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: 25



Work Step by Step

$(2, \frac{3}{2})$ lies on the curve. Thus, $$\frac{3}{2} = Ce^{2k}$$ $(4, 5)$ lies on the curve. Thus, $$5 = Ce^{4k}$$ Dividing eq 1. by eq 2. $$\frac{10}{3} = e^{2k}$$ $$k=\frac{ln(10/3)}{2}=0.602$$ Putting this in eq 1 gives, $$\frac{3}{2}=C(\frac{10}{3})$$ Or $$C=\frac{9}{20}=0.45$$ So, the function is $$y=0.45e^{0.602t}$$
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