1. Not finding help here? Sign up for a free 30min 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!

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:
    1/(b-a)×(ln((a-x)/(b-x))=kt+c

    when t=0 x=0
    ∴c=(ln(a/b))/(b-a)

    then when I rearrange I get:
    x=(e^((b-a)(kt+c))×(a-b))/((e^((b-a)(kt+c))-1)

    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:

    [itex]\frac{ln(x-a)-ln(x-b)}{(a-b)}=kt[/itex]

    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

    CAF123

    User Avatar
    Gold Member



    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:
    ln((b-x)/(a-x))/(b-a)=kt+b


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

    x=(e(b-a)(kt+c)-b)/(-e((b-a)(kt+c))-1)

    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

    CAF123

    User Avatar
    Gold Member



    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

    CAF123

    User Avatar
    Gold Member

    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

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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

    ehild

    User Avatar
    Homework Helper
    Gold Member

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

    ehild
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Integrating dx/dt=k(a-x)(b-x)
  1. Dx/dt = x(t)(1-x(t)) (Replies: 6)

Loading...