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: 101


(a)$12$ (b) $\begin{cases} 12&y(0)=12\\ \infty&y(0)>12\\ -\infty&y(0)<12 \end{cases}$ (c) $ -3$

Work Step by Step

(a) Since \begin{align*} \frac{d y}{d t}&=-4(y-12)\\ \int \frac{dy}{y-12} &= \int -4dt\\ \ln|y-12|&= -4t +c\\ y-12&= e^{-4t+c}\\ y&=12+ C e^{-4t} \end{align*} Then \begin{align*} \lim _{t\to \infty }y&=\lim _{t\to \infty }(12+ C e^{-4t})\\ &=12 \end{align*} (b) Since \begin{align*} \frac{d y}{d t}&=4(y-12)\\ \int \frac{dy}{y-12} &= \int 4dt\\ \ln|y-12|&= 4t +c\\ y-12&= e^{4t+c}\\ y&=12+ C e^{4t} \end{align*} Then \begin{align*} \lim _{t\to \infty }y&=\lim _{t\to \infty }(12+ C e^{4t})\\ &=\begin{cases} 12&y(0)=12\\ \infty&y(0)>12\\ -\infty&y(0)<12 \end{cases} \end{align*} (c) Since \begin{align*} \frac{d y}{d t}&=-4y-12\\ \int \frac{dy}{y+3} &= \int -4dt\\ \ln|y+3|&= -4t +c\\ y+3&= e^{-4t+c}\\ y&=-3+ C e^{-4t} \end{align*} Then \begin{align*} \lim _{t\to \infty }y&=\lim _{t\to \infty }(-3+ C e^{-4t} )\\ &=-3 \end{align*}
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