First Order Differentiation

**shaojie2k** · October 20th 2006, 04:02 PM

First order Differentiation

An infection disease spreads at the rate which is proportional to the product of the number infected and the number uninfected. Initially one half of the population is infected and the rate of spread is such that, were it to remain constant, the whole population would become infected in 24 days. Calculate the proportion of the population which is infected after 12 days.

Appreciate all ur help. Thank You!!!

**CaptainBlack** · October 21st 2006, 12:33 AM

Originally Posted by shaojie2k

First order Differentiation

An infection disease spreads at the rate which is proportional to the product of the number infected and the number uninfected. Initially one half of the population is infected and the rate of spread is such that, were it to remain constant, the whole population would become infected in 24 days. Calculate the proportion of the population which is infected after 12 days.

Appreciate all ur help. Thank You!!!

Let $x$ be the number of infecteds, and $y$ the population. Then you are told the rate of increase in the number of infecteds is:

$ \frac{dx}{dt} = k\ x(y-x) $

You are also told that when $t=0$ that:

$x=y/2$

and:

$\left. \frac{dx}{dt} \right|_{t=0}=\frac{y/2}{24}=\frac{y}{48}$

Hence at $t=0$ we have:

$ \frac{y}{48} = k (y/2)(-y/2) $

or $k=-1/(12y)$

Because $y$ is a constant the DE is of variables seperable type so:

$ \int_{x=0}^{x(t)} \frac{1}{x(y-x)} dx = \int_{\tau=0}^t k\ d\tau$

so:

$ \frac{\ln (x)}{y} - \frac{\ln (x-y)}{y} = k\ t+const $

or rearranging and exponentiating:

$\left[ \frac{x}{x-y} \right] = A\ \exp(k \ yt)$

Now the condition that $x=y/2$ when $t=0$ allows us to determin that $A=-1$ .

Now putting in the value of $k$ we have:

$\left[ \frac{x}{x-y} \right] = - \exp(-t/12)$ .

Finally after 12 days we have:

$\left[ \frac{x}{x-y} \right] = - \exp(-1)$ .

Then if we let $p$ be the proportion infected we have:

$\left[ \frac{p}{p-1} \right] = - \exp(-1)$ ,

which we may solve for $p$ to get $p\approx 0.269$

RonL

**shaojie2k** · October 21st 2006, 01:07 AM

may i ask how is this step 2d81d9d0d22764abbae5966c7a22fa61-1.png derive ?

**CaptainBlack** · October 21st 2006, 01:19 AM

Originally Posted by shaojie2k

may i ask how is this step 2d81d9d0d22764abbae5966c7a22fa61-1.png derive ?

You know that initialy y/2 are infected so there remain another y/2
uninfected and at the rate of growth at t=0 the entire population will be
infected in 24 days. So the number that would have to be infected per day
to achive this (y/2) / 24, which is equal to the rate of new infections at
t=0.

RonL

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