Explanation Theorem used: “Let the functions M , N , M y , and N x , be continuous in the rectangular region R : α < x < β , γ < y < δ . Then M ( x , y ) + N ( x , y ) y ′ = 0 is an exact differential equation in R if and only if M y ( x , y ) = N x ( x , y ) at each point of R ”. Calculation: The given equation is y + ( 2 x − y e y ) y ′ = 0 . Here, it is observed that M = y and N = 2 x − y e y . Differentiate M = y with respect to y . M y = d d y ( y ) = 1 Differentiate N = 2 x − y e y with respect to x . N x = d d x ( 2 x − y e y ) = 2 Since M y ≠ N x , it can be concluded that the given equation is not exact. Multiply the equation y + ( 2 x − y e y ) y ′ = 0 by the integrating factor y . ( y + ( 2 x − y e y ) y ′ ) y = 0 y 2 + ( 2 x y − y 2 e y ) y ′ = 0 Here, it is observed that M = y 2 and N = 2 x y − y 2 e y . Differentiate M = y 2 with respect to y . M y = d d y ( y 2 ) = 2 y Differentiate N = 2 x y − y 2 e y with respect to x . N x = d d x ( 2 x y − y 2 e y ) = 2 y By the theorem stated above, the equation is exact. There is a function ψ ( x , y ) such that ψ x ( x , y ) = y 2 and ψ y ( x , y ) = 2 x y − y 2 e y

10th Edition

ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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Chapter 2.6, Problem 21P

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To show: The equation y+(2x−yey)y′=0 is not exact but it becomes exact when the multiplied by the integrating factor μ(x,y)=y and then solve it.

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Chapter 2.6, Problem 20P

Chapter 2.6, Problem 22P

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Use the Euclidean algorithm to find two sets of integers (a, b, c) such that 55a65b+143c: Solution = 1. By the Euclidean algorithm, we have: 143 = 2.65 + 13 and 65 = 5.13, so 13 = 143 – 2.65. - Also, 55 = 4.13+3, 13 = 4.3 + 1 and 3 = 3.1, so 1 = 13 — 4.3 = 13 — 4(55 – 4.13) = 17.13 – 4.55. Combining these, we have: 1 = 17(143 – 2.65) - 4.55 = −4.55 - 34.65 + 17.143, so we can take a = − −4, b = −34, c = 17. By carrying out the division algorithm in other ways, we obtain different solutions, such as 19.55 23.65 +7.143, so a = = 9, b -23, c = 7. = = how ? come [Note that 13.55 + 11.65 - 10.143 0, so we can obtain new solutions by adding multiples of this equation, or similar equations.]

- Let n = 7, let p = 23 and let S be the set of least positive residues mod p of the first (p − 1)/2 multiple of n, i.e. n mod p, 2n mod p, ..., p-1 2 -n mod p. Let T be the subset of S consisting of those residues which exceed p/2. Find the set T, and hence compute the Legendre symbol (7|23). 23 32 how come? The first 11 multiples of 7 reduced mod 23 are 7, 14, 21, 5, 12, 19, 3, 10, 17, 1, 8. The set T is the subset of these residues exceeding So T = {12, 14, 17, 19, 21}. By Gauss' lemma (Apostol Theorem 9.6), (7|23) = (−1)|T| = (−1)5 = −1.

Let n = 7, let p = 23 and let S be the set of least positive residues mod p of the first (p-1)/2 multiple of n, i.e. n mod p, 2n mod p, ..., 2 p-1 -n mod p. Let T be the subset of S consisting of those residues which exceed p/2. Find the set T, and hence compute the Legendre symbol (7|23). The first 11 multiples of 7 reduced mod 23 are 7, 14, 21, 5, 12, 19, 3, 10, 17, 1, 8. 23 The set T is the subset of these residues exceeding 2° So T = {12, 14, 17, 19, 21}. By Gauss' lemma (Apostol Theorem 9.6), (7|23) = (−1)|T| = (−1)5 = −1. how come?

Chapter 2 Solutions

Elementary Differential Equations

Ch. 2.1 - In each of Problems 1 through 12: Draw a direction...Ch. 2.1 - In each of Problems 1 through 12: Draw a direction...Ch. 2.1 - Prob. 3P Ch. 2.1 - Prob. 4P Ch. 2.1 - Prob. 5P Ch. 2.1 - Prob. 6P Ch. 2.1 - In each of Problems 1 through 12: Draw a direction...Ch. 2.1 - Prob. 8P Ch. 2.1 - Prob. 9P Ch. 2.1 - Prob. 10P

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The equation y + ( 2 x − y e y ) y ′ = 0 is not exact but it becomes exact when the multiplied by the integrating factor μ ( x , y ) = y and then solve it.

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To show: The equation y+(2x−yey)y′=0 is not exact but it becomes exact when the multiplied by the integrating factor μ(x,y)=y and then solve it.

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