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Solving Differential Equations

  1. Feb 11, 2013 #1
    1. The problem statement, all variables and given/known data
    A psychology class is studying memory. Several objects are uncovered to view for a given amount of minutes and then covered again. At most 150 objects can be viewed and remembered. The class found that after 10 minutes the average student could remember 25 objects. The differential equation that models this study is given by

    dN/dt = k(L-N)

    Solve this differential equation to find an equation that will give the number of objects remembered at any time t.


    2. Relevant equations



    3. The attempt at a solution

    I'm a little unsure of what I'm doing, so some feedback would be great.

    dN/dt = k(L-N)

    dN = (kL - kN)dt

    So, I think I need to move the N on the right hand side to the left, but I'm not sure how to do this, as every action I take keeps other prevents me from isolating N and dN on the left hand side.

    Can I leave the N on the right hand side and integrate the left with respect to dN and the right with respect to dt?
     
  2. jcsd
  3. Feb 11, 2013 #2

    Dick

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    Move all of the N's to one side. dN/(k*(L-N))=dt. Now integrate both sides.
     
  4. Feb 11, 2013 #3
    Assuming K and L are constants.

    How about dividing by (L-N) on both sides from dN/dt = k(L-N) instead of distributing the K? This equation is separable.
     
  5. Feb 11, 2013 #4
    I hadn't considered moving t to one side by itself. Thanks.

    So, I should end up with:

    t = 1/k*ln(L-N)
     
  6. Feb 11, 2013 #5

    Dick

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    Almost. I'd write that (1/k)*ln(L-N). But I think you missed a '-' sign and you should put a +C someplace.
     
  7. Feb 11, 2013 #6

    Ray Vickson

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    Even easier: notice that since L is a constant we have dL/dt = 0, so dN/dt = d(N-L)/dt. Therefore, if x = N-L you have dx/dt = -kx.
     
  8. Feb 13, 2013 #7
    When I spoke with my professor she informed me that I needed to use a u-substitution in order to integrate.

    ∫dN/(L-N) = ∫k dt

    u = L-N

    du = -dN
    dN = -du

    ∫-du/u = ∫k dt

    -ln(u) = kt + C

    -ln(L-N) = ekt + C

    N = ekt + C+ L

    Then using the information given, I think that at t=0 N=150

    150 = eC + L

    So I have L = 150 - eC

    I also know that at t=10 that N=25

    25 = e10k +C + L

    I have three unknowns but only two equations. Unless I'm missing something, I don't think it's possible for me to solve for L, C and k.

    All suggestions are appreciated.
     
  9. Feb 13, 2013 #8

    Dick

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    I think the sign in front of k got messed up. When you've got -ln(L-N)=kt+C, it's better to multiply both sides by (-1) first so you've got ln(L-N)=(-kt+C). Now exponentiate. And when you've got something like e^(-kt+C) it's usually easier to write it as e^(-kt)*e^C. And since C is a constant, e^C is just another constant, so write it as just plain C. I.e. Ce^(-kt).

    On the initial value you thing, the statement of the question isn't crystal clear, but I think what they mean is at t=0, the student is just starting and hasn't memorized anything yet. So at t=0, N=0. At t=10, N=25. Finally for very large values of t the student should max out at 150 objects, so as t->infinity N->150. You do have three boundary conditions.
     
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