# Solving DE for the spread of a rumour.

• May 21st 2010, 05:32 AM
Rina
Solving DE for the spread of a rumour.
here is my problem and what I have done so far.
• May 21st 2010, 05:40 AM
HallsofIvy
The integral of $\displaystyle \frac{dx}{x(1- x)}$ is NOT "ln|x(1- x)|". First write it as "partial fractions": $\displaystyle \frac{1}{x(1-x)}= \frac{A}{x}+ \frac{B}{1- x}$, then integrate.
• May 21st 2010, 05:52 AM
Prove It
Quote:

Originally Posted by Rina
here is my problem and what I have done so far.

$\displaystyle \frac{dx}{dt} = x(1 - x)$

$\displaystyle \frac{dt}{dx} = \frac{1}{x(1 - x)}$.

Now using the method of partial fractions:

$\displaystyle \frac{A}{x} + \frac{B}{1 - x} = \frac{1}{x(1 - x)}$

$\displaystyle \frac{A(1 - x) + Bx}{x(1 - x)} = \frac{1}{x(1 - x)}$

$\displaystyle A(1 - x) + Bx = 1$

$\displaystyle A - Ax + Bx = 1$

$\displaystyle A + (B - A)x = 1 + 0x$.

Therefore $\displaystyle A = 1$ and $\displaystyle B - A = 0$, so $\displaystyle B = 1$.

Thus $\displaystyle \frac{1}{x(1 - x)} = \frac{1}{x} + \frac{1}{1 - x}$.

Back to the DE:

$\displaystyle \frac{dt}{dx} = \frac{1}{x(1 - x)}$

$\displaystyle \frac{dt}{dx} = \frac{1}{x}+ \frac{1}{1 - x}$

$\displaystyle t = \int{\left(\frac{1}{x} + \frac{1}{1 - x}\right)\,dx}$

$\displaystyle t = \ln{|x|} - \ln{|1 - x|} + C$

$\displaystyle t = \ln{\left|\frac{x}{1 - x}\right|} + C$

$\displaystyle t = \ln{\left|\frac{x - 1 + 1}{1 - x}\right|} + C$

$\displaystyle t = \ln{\left|\frac{-(1 - x)}{1 - x} + \frac{1}{1 - x}\right|} + C$

$\displaystyle t = \ln{\left|-1 + \frac{1}{1 - x}\right|} + C$

$\displaystyle t - C = \ln{\left|-1 + \frac{1}{1 - x}\right|}$

$\displaystyle e^{t - C} = \left|-1 + \frac{1}{1 - x}\right|$

$\displaystyle e^{-C}e^t = \left|-1 + \frac{1}{1 - x}\right|$

$\displaystyle \pm e^{-C}e^t = -1 + \frac{1}{1 - x}$

$\displaystyle A\,e^t = -1 + \frac{1}{1 - x}$, where $\displaystyle A = \pm e^{-C}$

$\displaystyle A\,e^t + 1 = \frac{1}{1 - x}$

$\displaystyle \frac{1}{A\,e^t + 1} = 1 - x$

$\displaystyle x = 1 - \frac{1}{A\,e^t + 1}$

$\displaystyle x = \frac{A\,e^t + 1 - 1}{A\,e^t + 1}$

$\displaystyle x = \frac{A\,e^t}{A\,e^t + 1}$.
• May 21st 2010, 05:58 AM
Rina
thank you.

and C is? .
• May 21st 2010, 06:50 AM
Prove It
Quote:

Originally Posted by Rina
the first A has nothing to do with a second A. It is confusing. should i have used a different letter for that constant at the end of the calculation.

They are the same $\displaystyle A$. Otherwise I would have used different letters.
• May 21st 2010, 06:54 AM
Rina
I am sorry that I am so stupit. It is not ez trust me

I am sorry, but I do not understand how did the first A, just a constant initially, all the sudden became +-e^(-c).
• May 21st 2010, 08:18 AM
Rina
If we use this DE solution, x=1/(1-e^(-t)) and try to find the proportion of the population that has heard the rumor at the time t=0. The solution gives us the result - 0.5; that is a half of population before the rumor has started spreading? It doesn't make sense. Am I interpreting it wrong?
• May 21st 2010, 02:56 PM
mr fantastic
Quote:

Originally Posted by Rina
I am sorry that I am so stupit. It is not ez trust me

I am sorry, but I do not understand how did the first A, just a constant initially, all the sudden became +-e^(-c).

C is arbitrary therefore -C is arbitrary therefore e^(-C) is arbitrary therefore it can be represneted by a new arbitrary symbol eg. A.
• May 21st 2010, 03:00 PM
mr fantastic
Quote:

Originally Posted by Rina
If we use this DE solution, x=1/(1-e^(-t)) and try to find the proportion of the population that has heard the rumor at the time t=0. The solution gives us the result - 0.5; that is a half of population before the rumor has started spreading? It doesn't make sense. Am I interpreting it wrong?

1. The given solution is x=1/(1+e^(-t)), not what you have said.

2. t = 0 => x = 1/2. All that means is that at t = 0 half the population have heard the rumour. Big deal.

3. The question asked you to show that the solution was x=1/(1+e^(-t)). So you don't actually have to solve the DE. Just substitute x=1/(1+e^(-t)) into it and show that the resulting left hand and right hand sides are equal to each other. The fact that you have been given no boundary condition suggests that this is the approach you were meant to take ... (And given what I have said in my second point, a possible boundary condition would have been the initial condition x(0) = 1/2).