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A non-fatal infectious disease divides a population into two groups: normal or ill. Assume that the average number of contacts that each ill individual has with normal individuals is $a$ multiply the size of the normal population (i.e. $aN$), and for each contact the probability of transmission is $k$. Let the average duration of the disease be $\frac{1}{c}$. Assume that the birth rate is $b$ for both normal and ill populations, and that all babies are born normal. The death rate (independent with disease) of normal group is $d_{n}$, and of ill group is $d_{i}$.

  1. Build differential equations describing the size of each group.
  2. Assuming the total population is constant, rewrite the differential equations to obtain an unique differential equation.
  3. Find the equilibria and determine if they are stable or not.

What I have done:

  1. $$ \ \frac{dN}{dt}=b(I+N)+cI-d_{n}N-aNIk$$ $$\frac{dI}{dt}=aNIk-cI-d_{i}I-bI$$ $I,N$ are the size of normal and ill

  2. Since the population is constant, we have $b(I+N)=d_{n}N+d_{i}I$, then we get $I=\frac{d_{n}-b}{b-d_{i}}N$.

Am I right? And I have no idea what "rewrite the differential equations to obtain an unique differential equation" means, and how can I get the unique equation?

Can anyone help me with this problem? Thanks very much!

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There is another way to express the idea of a constant population.

$$I+N=P_{total} \implies \frac{d}{dt}(I+N)=0$$

Using this, you can deduce $\frac{dI}{dt}=-\frac{dN}{dt}$ and $I=P_{total}-N$. Thus, you can rewrite $I$ in terms of $N$ (or $N$ in terms of $I$).

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  • $\begingroup$ I got it, thanks very much. And are there any wrong with my first two equations? $\endgroup$ – user133140 Feb 24 '15 at 5:55
  • $\begingroup$ @user133140 I'm not sure about the $cI(t)$ term. I would be tempted to replace it by $I(t-\frac{1}{c})$, which would turn this into a delay-differential equation. However, delay-differential equations are generally much harder to solve, so if this is an undergraduate question, your answer is probably good enough. $\endgroup$ – hasnohat Feb 24 '15 at 6:05
  • $\begingroup$ This is undergraduate question, thanks again, it's really helpful to me. $\endgroup$ – user133140 Feb 24 '15 at 6:13

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