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Homework Help: Integrating dx/dt=k(a-x)(b-x)

  1. Aug 23, 2013 #1
    1. The problem statement, all variables and given/known data

    I need to integrate dx/dt=k(a-x)(b-x) and then rearrange to find x

    3. The attempt at a solution

    1/((a-x)(b-x))dx = k dt

    Integrating dx using partial fractions:

    when t=0 x=0

    then when I rearrange I get:

    Then I change the x to y and t to x so that I can graph it, but it doesn't give me the curve that I want, I have attached the graph. I am was assuming that it would give something like the black curve rather than the red one that it gave. Am I doing something wrong?
    Last edited: Aug 28, 2013
  2. jcsd
  3. Aug 24, 2013 #2
    What are the values of a,b, and k?
  4. Aug 24, 2013 #3
    positive integers
  5. Aug 24, 2013 #4
    Hey jhast1,

    I tried solving your equation, and when you say
    "Integrating dx using partial fractions: 1/(b-a)×(ln((a-x)/(b-x))=kt+c"

    I've got something which looks more like:


    But maybe I just made a mistake..
  6. Aug 24, 2013 #5
    Here is my working.
    Last edited: Aug 28, 2013
  7. Aug 24, 2013 #6


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    This is nearly correct, but you have the argument of the log flipped. From looking at your attached work, it appears you integrated dx/(a-x) and dx/(b-x) incorrectly and this will be the source of your error.

    Your work assumes a≠b and this is not stated, however, given the set up of the question (i.e it is not written as k(a-x)2) then I suppose you can assume this is not the case. Were there any conditions on a and b given after the question?
  8. Aug 24, 2013 #7
    ok so I did that part again and i got:

    Then I made x the subject and got this but I'm still not sure if it is right.


    where c=ln(a/b)/(b-a)

    Edit: a and b are not equal
    Last edited: Aug 24, 2013
  9. Aug 24, 2013 #8
    Looks like I get [tex]x=\frac { a{ e }^{ r }-b }{ { e }^{ r }-1 } \\ r=(b-a)(kt+c)[/tex]
  10. Aug 24, 2013 #9


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    Where is this initial condition specified?

    I think there are two small errors above - you missed a coefficient in front of the exp term and there is a stray minus.
  11. Aug 24, 2013 #10
    In the original post. t=0 when x=0.
  12. Aug 24, 2013 #11
    The conditions are given in the question I have:
    a is the amount of substance A and b is the amount of substance B, at t=0. x is the amount of product that these substances have produced at time t.

    I see where I missed those things, however this does still not give me the curve that I'm expecting. Unless it's supposed to give me this curve.
  13. Aug 24, 2013 #12


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    The statement 't=0 when x=0' appeared in the attempt at a solution section. Since it did not appear in the Problem statement section, I wondered where that initial condition came from.
  14. Aug 24, 2013 #13


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    I think you have a sign wrong.
    Shouldn't that be c=ln(b/a)/(b-a)? Substituting that in the (corrected) previous equation allows a lot of simplification.
    Btw, a check you can do on the answer is to let b tend to a and see if you get the right answer for the case b = a.
  15. Aug 24, 2013 #14


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    it should be 1/(a-b)×(ln((a-x)/(b-x))=kt+c

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