# First Order D.E

1. Jun 12, 2015

### AAO

1. The problem statement, all variables and given/known data

A carrier of an infectious disease joins a herd of 500 initially uninfected cattle. At any
instant in time, the rate at which the disease spreads dx/dt is known to be proportional to
the product of:
(i) the number of infected cattle x(t); and
(ii) the number of uninfected cattle.
If the number of cattle infected after 4 days is 50, how many will have been infected after
6 days?

2. Relevant equations

3. The attempt at a solution
I interpreted the problem as the following D.E:
dx/dt=k*x*(500-x); where k is a constant to be determined

And the initial condition: x(0)=0
And the given information: x(4)=50

The equation looks to be separable:
dx/(500*k*x-k*x^(2)) = dt

but I cannot integrate the left hand side.

Is my interpreted D.E correct? If yes how to integrate the left hand side?

Appreciate your help.

Last edited: Jun 12, 2015
2. Jun 12, 2015

### Orodruin

Staff Emeritus
Your initial condition here is saying that 500 kettle are infected at t=0 ...

3. Jun 12, 2015

### AAO

Thanks I corrected it. Any hints about the solution to this DE?

I tried to use Matlab:
>> syms x(t) k
>> x(t) = dsolve(diff(x,t) == k*x(t)*(500-x(t)) , x(0)==0)

But I got
x(t) = 0

!!

4. Jun 12, 2015

### Orodruin

Staff Emeritus
Try splitting the integrand into two by partial fraction decomposition.

Edit: Also, your initial condition is not x(0) = 0. Obviously, if there are no infected cattle, there will be no spread of the disease.

5. Jun 12, 2015

### RUber

How many infected cattle are there at t=0?
If it is zero, transmission rate will surely be zero.

6. Jun 12, 2015

### Ray Vickson

x(0) is not zero. The problem said that one infected cow joined the others, so (presumably), x(0) = 1.

7. Jun 13, 2015

### AAO

Many Thanks All for your great help.

I managed to get the answer assuming that x(0)=1, and performing the integration using partial fraction.

The final answer is:
500
---------------------------
exp(log(499) - t) + 1

I also confirmed this using Matlab.

Once again, thank you very much.

8. Jun 13, 2015

### LCKurtz

Shouldn't your original DE have been$$\frac {dx}{dt}= kx(501-x),~x(0) = 1$$

9. Jun 13, 2015

### AAO

I believe you are right LCKurtz, I should have done this. Thanks a lot for the hint.

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