 − 1 ( n + 2 ) ( n + 1 ) c n The recursive relation is c n + 2 = − 2 n + 2 c n + 1 − 1 ( n + 2 ) ( n + 1 ) c n . Hence, y = ( c 0 + c 1 x + ( − c 1 − 1 2 c 0 ) x 2 + ( 1 2 c 1 + 1 3 c 0 ) x 3 + ( − 1 6 c 1 − 1 8 c 0 ) x 4 + ( 1 24 c 1 + 1 30 c 0 ) x 5 + ( − 1 120 c 1 − 1 144 c 0 ) x 6 + ⋯ ) = ( c 0 − 1 2 c 0 x 2 + 1 3 c 0 x 3 − 1 8 c 0 x 4 + 1 30 x 5 − 1 144 c 0 x 6 + ⋯ + c 1 x − c 1 x 2 + 1 2 c 1 x 3 − 1 6 c 1 x 3 − 1 6 c 1 x 4 + 1 24 c 1 x 5 − 1 120 x 6 + ⋯ ) = ( c 0 + <

University Calculus: Early Transcendentals (4th Edition)

4th Edition

ISBN: 9780134995540

Author: Joel R. Hass, Christopher E. Heil, Przemyslaw Bogacki, Maurice D. Weir, George B. Thomas Jr.

Publisher: PEARSON

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Chapter 17.5, Problem 2E

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To find: The general solution of the differential equation y″+2y′+y=0 by using power series.

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Chapter 17.5, Problem 1E

Chapter 17.5, Problem 3E

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Chapter 17 Solutions

University Calculus: Early Transcendentals (4th Edition)

Ch. 17.1 - In Exercises 1−30, find the general solution of...Ch. 17.1 - In Exercises 1 – 30, find the general solution of...Ch. 17.1 - In Exercises 1 – 30, find the general solution of...Ch. 17.1 - Prob. 4E Ch. 17.1 - Prob. 5E Ch. 17.1 - Prob. 6E Ch. 17.1 - Prob. 7E Ch. 17.1 - Prob. 8E Ch. 17.1 - Prob. 9E Ch. 17.1 - Prob. 10E

Ch. 17.1 - Prob. 11E Ch. 17.1 - Prob. 12E Ch. 17.1 - Prob. 13E Ch. 17.1 - Prob. 14E Ch. 17.1 - In Exercises 1 – 30, find the general solution of...Ch. 17.1 - Prob. 16E Ch. 17.1 - Prob. 17E Ch. 17.1 - Prob. 18E Ch. 17.1 - Prob. 19E Ch. 17.1 - Prob. 20E Ch. 17.1 - Prob. 21E Ch. 17.1 - Prob. 22E Ch. 17.1 - Prob. 23E Ch. 17.1 - Prob. 24E Ch. 17.1 - Prob. 25E Ch. 17.1 - Prob. 26E Ch. 17.1 - Prob. 27E Ch. 17.1 - Prob. 28E Ch. 17.1 - Prob. 29E Ch. 17.1 - Prob. 30E Ch. 17.1 - Prob. 31E Ch. 17.1 - Prob. 32E Ch. 17.1 - Prob. 33E Ch. 17.1 - Prob. 34E Ch. 17.1 - Prob. 35E Ch. 17.1 - Prob. 36E Ch. 17.1 - Prob. 37E Ch. 17.1 - Prob. 38E Ch. 17.1 - Prob. 39E Ch. 17.1 - Prob. 40E Ch. 17.1 - Prob. 41E Ch. 17.1 - Prob. 42E Ch. 17.1 - Prob. 43E Ch. 17.1 - Prob. 44E Ch. 17.1 - Prob. 45E Ch. 17.1 - Prob. 46E Ch. 17.1 - In Exercises 41−55, find the general solution. 47....Ch. 17.1 - Prob. 48E Ch. 17.1 - Prob. 49E Ch. 17.1 - Prob. 50E Ch. 17.1 - Prob. 51E Ch. 17.1 - Prob. 52E Ch. 17.1 - Prob. 53E Ch. 17.1 - Prob. 54E Ch. 17.1 - Prob. 55E Ch. 17.1 - In Exercises 56−60, solve the initial value...Ch. 17.1 - Prob. 57E Ch. 17.1 - Prob. 58E Ch. 17.1 - Prob. 59E Ch. 17.1 - Prob. 60E Ch. 17.1 - Prob. 61E Ch. 17.1 - Prob. 62E Ch. 17.1 - Prob. 63E Ch. 17.1 - Prob. 64E Ch. 17.1 - Prob. 65E Ch. 17.1 - Show that if a, b, and c are positive constants,...Ch. 17.2 - Prob. 1E Ch. 17.2 - Prob. 2E Ch. 17.2 - Prob. 3E Ch. 17.2 - Prob. 4E Ch. 17.2 - Prob. 5E Ch. 17.2 - Prob. 6E Ch. 17.2 - Prob. 7E Ch. 17.2 - Prob. 8E Ch. 17.2 - Prob. 9E Ch. 17.2 - Prob. 10E Ch. 17.2 - Prob. 11E Ch. 17.2 - Prob. 12E Ch. 17.2 - Prob. 13E Ch. 17.2 - Prob. 14E Ch. 17.2 - Prob. 15E Ch. 17.2 - Prob. 16E Ch. 17.2 - Prob. 17E Ch. 17.2 - Prob. 18E Ch. 17.2 - Prob. 19E Ch. 17.2 - Prob. 20E Ch. 17.2 - Prob. 21E Ch. 17.2 - Prob. 22E Ch. 17.2 - Prob. 23E Ch. 17.2 - Prob. 24E Ch. 17.2 - Prob. 25E Ch. 17.2 - Prob. 26E Ch. 17.2 - Prob. 27E Ch. 17.2 - Prob. 28E Ch. 17.2 - Prob. 29E Ch. 17.2 - Prob. 30E Ch. 17.2 - Prob. 31E Ch. 17.2 - Prob. 32E Ch. 17.2 - Prob. 33E Ch. 17.2 - Prob. 34E Ch. 17.2 - Prob. 35E Ch. 17.2 - Prob. 36E Ch. 17.2 - Prob. 37E Ch. 17.2 - Prob. 38E Ch. 17.2 - Prob. 39E Ch. 17.2 - Prob. 40E Ch. 17.2 - Prob. 41E Ch. 17.2 - Prob. 42E Ch. 17.2 - Prob. 43E Ch. 17.2 - Prob. 44E Ch. 17.2 - Prob. 45E Ch. 17.2 - Prob. 46E Ch. 17.2 - Prob. 47E Ch. 17.2 - Prob. 48E Ch. 17.2 - Prob. 49E Ch. 17.2 - Prob. 50E Ch. 17.2 - Prob. 51E Ch. 17.2 - Prob. 52E Ch. 17.2 - Prob. 53E Ch. 17.2 - Prob. 54E Ch. 17.2 - Prob. 55E Ch. 17.2 - Prob. 56E Ch. 17.2 - Prob. 57E Ch. 17.2 - Prob. 58E Ch. 17.2 - Prob. 59E Ch. 17.2 - Prob. 60E Ch. 17.3 - Prob. 1E Ch. 17.3 - Prob. 2E Ch. 17.3 - Prob. 3E Ch. 17.3 - Prob. 4E Ch. 17.3 - Prob. 5E Ch. 17.3 - Prob. 6E Ch. 17.3 - Prob. 7E Ch. 17.3 - Prob. 8E Ch. 17.3 - Prob. 9E Ch. 17.3 - Prob. 10E Ch. 17.3 - Prob. 11E Ch. 17.3 - Prob. 12E Ch. 17.3 - Prob. 13E Ch. 17.3 - Prob. 14E Ch. 17.3 - Prob. 15E Ch. 17.3 - Prob. 16E Ch. 17.3 - Prob. 17E Ch. 17.3 - Prob. 18E Ch. 17.3 - Prob. 19E Ch. 17.3 - Prob. 20E Ch. 17.3 - Prob. 21E Ch. 17.3 - Prob. 22E Ch. 17.3 - Prob. 23E Ch. 17.3 - Prob. 24E Ch. 17.3 - Prob. 25E Ch. 17.3 - Prob. 26E Ch. 17.4 - In Exercises 1–24, find the general solution to...Ch. 17.4 - Prob. 2E Ch. 17.4 - Prob. 3E Ch. 17.4 - Prob. 4E Ch. 17.4 - Prob. 5E Ch. 17.4 - Prob. 6E Ch. 17.4 - Prob. 7E Ch. 17.4 - Prob. 8E Ch. 17.4 - Prob. 9E Ch. 17.4 - Prob. 10E Ch. 17.4 - Prob. 11E Ch. 17.4 - Prob. 12E Ch. 17.4 - Prob. 13E Ch. 17.4 - Prob. 14E Ch. 17.4 - Prob. 15E Ch. 17.4 - Prob. 16E Ch. 17.4 - Prob. 17E Ch. 17.4 - Prob. 18E Ch. 17.4 - Prob. 19E Ch. 17.4 - Prob. 20E Ch. 17.4 - Prob. 21E Ch. 17.4 - Prob. 22E Ch. 17.4 - In Exercises 1–24, find the general solution to...Ch. 17.4 - Prob. 24E Ch. 17.4 - Prob. 25E Ch. 17.4 - Prob. 26E Ch. 17.4 - Prob. 27E Ch. 17.4 - Prob. 28E Ch. 17.4 - Prob. 29E Ch. 17.4 - Prob. 30E Ch. 17.5 - Prob. 1E Ch. 17.5 - Prob. 2E Ch. 17.5 - Prob. 3E Ch. 17.5 - Prob. 4E Ch. 17.5 - Prob. 5E Ch. 17.5 - Prob. 6E Ch. 17.5 - Prob. 7E Ch. 17.5 - Prob. 8E Ch. 17.5 - Prob. 9E Ch. 17.5 - Prob. 10E Ch. 17.5 - Prob. 11E Ch. 17.5 - Prob. 12E Ch. 17.5 - Prob. 13E Ch. 17.5 - Prob. 14E Ch. 17.5 - Prob. 15E Ch. 17.5 - Prob. 16E Ch. 17.5 - Prob. 17E Ch. 17.5 - Prob. 18E

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The general solution of the differential equation y ″ + 2 y ′ + y = 0 by using power series.

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Chapter 17 Solutions