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Help with initial value problem

  1. Jan 30, 2012 #1
    OK, so clearly I am missing something, because I know this is supposed to be a simple problem. It reads:

    solve the following initial value problem:
    dy/dt=-y+5
    y(0)=y_naught

    my process is as follows:
    dy/(5-y)=dt
    integrate
    ln(5-y)=t+C
    exponential both sides
    5-y=(e^t)(e^c)
    y=5-(e^t)(e^c)

    solve for constant:
    y_naught=5-e^c
    e^c=5-y_naught

    final answer:
    y=5-(5-y_naught)e^t


    My book disagrees with this answer slightly, can anyone see where I've stumbled in the process?
     
  2. jcsd
  3. Jan 30, 2012 #2

    tiny-tim

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    hi cameuth! :smile:
    nooo :redface:

    try dy/(y-5) = -dt :wink:
     
  4. Jan 30, 2012 #3
    I'm sorry... but can you explain why that is. Intuitively what I see is
    dy/dt=5-y
    divide by (5-y) multiply by dt
    dy/5-y=dt

    I'm sorry again that I'm such a beginner at this, I just don't understand why you did what you did. I see that it gets me the right answer your way, but not why we go that path.
     
  5. Jan 30, 2012 #4

    Char. Limit

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    Because the integral of 1/(5-y) isn't ln(5-y). Chain rule.
     
  6. Jan 31, 2012 #5

    tiny-tim

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    (just got up :zzz: …)

    yes … you missed out a minus :wink:
     
  7. Jan 31, 2012 #6
    sorry for the late reply. I got the answer guys, and y'all were a ton of help. Seriously, thanks. It's been awhile since six months since I've done cal 3 and so my integrating has some rust to knock off.
     
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