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(Exponential growth/decay) If f'(x) = -f(x) and f(1) = 1 then

  1. Feb 26, 2012 #1
    1. The problem statement, all variables and given/known data

    If f'(x) = -f(x) and f(1) = 1, then f(x) = ?

    (A) [itex]\frac{1}{2}[/itex]e-2x+2.
    (B) e-x-1.
    (C) e1-x.
    (D) e-x.
    (E) -ex.

    2. Relevant equations

    P(t) = P0ekt.
    [itex]\frac{dP}{dt}[/itex] = kP.

    3. The attempt at a solution

    I was thinking f(1) = t and 1 = P0, therefore:

    P(1) = ek.

    Then, I exponetiated both sides and got:

    ln(P(1)) = k.

    I'm probably going off in a completely different direction, but this is what I did so far.
     
  2. jcsd
  3. Feb 26, 2012 #2

    Dick

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    You don't even have to solve the differential equation. You just have to figure out which one of them work. That's the beauty of multiple choice.
     
  4. Feb 26, 2012 #3
    I did try that, and the only one I got remotely close to that is letter C, which I got -1 but I set it up as:

    -(e1-x)|x = 1 = -1.

    The negative is there because -f(x), is that correct or do I leave it out?

    EDIT: Oh my, never mind that. I am finding f(1), which if I evaluate letter C at x = 1, I will get 1. Thank you for leading me on the right direction, Dick!
     
    Last edited: Feb 26, 2012
  5. Feb 26, 2012 #4
    You are absolutely right, I misread it. Thank you. But say I didn't have the multiple choice, how would I come about doing this problem, or is there no way?
     
  6. Feb 26, 2012 #5

    Dick

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    Sure. Solve the differential equation. df/dt=(-f). Separate the variables, then use the initial condition.
     
  7. Feb 26, 2012 #6
    Okay, this is what I did:

    [itex]\frac{df}{dt}[/itex] = -f

    [itex]\int\frac{df}{-f}[/itex] = [itex]\int dt[/itex]

    -ln|f| = t

    -f = et and obtain:

    f = -et.

    Where did I go wrong?
     
  8. Feb 26, 2012 #7

    Dick

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    You started out going wrong by forgetting the constant of integration -ln(f)=t+C. Then it got worse when you decided e^(-ln(f))=(-f). That's not right. I'd suggest moving the minus sign over to the t side before you exponentiate.
     
    Last edited: Feb 26, 2012
  9. Feb 26, 2012 #8
    Wow, of course. Careless mistake.

    Alas! I got it!
    Thank you so very much, you were of much help.

    Have a good night, my friend.
     
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