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Differential Equation dx/dt = k(a-x)(b-x)

  1. Sep 23, 2008 #1
    1. The problem statement, all variables and given/known data

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

    x(0)=0

    3. The attempt at a solution

    dx/((a-x)(b-x)) = kdt

    Integral (dx/(ab-bx-ax+x^2)) = kt +C

    x/ab - (1/b)ln|x| - (1/a)ln|x| - (x^-1) = kt + C

    however, if i try to substitute x(0) = 0, i get the ln 0

    please help i have no idea if i am doing this right
     
  2. jcsd
  3. Sep 23, 2008 #2

    Dick

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    Science Advisor
    Homework Helper

    I don't think the integration dx is correct at all. Use partial fractions.
     
  4. Sep 24, 2008 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    NO, 1/(a+ bx+ cx2) is NOT equal to 1/a+ 1/bx+ 1/cx2.

    Rule of thumb: since it is harder to factor than to multiply, NEVER multiply terms that are already factored! (Well, almost never.)

    Integrate dx/((a-x)(b-x))= dx/((x-a)(x-b) using partial fractions.
     
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