# Calculus 2 - Introduction to Differential Equations

1. Sep 17, 2011

### GreenPrint

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

One model for the spread of rumor is that the rate of spread is proportional to the product of the fraction y of the population who have heard the rumor and the fraction who have not heard the rumor.
i) Write a differential equation that is satisfied by y
ii) Solve the differential equation.

2. Relevant equations

3. The attempt at a solution

i) dy/dt = ky(1-y)
ii) I was having trouble solving this equation
When I took the integral I got
ln| (y^2)/(y - y^2) | + c = kt
I checked that ln| (y^2)/(y - y^2) | + c was the correct integral of 1/(y-y^2) with wolfram alpha and it was
I than began to solve for y and got a quadratic equation but than stopped solving because I checked the answer key...

"Logistic Eqn dp/dt = kp(1-p/k) so k=1, P=y
Now the solution to (i) is y(t) = k/(1 + A e^(-kt))
y(t) = 1/(1+Ae^(-kt))
At t = 0 y_0 = 1/(1+A) => 1+ A = 1/y_0
A = 1/y_0 - 1
y = 1/(1+(1/y_0 - 1)e^(-kt) ) = y_0/(y_0 + (1-y_0)e^(-kt))"

I don't understand what this equation is y(t) = k/(1 + A e^(-kt)) and have never seen it before and don't think that if I solved the equation ln| (y^2)/(y - y^2) | + c = kt I would of gotten this... I don't understand... also y_0 is y sub zero or y at time equals zero...

thanks for any help!

Last edited: Sep 17, 2011
2. Sep 17, 2011

### dynamicsolo

Since you presumably used the "method of partial fractions", it would be easier to deal with this if you wrote ln| y/(1 - y) | + c = kt .

You now want to exponentiate both sides, that is, make each side a power of e , and then solve for y . You will not have a quadratic equation, but you do have something which will yield to algebraic manipulation...

["Wolfram" is solving the initial-value problem with y = y0 at t = 0 , which lets you replace c .]

3. Sep 17, 2011

### GreenPrint

So when I did so I got

y(t) = e^(kt)/(C+e^(kt))
Is this correct
y(0) = 1/(C+1)
I don't see how I am suppose to solve for C in any way

4. Sep 17, 2011

### HallsofIvy

Staff Emeritus
No, that is not correct. Why don't you precisely what you got when you integrated, before solving for y.

5. Sep 17, 2011

### dynamicsolo

Exponentiating ln| y/(1 - y) | + c = kt should give

$$e^{ln |\frac{y}{1-y}| + c} = e^{kt} \Rightarrow \frac{y}{1-y} \cdot e^{c} = e^{kt} \Rightarrow y = \frac{e^{kt}}{A \cdot (1 + e^{kt})} ,$$

where $e^{c}$ is conventionally replaced by an arbitrary multiplicative constant A . Since you are finding the "general solution" to the differential equation, A is unspecified. This "logistic function" is more usually written as $y = \frac{1}{A \cdot (1 + e^{-kt})}$ , which has two horizontal asymptotes, one at y = 0 , the other at y = 1 .

Last edited: Sep 17, 2011
6. Sep 17, 2011

### GreenPrint

Thank you very much for your help.