Explanation Given: The differential equation is sec x y ′ ( x ) = y 3 , y ( 0 ) = 3 . Calculation: Separate the x and y variables from the given equation. sec x y ′ ( x ) = y 3 sec x d y d x = y 3 d y y 3 = d x sec x y − 3 d y = cos x d x Therefore, the given equation is separable. Solve the equation y − 3 d y = cos x d x for y . Integrate both side of the equation. ∫ y − 3 d y = ∫ cos x d x y − 2 2 = sin x + c 1 y 2 = 2 sin x + C That is, the solution is y = ± ( − 2 sin x + C ) − 1 2 (1) Substitute the value of y ( 0 ) = 3 in above equation to evaluate the value of constant C

Calculus: Early Transcendentals, 2nd Edition

2nd Edition

ISBN: 9780321965165

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

Publisher: PEARSON

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Chapter D1.3, Problem 24E

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To determine: Whether the given equation are separable or not and to find the solution it is separable.

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Chapter D1.3, Problem 23E

Chapter D1.3, Problem 25E

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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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T o determine: Whether the given equation are separable or not and to find the solution it is separable.

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