Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 6 - Differential Equations - Review Exercises - Page 432: 46

Answer

$$y = \frac{8}{{1 + \left( {5/3} \right){e^{ - 1.76t}}}}$$

Work Step by Step

$$\eqalign{ & \frac{{dy}}{{dt}} = 1.76y\left( {1 - \frac{y}{8}} \right),{\text{ }}\left( {0,3} \right) \cr & {\text{The differential equation is in the form }}\frac{{dy}}{{dt}} = ky\left( {1 - \frac{y}{L}} \right) \cr & \underbrace {\frac{{dy}}{{dt}} = 1.76y\left( {1 - \frac{y}{8}} \right)}_{\frac{{dy}}{{dt}} = ky\left( {1 - \frac{y}{L}} \right)} \to k = 1.76,{\text{ }}L = 8 \cr & {\text{The general solution is }}y = \frac{L}{{1 + b{e^{ - kt}}}}{\text{ }}\left( {{\text{Example 6, page 419}}} \right) \cr & {\text{Therefore, substituting }}k = 1.76{\text{ and }}L = 8 \cr & \underbrace {y = \frac{L}{{1 + b{e^{ - kt}}}}}_ \downarrow \cr & y = \frac{8}{{1 + b{e^{ - 1.76t}}}}{\text{ }}\left( {\bf{1}} \right) \cr & {\text{Using the initial condition }}\left( {0,3} \right) \cr & 3 = \frac{8}{{1 + b{e^0}}} \cr & b = \frac{5}{3} \cr & {\text{Substituting }}b{\text{ into }}\left( {\bf{1}} \right) \cr & y = \frac{8}{{1 + \left( {5/3} \right){e^{ - 1.76t}}}} \cr} $$
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