#### Answer

$c'(t)=0$; 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.