Explanation Given: The differential equation is y ′ ( x ) = x + 1 y + 4 with initial condition y ( 3 ) = 5 . Calculation: Separate the x and y variables, y ′ ( x ) = x + 1 y + 4 d y d x = x + 1 y + 4 y + 4 d y = x + 1 d x Integrate both side of the equation. ∫ y + 4 d y = ∫ x + 1 d x ∫ ( y + 4 ) 1 2 d y = ∫ ( x + 1 ) 1 2 d x 2 3 ( y + 4 ) 3 2 = 2 3 ( x + 1 ) 3 2 + c (1) Substitute the value y ( 3 ) = 5 to evaluate the value of constant c . 2 3 ( 5 + 4 ) 3 2 = 2 3 ( 3 + 1 ) 3 2 + c 2 3 ( 9 ) 3 2 = 2 3 ( 4 ) 3 2 + c 18 = 16 3 + c c = 18 − 16 3 Thus, the value of c = 38 3

Calculus: Early Transcendentals, 2nd Edition

2nd Edition

ISBN: 9780321965165

Author: William L. Briggs, Lyle Cochran, Bernard Gillett

Publisher: PEARSON

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Differential Equations

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Chapter D1.3, Problem 31E

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Chapter D1.3, Problem 30E

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Calculus: Early Transcendentals, 2nd Edition

Ch. D1.1 - Prob. 1E Ch. D1.1 - Prob. 2E Ch. D1.1 - Prob. 3E Ch. D1.1 - If the general solution of a differential equation...Ch. D1.1 - Does the function y(t) = 2t satisfy the...Ch. D1.1 - Does the function y(t) = 6e3t satisfy the initial...Ch. D1.1 - Verifying general solutions Verify that the given...Ch. D1.1 - Verifying general solutions Verify that the given...Ch. D1.1 - Verifying general solutions Verify that the given...Ch. D1.1 - Verifying general solutions Verify that the given...

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