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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


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    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


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    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


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    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.
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