Answer
$c'(t)=0$
This implies that $c(t)$ is constant.
Work Step by Step
Here, we have $\dfrac{dx}{dt}=(a-by) x$ and $\dfrac{dy}{dt}=(-c+dx) x$
By taking the derivative of both sides of the given equation, we get
$c'(t)=(\dfrac{a}{y(t)}-b) y'(t)-(\dfrac{x}{x(t)}-d) x'(t)$
or,
$c'(t)=(a-by)(-c+dx) +(c-dx) (a-by) $
or, $c'(t)=(a-by)(-c+dx) -(-c+dx) (a-by) $
Hence, $c'(t)=0$
This implies that $c(t)$ is constant.
