Results 1 to 2 of 2

Thread: Easy differential equation

  1. #1
    Junior Member
    Sep 2007

    Easy differential equation

    Can someone help me with the problem below? I'm 99% positive i screwed up part B

    Suppose two students memorize lists according to the model
    $\displaystyle \frac{dL}{dt} = 2(1-L)$, where $\displaystyle 0 \le L \le 1$
    (A) If one of the students knows one-half of the list at time $\displaystyle t=0$ and the other knows none of the list, which student is learning most rapidly at this instant?

    (B) Will the student who starts out knowing none of the list ever catch up to the student who starts out knowing one-half of the list?



    Student A:
    $\displaystyle \frac{dL}{dt} = 2(1-0.5) = 1$
    Student B:
    $\displaystyle \frac{dL}{dt} = 2(1-0) = 2$
    Therefore, student B is learning at a faster rate than student A

    $\displaystyle \frac{dL}{dt} = 2(1-L)$

    $\displaystyle \frac{dL}{1-L}= 2 dt$

    $\displaystyle -ln |1-L| = 2t +c $

    $\displaystyle L = e^{2t +c} = ce^{2t}$

    Using the given information $\displaystyle L_A(0) = 0.5$; $\displaystyle L_B(0) = 0$:
    $\displaystyle L_A = 0.5e^2t$
    $\displaystyle L_A = 0$

    No, student B will NOT catch up.

    ^I was going to set $\displaystyle L_A$ equal to $\displaystyle L_B$, but then I'll end up with trying to solve $\displaystyle ln(0)$, which is not possible, seeing as it approaches negative infinity...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Jun 2010
    CT, USA
    In your solution, you made a mistake going from

    $\displaystyle -\ln |1-L| = 2t +c$ to

    $\displaystyle L = e^{2t +c} = ce^{2t}$

    Instead, it should be this (I can't skip that many steps safely!):

    $\displaystyle \ln|1-L|=-2t-c$

    $\displaystyle 1-L=e^{-2t}e^{-c}$

    $\displaystyle L=1-e^{-2t}e^{-c}.$

    Now try plugging in the initial conditions.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: May 8th 2011, 12:27 PM
  2. Differential equations problem (easy?)
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: Jan 13th 2010, 10:24 AM
  3. Replies: 1
    Last Post: May 8th 2009, 01:12 PM
  4. Replies: 9
    Last Post: Apr 23rd 2009, 10:46 PM
  5. Changing equation into ODE... easy?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Oct 29th 2007, 05:27 PM

Search Tags

/mathhelpforum @mathhelpforum