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Math Help - First Order ODE

  1. #1
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    First Order ODE

    \frac{dz}{dt} + 3e^{t+z} = 0

    Here's what I have done:

    \frac{dz}{dt} = -3e^{t+z}

    \ln(dz) - \ln(dt) = \ln(-3) + ln(e^{t+z})

    Am I approaching this correctly? Something is obviously wrong since the ln(-3) is undefined, which makes me think that I am doing this wrong.

    Thanks!
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  2. #2
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    re: First Order ODE

    I see it like this

    \frac{dz}{dt} = -3e^{t+z}

    \frac{dz}{dt} = -3e^{t}e^{z}

    e^{-z}~dz= -3e^{t}~dt

    \int e^{-z}~dz= -3 \int e^{t}~dt

    etc..
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  3. #3
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    re: First Order ODE

    Thanks for that. This is my very first differential equation (my deflowering, so to speak), so this is all new to me.
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  4. #4
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    re: First Order ODE

    But wait...

    Carrying on from where you left off:

    e^{-z} = -3e^t + C

    -z = \ln(-3e^t + C)

    I'm back to my same dilemma. I have the logarithm of a negative number. How do I handle that? Thanks again.
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  5. #5
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    re: First Order ODE

    Quote Originally Posted by joatmon View Post

    -z = \ln(-3e^t + C)

    I'm back to my same dilemma. I have the logarithm of a negative number. How do I handle that?
    But do you? consider the case where t=0 and C=4. Is the logarithm negative?
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  6. #6
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    re: First Order ODE

    I thought of that, too, that the constant can always push the equation positive. Where I am confused is that when I run this problem through Wolfram (or my calculator) to check my work, it shows the answer as \ln(3e^t + C). I'm just trying to understand whether I am getting the right answer or not, and I have two sources (that I trust more than my own work) that have a slightly different answer.

    Any thoughts? Thanks again.
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  7. #7
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    Re: First Order ODE

    Quote Originally Posted by joatmon View Post
    But wait...

    Carrying on from where you left off:

    e^{-z} = -3e^t + C

    -z = \ln(-3e^t + C)

    I'm back to my same dilemma. I have the logarithm of a negative number. How do I handle that? Thanks again.
    I don't quite get the same result. I have

    -e^{-z}=-3e^{t}+C, and so

    e^{-z}=3e^{t}+C, where I've redefined my constant. Then

    -z=\ln(3e^{t}+C), and hence

    z=-\ln(3e^{t}+C).
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