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, Books A La Carte Edition (3rd Edition)

3rd Edition

ISBN: 9780134770512

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

Publisher: PEARSON

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Chapter 9.4, Problem 36E

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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 9.4, Problem 35E

Chapter 9.4, Problem 37E

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Chapter 9 Solutions

Calculus: Early Transcendentals, Books A La Carte Edition (3rd Edition)

Ch. 9.1 - What are the orders of the equations in Example 2?...Ch. 9.1 - What is the solution of the initial value problem...Ch. 9.1 - Solve the initial value problem in Example 4a with...Ch. 9.1 - Suppose the initial conditions in Example 5a are...Ch. 9.1 - In Example 7, if the height function were given by...Ch. 9.1 - Consider the differential equation y(t) + 9y(t) =...Ch. 9.1 - If the general solution of a differential equation...Ch. 9.1 - Does the function y(t) = 2t satisfy the...Ch. 9.1 - Does the function y(t) = 6e3t satisfy the initial...Ch. 9.1 - The solution to the initial value problem y(t) = 2...

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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.