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Differential Equations model

  1. Oct 8, 2013 #1
    1. The problem statement, all variables and given/known data

    Model for learning in the form of a differential equation:

    [itex]\frac{dP}{dt}[/itex]= k(M-P)

    Where P(t) measures the performance of someone learning a skill after training time (t), M is the maximum level of performance, and k is a positive constant. Solve this differential equation to find an expression for P(t). What is the limit of this expression?


    2. Relevant equations



    3. The attempt at a solution

    I think I am doing this problem the correct way. However, my textbook uses a different method. Would you be able to confirm if I am do this correctly?

    dP=k(M-P)dt
    [itex]\frac{dP}{M-P}[/itex] = kdt

    [itex]\int\frac{dP}{M-P}[/itex] = k [itex]\int dt [/itex]

    for [itex]\int\frac{dP}{M-P}[/itex] let u=M-P, du=-dP

    -[itex]\int\frac{1}{u}[/itex] du= -ln|M-P|

    -ln|M-P|=kt+C
    ln|M-P|=-kt-c
    e^(-kt-c)=|M-P|
    [itex]\pm[/itex]e^(-kt)*e^(-c) =M-P
    [itex]\pm[/itex] e^(-c) is a constant so call it A
    M-Ae^(-kt)=P(t)
    as t→∞ P→M

    The book does something different. At the very start they say: [itex]\int\frac{dP}{P-M}[/itex] = [itex]\int -kdt [/itex] and get a final answer of: P(t)= M+Ae^(-kt). Are these answers equivalent because we can just say -A is another constant?
     
  2. jcsd
  3. Oct 9, 2013 #2
    So then it would be: let -A=D

    M+De^(-kt)=P(t)
     
  4. Oct 9, 2013 #3

    Mark44

    Staff: Mentor

    Per PF rules, you shouldn't bump a thread earlier than 24 hours after you post it. If you need to make a change, just edit your post.
     
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