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A Differential Equation

  1. Jun 2, 2005 #1
    This is out of a P3 (OCR) text book so it should be dead easy but I just can't see the answer. I know it is not a simple ln(N(500-N)) because the differential is not on top. Anyway here it is:

    (5000)dN/dt=N(500-N)

    they also say when N = 100, dN/dt = 8 (which is obvious)

    the answer in the back of the book is:

    t=(10)ln(N/(500-N))+k

    Cheers
     
    Last edited: Jun 2, 2005
  2. jcsd
  3. Jun 2, 2005 #2
    Hmm, most people here won't know about the exam boards we have :smile:. Have you done anything on the question and if so, could you post it?

    Edit: what did you mean by the differential's not on top?

    I'm stuck trying to integrate [tex]\int \frac{dN}{N(500 - N)}[/tex]. Argh :/.
     
    Last edited: Jun 2, 2005
  4. Jun 2, 2005 #3
    The numerator

    This is a variable seperable right?

    so you take the N(500-N) to the left side, split the dN and the dt to get:

    int[ 1/{N(500-N)} ] dN = int[ 1/5000 ] dt

    Now by chain rule (I think), if the differential is on the top of the fraction you can just say it is the log of the denominator. eg int [1/x] dx = ln|x|+c

    However, the differential is not on the top in this case so I am screwed.
     
  5. Jun 2, 2005 #4

    shmoe

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    Partial fractions will work.
     
  6. Jun 2, 2005 #5
    oh yeah sometimes you can be so obsessed looking for something really complicated and miss something simple. It does work. Thanx
     
  7. Jun 2, 2005 #6
    break the numerator into 2 pieces. I don't remember how you say that in ze english, partial fraction decomposition or something.
     
  8. Jun 2, 2005 #7
    Don't worry sorted. You just needed to say partial fractions. Thanx anyway.
     
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