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Homework Help: Simple DE.

  1. Dec 1, 2009 #1
    1. The problem statement, all variables and given/known data
    The growth of bacteria is proportional to the number present but is being reduced at a constant rate for experimentation. Find the equation for N

    2. Relevant equations

    d(N)/dt = KN-R , N = number of bacteria, K = proportionality const., R = reduction rate.

    3. The attempt at a solution

    Above was my guess to what the differential equation should look like...it's a separable equation. My question: Is the above formula the right formula based on my understanding of the question? With that equation I can't seem to solve it.
  2. jcsd
  3. Dec 1, 2009 #2
    Seems reasonable. Try doing the regular thing, which is to separate t and N to different sides of the equation like this

    [tex] \frac{dN}{KN-R} = dt [/tex]

    and then integrate. You also need to assume that KN>R. What does this mean?
  4. Dec 1, 2009 #3
    I tried that but I just can't seem to get it right. I don't get the same solution as the book. By the way the question is No. 22 of Section 2 in Chapter 8 of Boas' Math Methods for those that have the book.

    I integrate those and get,

    (1/K)*ln(KN-R) = t + ln(C) , C is the initial number of bacteria...I put in ln(C) for simplification.
  5. Dec 1, 2009 #4


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    Okay, so you have ln(KN- R)= KT+ Kln(C) and from that [itex]KN-R= (Ce^K) e^{KT}[/itex]. [itex]KN= R+ (Ce^K)e^{KT}[/itex] and, finally, [itex]N= R/K+ (Ce^K/K)e^{KT}[/itex]. What is the answer in the book?
    Last edited by a moderator: Dec 2, 2009
  6. Dec 1, 2009 #5
    The answer in the book is,

    [itex] N = Ce^{Kt} - (R/K)(e^{Kt}-1) [/itex]
  7. Dec 2, 2009 #6
    So, how do you solve the DE?

    [tex] \int_{N(t_0)}^{N(t)} \frac{dN}{KN-R} = \int_{t_0}^t dt [/tex]

    where t_0 is arbitrary (choose zero)
  8. Dec 2, 2009 #7
    Thanks, I just got it. Can you tell me why I needed the limits? The other one before this didn't have the R (reduction) so the equation was just

    [tex] N = N_0e^{Kt} [/tex]

    When I was solving that I didn't need limits.
  9. Dec 2, 2009 #8
    The limits contain the constant of integration in a more transparent way; you don't need to be guessing the correct forms. You could have just as well derived the case R=0 with the limits, probably with less work too.
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