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Math Help - Differential Equation options question?

  1. #16
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    Re: Differential Equation options question?

    Quote Originally Posted by canyouhelp View Post
    Um. I had I and III to begin with and I was told that was wrong...
    that's because it's not I and III, it's only one of them. I'm trying not to tell you which one so you can figure it out.
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  2. #17
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    Re: Differential Equation options question?

    Quote Originally Posted by romsek View Post
    that's because it's not I and III, it's only one of them. I'm trying not to tell you which one so you can figure it out.
    Ok I guess you've suffered enough. You really should be able to solve this though.

    solving I

    $\dfrac{dP_L}{dt}=k(M-P_L)$

    $\dfrac{dP_L}{M-P_L}=k dt$

    $-\ln\left|M-P_L\right| = k t + C_1$

    $M-P_L=Ce^{-kt}$ Note that $M\geq P_L$ so I can omit the $||'s$. $C=e^{C_1}$

    $P_L=M-Ce^{-kt}$

    This is exactly the behavior we want from our learning function.

    now III

    $\dfrac{dP_L}{dt}=k\sqrt{M-P_L}$

    $\dfrac{dP_L}{\sqrt{M-P_L}}=k dt$

    $-2\sqrt{M-P_L}=kt + C$

    $\sqrt{M-P_L}=-\dfrac{kt+C}{2}$

    $M-P_L=\left(-\dfrac{kt+C}{2}\right)^2$

    $P_L=M-\left(\dfrac{kt+C}{2}\right)^2$

    This is an inverted parabola that reaches a maximum of M (or less) but then begins to decrease as time continues. This is not the behavior we want.

    So only I models the situation correctly.
    Last edited by romsek; March 31st 2014 at 08:25 PM.
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  3. #18
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    Re: Differential Equation options question?

    Quote Originally Posted by romsek View Post
    Ok I guess you've suffered enough. You really should be able to solve this though.

    solving I

    $\dfrac{dP_L}{dt}=k(M-P_L)$

    $\dfrac{dP_L}{M-P_L}=k dt$

    $-\ln\left|M-P_L\right| = k t + C_1$

    $M-P_L=Ce^{-kt}$ Note that $M\geq P_L$ so I can omit the $||'s$. $C=e^{C_1}$

    $P_L=M-Ce^{-kt}$

    This is exactly the behavior we want from our learning function.

    now III

    $\dfrac{dP_L}{dt}=k\sqrt{M-P_L}$

    $\dfrac{dP_L}{\sqrt{M-P_L}}=k dt$

    $-2\sqrt{M-P_L}=kt + C$

    $\sqrt{M-P_L}=-\dfrac{kt+C}{2}$

    $M-P_L=\left(-\dfrac{kt+C}{2}\right)^2$

    $P_L=M-\left(\dfrac{kt+C}{2}\right)^2$

    This is an inverted parabola that reaches a maximum of M (or less) but then begins to decrease as time continues. This is not the behavior we want.

    So only I models the situation correctly.
    Oh wow, thank so much! I had class today so I hadn't had time to work out I and III (for the umpteenth time) and respond. I really appreciate your help!
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