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 431: 26

Answer

$y=6.349e^{-0.462t}$

Work Step by Step

$(1, 4)$ lies on the curve. Thus, $$4 = Ce^{k}$$ $(4, 1)$ lies on the curve. Thus, $$1 = Ce^{4k}$$ Dividing this equation by eq 1 gives, $$e^{3k}=\frac{1}{4}$$ Or,$$ k=-\frac{ln(4)}{3}=-0.462$$ By eq 1 we have $$C=4e^{-k}=4e^{0.462}=6.349$$ So, the function is $$y=6.349e^{-0.462t}$$
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