Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - Chapter Review Exercises - Page 388: 99

Answer

See details below.

Work Step by Step

The general solution of $y'=-2 y+8 =-2(y-4)$ is $$y=4+c e^{-2t}.$$ When $y(0)=3$, then $3=4+c$, i.e. $c=3-4=-1$. In this case $$y=4- e^{-2t}.$$ When $y(0)=4$, then $4=4+c$, i.e. $c=0$. In this case $$y=4.$$ See the graphs below.
Small 1575116861
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.