Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.6 Models Involving y'=k(y-b) - Exercises - Page 359: 2

Answer

See the details below.

Work Step by Step

Since $y=12+Ce^{-3t}$, then by differentiation, we have $$y'=-3Ce^{-3t}=-3(y-12).$$ Now, when $y(0)=20$, then $20=12+C$, i.e. $C=20-12=8$. In this case $$y=12+8 e^{-3t}.$$ When $y(0)=0$, then $0=12+C$, i.e. $C=0-12=-12$. In this case $$y=12-12e^{-3t}.$$ See the graphs below.
Small 1574093665
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.