Explanation Verification: The differential equation is y ′ ( t ) = k y + b . Apply the condition k y + b < 0 on the given differential equation. That is, the differential equation is negative. Solve the above differential equation as follows, y ′ ( t ) = k y + b d y d t = k y + b d y = ( k y + b ) d t d y k y + b = d t Integrate on both sides of the above expression as follows, ∫ d y k y + b = ∫ 1 d t 1 k ln ( k y + b ) = t + c k × 1 k ln ( k y +<

Calculus: Early Transcendentals (2nd Edition)

2nd Edition

ISBN: 9780321947345

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

Publisher: PEARSON

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Chapter D1.4, Problem 32E

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To verify: The differential equation y′(t)=ky+b has a general solution y(t)=Cekt−bk

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Chapter D1.4, Problem 31E

Chapter D1.4, Problem 33E

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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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To verify: The differential equation y ′ ( t ) = k y + b has a general solution y ( t ) = C e k t − b k

Solution Summary: The author explains that the differential equation yprime(t)=ky+b has a general solution.

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To verify: The differential equation y′(t)=ky+b has a general solution y(t)=Cekt−bk

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