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Diff Eqs: can't get a reasonable answer

  1. Jul 17, 2015 #1
    1. The problem statement, all variables and given/known data

    C1epJzI.png

    Imgur link: http://i.imgur.com/C1epJzI.png

    My issues arise in part (b) but cause me to doubt my solution to part (a) as well.

    2. Relevant equations


    3. The attempt at a solution

    [tex]\frac{dx}{dt} = k(a-x)(b-x)[/tex]

    [tex]\int \frac{1}{(a-x)(b-x)} dx = k \int dt[/tex]

    Use partial fraction decomposition on LHS

    [tex]\frac{1}{(a-x)(b-x)} = \frac{A}{a-x}+\frac{B}{b-x}[/tex]

    [tex]A(b-x) + B(a-x) = 1[/tex]

    Let x = b

    [tex]B(a-b) = 1 \ \ B = \frac{1}{a-b}[/tex]

    Let x = a

    [tex]A(b-a) = 1 \ \ A = \frac{1}{b-a} = \frac{-1}{a-b}[/tex]

    [tex]\int \frac{1}{(a-x)(b-x)} dx = \int \frac{1}{(a-b)(b-x))} - \frac{1}{(a-b)(a-x)} dx[/tex]

    [tex] = \frac{1}{a-b} \left[ -\ln{|b-x|} + \ln{|a-x|} \right] = \frac{1}{a-b} \ln{\frac{|a-x|}{|b-x|}} = k(t+C)[/tex]

    [tex]\ln{\frac{|a-x|}{|b-x|}} = (a-b)k(t+C)[/tex]

    [tex]\frac{|a-x|}{|b-x|} = e^{(a-b)kC}e^{(a-b)kt}[/tex]

    Now let's skip a few steps and get

    [tex]\frac{a-x}{b-x} = Ke^{(a-b)kt}[/tex]

    already here I can see problems with part (b), as when a = b, [itex]\frac{a-x}{b-x} = K[/itex]

    But if we carry on with some algebra from the last step I arrive at

    [tex]x(t) = \frac{bKe^{(a-b)kt}-a}{Ke^{(a-b)kt}-1}[/tex]

    and if x(0) = 0 then

    [tex]K = \frac{a}{b}[/tex]

    and then

    [tex]x(t) = \frac{a(e^{(a-b)kt}-1)}{\frac{a}{b}e^{(a-b)kt}-1}[/tex]

    Now, if a = b x(t) = 0...

    Where have I gone wrong?

    edit: or is my answer to part (a) correct, and I need to return to the very beginning of the sequence and assume a = b in order to solve (b), thus, the solution to part (a) only holding for the condition that [itex]a \neq b[/itex]?
     
  2. jcsd
  3. Jul 17, 2015 #2

    fzero

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    I didn't go through all of the algebra, but your solution to (a) looks ok. If ##a=b##, the partial fraction decomposition is not valid, but you can instead integrate ##1/(a-x)^2## directly. The solution will have a different functional form involving a reciprocal of ##t## rather than an exponential function.
     
  4. Jul 17, 2015 #3
    Thanks, I'm still trying to get an intuitive handle on diff eqs. That's why I was getting frustrated. Because my technical skills in everything required in this problem are fine, so not being able to find a mistake, I thought maybe I'd missed something big at the beginning...which it turns out I did I guess.

    Domains become really important to consider when modelling with diff eqs then right?
     
  5. Jul 18, 2015 #4

    Ray Vickson

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    You have gone wrong putting ##a = b## in your ##x(t)## formula because you will have the dreaded (and illegal) form ##0/0##. Either you need to put ##a = b## before solving the DE, or else you need to look at ##\lim_{b \to a} x(t)## after you have solved it. When you do it the latter way you do not get ##x(t) = 0##.
     
    Last edited: Jul 18, 2015
  6. Jul 18, 2015 #5
    Kostoglotov,
    I have arrived at your same solution. Partial fractions are still valid since we are asssuming a does not equal b in part a. Try letting a=b and integrating just using a substitution for part b. It wouldn't make sense to let a=b with the equation just derived since x(t)= 0 for all t.

    Edit: Ahh Ray beat me to the punch. I second that, re do from the beginning letting a=b
     
  7. Jul 18, 2015 #6

    epenguin

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    If you go back to your equations for A and B you already see it's trouble if a = b.

    In fact no one would ask you to expand 1/(x - a)2 as a partial fraction - there is no way to make it identical to any constant divided by (x - a).
     
  8. Jul 18, 2015 #7

    Zondrina

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    The domain is always important. Consider the homogeneous Cauchy-Euler equation:

    $$ax^2y'' + bxy' + cy = 0$$

    In standard form:

    $$y'' + \frac{b}{a x} y' + \frac{c}{a x^2}y = 0$$

    ##x = 0## would be a regular singular point because the solution may be undefined at ##x = 0##. We would then need to seek solutions for ##x > 0## and ##x < 0## individually.

    To derive a solution ##\forall x \neq 0##, then the absolute value would be required.
     
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