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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


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    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.

    \int (\frac{1}{N-n} + \frac{1}{n})dN = \int Nkdt

    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


    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.
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