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Differential equation

  1. Sep 26, 2006 #1
    Given that [tex] \frac{dx}{dt} = k(a-x)(b-x) [/tex]:

    (a) Assuming [tex] a \neq b [/tex], find [tex] x [/tex] as a function of [tex] t [/tex]. Use the fact that the initial concentration of [tex] C [/tex] is 0.
    (b) Find [tex] x(t) [/tex] assuming that [tex] a = b [/tex]. How does this expression for [tex] x(t) [/tex] simplify if it is known that [tex] [C] = \frac{a}{2} [/tex] after 20 seconds.

    (a): So [tex] \frac{dx}{(a-x)(b-x)} = kdt [/tex]. After integrating by partial fractions and using the initial condition, I got [tex] x(t) = \frac{a-abe^{akt-bkt}}{1-\frac{a}{b}e^{akt-bkt}} [/tex].

    (b). When I set [tex] a = b [/tex] I got an undefined expression, leading me to believe that part(a) is incorrect.

    What did I do wrong?

    Thanks
     
    Last edited: Sep 26, 2006
  2. jcsd
  3. Sep 26, 2006 #2

    StatusX

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    You can't just set a=b, you have to take the limit. Alternatively, go back and integrate again with the knowledge that a=b. This will change how the partial fractions expansion goes.
     
  4. Sep 27, 2006 #3

    HallsofIvy

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    If a= b the method you used to solve the equation assuming a [itex]\ne[/itex] b does not work. Go back to the original equation, set a= b, and solve again.
     
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