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.

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.