## Calculus (3rd Edition)

Published by W. H. Freeman

# 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.

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