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Math Help - Separation of variables to solve growth equation

  1. #1
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    Separation of variables to solve growth equation

    How would I use separation of variable to find the general solution to this differential equation? I don't really even know where to start.

    W'(t) = \mu_0 e^{-Dt}W
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  2. #2
    MHF Contributor chisigma's Avatar
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    Generally specking if we wtite the equation as...

    w^{'} = a(t)\cdot w (1)

    ... we can separate the variables sterting from (1) in such way...

    \frac {dw}{dt} = a(t)\cdot w \rightarrow \frac{dw}{w} = a(t)\cdot dt \rightarrow \ln w = \int a(t)\cdot dt + \ln c (2)

    ... where c>0 is an 'arbitrary constant'...

    Kind regards

    \chi \sigma
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  3. #3
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    Now I'm really confused.

    Where did a(t) come from? Or are you just speaking generally? Should I be assigning aspects of the equation to those variables? e.g:

    a(t) = \mu_0 e^{-Dt}

    Any extra help would be greatly appreciated. I really can't seem to grasp the separation of variable method. Do you know of any good examples that step through working?
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  4. #4
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    Ah, I think I just had an epiphany thanks to Midnight Tutor (MidnightTutor.com Free Online Calculus and Chemistry Tutorials).

    W'(t) = \mu_0 e^{-Dt}W

    \frac{dW}{dt} = \mu_0 e^{-Dt}W

    \int{\frac{dW}{dt}} = \int{\mu_0 e^{-Dt}W}

    lnW = \frac{-\mu_0e^{-Dt}}{D}

    Is this the general solution? or does it need to be taken somewhere from there?
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  5. #5
    MHF Contributor chisigma's Avatar
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    Setting a(t)= \mu_{0} \cdot e^{-Dt} and applying the procedure I have indicated you arrive to write...

    \ln w = - \frac{\mu_{0}}{D} \cdot e^{-Dt} + \ln c (1)

    ... where c>0 is an arbitrary constant. Exponentiating both terms of (1) you obtain...

    w= c\cdot e^{-\frac{\mu_{0}}{D} e^{-Dt}} (2)

    That's the general solution!...

    Kind regards

    \chi \sigma
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  6. #6
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    Ah, okay. Now I'm starting to get it. Thanks for the help.
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