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Math Help - dP/dt =, how do you find the solution?

  1. #1
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    dP/dt =, how do you find the solution?

    dP/dt = te^(t+P) ; where P = 1 when t = 0.

    I got to the part where:

    dP/(e^P) = te^t dt

    Found the integral:

    -e^(-P) = te^t - e^t + c

    Not sure how to find the solution. (we also can't ln a negative).
    Last edited by AlphaRock; April 14th 2010 at 12:33 PM.
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  2. #2
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    Have you consdiered substituting the given values for p and t?

    Note: Is it "p" or "P". Normally, it's bad form to be switching about.
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  3. #3
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    Quote Originally Posted by TKHunny View Post
    Have you consdiered substituting the given values for p and t?

    Note: Is it "p" or "P". Normally, it's bad form to be switching about.
    Yes. I got c = 0.

    Placing that c in the equation:

    -e^(-P) = te^t - e^t + 0.

    Then when I lned both sides, I got
    -P = ln(e^t - te^t)
    P = - ln(e^t - te^t)

    which I don't think is right. How do we find the answer if this isn't the answer?
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  4. #4
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    How did you get C = 0? Please try that again. e^{0}\;=\;1\;\ne\;e

    It may help to write it as -e^{-P}\;=\;e^{t}\cdot (t-1) + C

    Hmm... "ln" really isn't a verb. Try not to speak pigeon math.
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  5. #5
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    Quote Originally Posted by TKHunny View Post
    How did you get C = 0? Please try that again. e^{0}\;=\;1\;\ne\;e

    It may help to write it as -e^{-P}\;=\;e^{t}\cdot (t-1) + C

    Hmm... "ln" really isn't a verb. Try not to speak pigeon math.

    -e^(-P) = te^t - e^t + c.

    P(0) = 1
    Therefore,
    -e^(-1) = 0e^t - e^0 + c.
    -1 = -1 + c
    -1 + 1 = c
    0 = c
    c= 0
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  6. #6
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    You seem to be saying that e^{-1}\;=\;1.

    Please reconsider, particularly since later you have e^{0}\;=\;1. You can't have it both ways!

    Be more careful.
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