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First order differential

  1. Apr 28, 2015 #1
    I dont understand this first order differential equation:
    https://lh5.googleusercontent.com/UUpQF4YjmjJRPvFuzGg2MhpMMMDyi2KFZPCKMKVIXGREc1owvXDzGR0bcA=s600 [Broken]
    How is it possible to get an exponent as answer?
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Apr 28, 2015 #2

    SteamKing

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    What if Eq. (2) were re-written as follows:

    dV/dt = -(E/R) * V

    then, separating the variables:

    dV/V = -(E/R) dt

    Care to take it from here?
     
  4. Apr 28, 2015 #3
    I would like to see what comes up next! :)
     
  5. Apr 28, 2015 #4

    SteamKing

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    Why can't you solve this DE now? All you have to do is integrate both sides of the equation. :wink:
     
  6. Apr 28, 2015 #5
    I have no idea how to get constant in front of exponent.
    https://lh4.googleusercontent.com/5VChRnbmBvEIkRzaZU7vGVBjf1r09ypiNOhcbq4C0LlVdP_3Tqdu_Pvtrg=s600 [Broken]

    EDIT: Forgot to add "-C" in exponent function which I got drom integration.
     
    Last edited by a moderator: May 7, 2017
  7. Apr 28, 2015 #6

    SteamKing

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    If you have ln V = Mess, what do you do to the LHS of the equation to get V? i.e., how is the natural log of a number related to the constant e?
     
    Last edited by a moderator: May 7, 2017
  8. Apr 28, 2015 #7
    Isn't that correct what I wrote? V is e^(MESS).
     
  9. Apr 28, 2015 #8

    SteamKing

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    It is. I'm sorry for not fully recognizing that.

    However, when you did the integration to obtain ln V, you forgot to include the constant of integration on the RHS, thus:

    ∫ dV/V = ∫ (-E/R) dt

    ln V = (-E/R)*t + C

    Let's say at t = 0, V = V0, then

    ln V0 = C, so

    ln V = (-E/R)*t + ln V0

    exponentiating both sides gives:

    eln V = e[(-E/R)*t + ln V0]

    which can be simplified:

    V = e(-E/R)*t * eln V0 {using the law of exponents}

    V = V0 * e(-E/R)*t

    if T = R/E, then

    V = V0 * e(-t / T)
     
  10. Apr 28, 2015 #9
    Thanks for clear explanation! Only thing I dont understand why constant appears only on one side of equation?
     
  11. Apr 28, 2015 #10

    SteamKing

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    Technically, integrating both sides of the equation will result in a constant of integration for each integral:

    ∫ dV/V = ∫ (-E/R) dt

    ln V + C1 = (-E/R) * t + C2

    The two separate constants of integration can be combined into one constant:

    ln V = (-E/R) * t + C2 - C1 = (-E/R) * t + C, where C = C2 - C1

    and the solution proceeds as described in Post #8.
     
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