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: Differential Equation Separable Variables

  1. May 17, 2010 #1
    Hi all, I have a first order separable differential equation that I find a little difficult to solve. The question is from an old maths exam paper from my country. The book I obtained it from only has the answer, but I'd really like to know how to obtain the correct general solution. Okay, here is the question :

    1. A large population of N individuals contains, at time t = 0, just one individual with a contagious disease. Assume that the spread of the disease is governed by the equation

    [tex]\frac{dN}{dt} = k (N-n)n[/tex]



    where [tex] n(t) [/tex] is the the number of infected individuals after a time t days and k is a constant.


    a) Find n explicitly as a function of t.

    Part a) gives me a hint telling me to rewrite [tex]\frac{1}{[(N-n)n]}[/tex] as the partial fractions [tex]\frac{1}{N}(\frac{1}{N-n} + \frac{1}{n})[/tex]. I manage to work this out, and I obtain this :

    [tex]\int \frac{1}{N}(\frac{1}{N-n} + \frac{1}{n})dN = \int kdt[/tex]

    But I got stuck here as I'm not sure how to go about this as all attempts give me more than one form of n, one of n and one of e to the power of n. I had a look at the back of the book and the answer gives :

    [tex]n = \frac{Ne^{Nkt}}{N - 1 + e^{Nkt}}[/tex]

    So this gave me some help, so in the end I had this :

    [tex]\int (\frac{1}{N-n} + \frac{1}{n})dN = \int Nkdt[/tex]

    But I am not sure how to get just one n term because I don't get anywhere near the result that I am looking for when I integrate both sides.

    I appreciate any helpful advice on this. Thanks in advance for any correspondence.
     
  2. jcsd
  3. May 17, 2010 #2

    Cyosis

    User Avatar
    Homework Helper

    First I assume [tex]\frac{dN}{dt} = k (N-n)n[/tex] should be [tex]\frac{dn}{dt} = k (N-n)n[/tex]

    What is N, and why are you differentiating and integrating over N? From your problem statement N seems to be a constant.

    [tex]
    \int (\frac{1}{N-n} + \frac{1}{n})dN = \int Nkdt
    [/tex]



    You are making a mess when it comes to N and n. One is a constant the other is a function.
     
    Last edited: May 17, 2010
  4. May 17, 2010 #3
    N is the number of individuals in the population.

    No, I took that right from the book, this is what is confusing me, should I assume the book is wrong and it should be [tex]\frac{dn}{dt}[/tex]?
     
  5. May 17, 2010 #4

    Mark44

    Staff: Mentor

    I believe that the original differential equation should be
    [tex]\frac{dn}{dt} = k (N-n)n[/tex]

    In this equation N and k are constants, and n = n(t) is the number of infected individuals.
     
  6. May 17, 2010 #5
    Thanks Mark and Cyosis, you two have been a terrific help! I did what you both suggested and used [tex]\frac{dn}{dt}[/tex] this time, and it gave me the correct answer. I really appreciate the help, I was so frustrated that I couldn't solve this question but now that it has been explained to me that N is a constant it makes so much sense. I'm away to change the case of that n now :smile:, Thanks again.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook