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Homework Help: First Order ODE problem (ipod battery)

  1. Oct 5, 2009 #1
    1. The problem statement, all variables and given/known data

    Cappy McGreenhorn is looking to buy a new MP3 player. His primary concern is the
    battery life of the MP3 player. He has decided on either the aPod or the bPod. The
    percentage of battery life A(t) (after t hours) for the aPod is governed by:

    (100 - t2)dA/dt + 4tA = 0; with A(0) = 100:

    The percentage of battery life B(t) (after t hours) for the bPod is governed by:
    100dB/dt - B2 + 10000 = 0; with B(0) = 100 tanh(12):

    (a) Which MP3 player should Cappy buy in order to get the longest playback time?
    (b) What is the battery charge on the bPod at t = 6?
    (c) When does the aPod hold a 50% battery charge?

    2. Relevant equations

    HINT: Do not use partial fractions to do the integration. Use the fact that:
    d/dx(arctanh(x)) =1/(1 - x2)

    3. The attempt at a solution

    I worked out Apod to be (1/100) (100-t2)2
    But I think this is wrong.
    The second part I used separable vales and ended up with

    B= 100tanh(1/100 (t + 12))
    However, I know that my second solution must be wrong since you cannot have negative
    time and when the battery = 0, my time will be -12 hours. :(

    Is it possible to use integrating factors to do the first order ode problem for bpod?
  2. jcsd
  3. Oct 6, 2009 #2
    Also for my hours i got 10 and 12 hours respectively. But I am still unsure about the Bpod :'(
  4. Oct 6, 2009 #3
    I have been trying to solve the Bpod but I can't seem to get it. That has to be the strangest initial condition I have ever seen. (For a textbook problem.) Obviously it is simply 100 because the tanh(12) is 1. (Assuming radians are being used.)

  5. Oct 7, 2009 #4


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    So dA/A= 4tdt/(100- t2). Let t= 100- t2 on the right and integrate

    Is that "B2" supposed to be B2? If so the equation is not linear and you cannot find the "integrating factor".

    100dB/(10000- B2)= -dt

    Good hint! So that was B2. Let u= B/100 so that 10000- B2= 10000- 10000u2= 10000(1- u2) and du= 100dB. Your equation becomes (1/10000)du/(1- u^2)= -dt or du/(1- u^2)= -10000 dt. Did you miss the negative here? As the hint tells you that integrates to arctanh(u)= arctanh(B/100)= -10000t+ C.3 Now, at t=0, you have arctanh(tanh(12)= C or C= 12.

  6. Oct 7, 2009 #5
    Let t=100-t^2? I tried doing this but I couldn't make sense of it... how would I integrate it?

    Also, thanks so much for posting your help.
  7. Oct 8, 2009 #6


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    I meant, of course, "let u= 100- t2". Then du= -2t dt so t dt=-(1/2)du. The equation becomes dA/A= -2du/u. That should be easy to integrate.
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