# 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.

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!!!

• thanks! that makes a lot of sense! was looking at it completely wrong i think
– Lucy
Commented Dec 4, 2013 at 16:33
• Glad to help out! If you like my answer enough, could you consider "accepting" it? (the green "check" mark). Commented Dec 4, 2013 at 17:22
• @ Lucy Ferrabee: thanks for the "acceptance"! Commented Dec 4, 2013 at 18:12