1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Calculus 2 - Introduction to Differential Equations

  1. Sep 17, 2011 #1
    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. jcsd
  3. Sep 17, 2011 #2

    dynamicsolo

    User Avatar
    Homework Helper

    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 .]
     
  4. Sep 17, 2011 #3
    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
     
  5. Sep 17, 2011 #4

    HallsofIvy

    User Avatar
    Science Advisor

    No, that is not correct. Why don't you precisely what you got when you integrated, before solving for y.
     
  6. Sep 17, 2011 #5

    dynamicsolo

    User Avatar
    Homework Helper

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

    [tex]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})} , [/tex]

    where [itex]e^{c} [/itex] 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 [itex] y = \frac{1}{A \cdot (1 + e^{-kt})} [/itex] , which has two horizontal asymptotes, one at y = 0 , the other at y = 1 .
     
    Last edited: Sep 17, 2011
  7. Sep 17, 2011 #6
    Thank you very much for your help.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook