Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 9: First-Order Differential Equations - Section 9.5 - Systems of Equations and Phase Planes - Exercises 9.5 - Page 556: 12


$c'(t)=0$; and $c(t)$ is constant.

Work Step by Step

Since, $\dfrac{dx}{dt}=(a-by) x$ Also, $\dfrac{dy}{dt}=(-c+dx) x$ Now, we have $c'(t)=[\dfrac{a}{y(t)}-b] (y'(t))-[\dfrac{x}{x(t)}-d] (x'(t))$ This implies that $c'(t)=(a-by)(-c+dx) +(c-dx) (a-by) $ $\implies c'(t)=(a-by)(-c+dx) -(-c+dx) (a-by) $ Thus, we find that $c'(t)=0$ and $c(t)$ is constant.
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