## Thomas' Calculus 13th Edition

$c'(t)=0$; 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.