Originally Posted by

**terrorsquid** I have the following question from a past paper which I seemed to be rusty on and just wanted to go through it:

A virus spreads slowly through a population of animals. Initially no animals are infected, but after 10 days 25% of the population in infected. Let $\displaystyle P(t)$ denote the proportion of the population that is infected. Then

$\displaystyle \frac{dP}{dt} = k(1-P)$

(i) Solve the differential equation to get an expression for P(t).

(ii) Find the value of the constant k.

(iii) How long will it take for 75% of the population to become infected?

My attempt:

(i)

$\displaystyle \integral \frac{dP}{dt} = \integral k(1-P)$

$\displaystyle dP=k(1-P)~dt$

$\displaystyle \int \frac{1}{1-P}~dP = \int k~dt$

$\displaystyle -ln|1-P| = kt~dt$

$\displaystyle -1+P = e^{kt}$

$\displaystyle P(t) = e^{kt} +1$

This seems wrong as at $\displaystyle P(0)$, no animals should be infected, but the equation I derived gives a value of 2. What have I done wrong?

Also, why don't I have a coefficient in front of the $\displaystyle e^{kt}$ term? I have also never seen a constant next to the normal $\displaystyle P_0e^{kt}$ term; is that normal?

I will wait till we have figured out part (i) before I attempt the next two parts.

Thanks.