system of differential equations for spread of infection query I have been given a set of equations describing the spread of infection in a population:
$${\frac{dS}{dt}}=-aIS+bI+c-cS$$
$${\frac{dI}{dt}}=aIS-bI-cI$$
where S and I are susceptible and infected populations, respectively, and a,b,c are positive constants.
I want to derive a differential equation for $Q(t)=S(t)+I(t)$
I started by assuming ${\frac{dQ}{dt}}={\frac{d}{dt}}(S+I)={\frac{dS}{dt}}+{\frac{dI}{dt}}=c-cS-cI$
hence, Q(t) would equal the integral of (c-cS-cI) with respect to t, but I'm not sure how to do this and even if it's right. Am I on the right track or is this completely wrong? I need to solve the equation I get for Q(t) to prove that Q(t) tends to 1 as t tends to infinity. Any help would be much appreciated.
 A: Your differential equation is OK, and in terms of $Q$ it looks like
$dQ / dt = c - cQ, \tag{1}$
and it's legitimate to write
$Q(t) - Q(t_0) = \int_{t_0}^t (c -cQ(s))ds, \tag{2}$
but trying to find $Q$ from (2) is not the easiest way to go, in my perhaps not-so-humble opinion.  Easier to write
$dQ / dt + cQ = c, \tag{3}$
and solve (3) for the simple, linear, first order differential equation it is.  One easy way to do this is to multiply (3) through by $e^{ct}$:
$e^{ct}dQ / dt + ce^{ct}Q = ce^{ct}, \tag{4}$
then observe that (4) may be written
$d / dt (e^{ct}Q(t)) = d / dt (e^{ct}), \tag{5}$
so if we integrate both sides we obtain
$\int_{t_0}^t (d / ds (e^{cs}Q(s))ds = \int_{t_0}^t (d / ds (e^{cs}))ds, \tag{6}$
or
$e^{ct}Q(t) - e^{ct_0}Q(t_0) = e^{ct} - e^{ct_0} \tag{7}$
which we readily solve for $Q(t)$:
$Q(t) = e^{-c(t - t_0)}(Q(t_0) - 1) + 1; \tag{8}$
it is easily seen that such $Q(t)$ takes the value $Q(t_0)$ at $t = t_0$ and, for $c > 0$, we have
$Q(t) \to 1 \; \text{as} \; t \to \infty, \tag{9}$
both of which are required properties of the solution.  The desired integral form (2) is also satisfied by such $Q(t)$.
Hope this helps!  Cheers,
and as always,
Fiat Lux!!!
