Differential Equations question

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  1. #1
    Cheesy Poofs! PJYelton's Avatar
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    Differential Equations question

    I just started a new class that has Differential Equations as a prerequisite, so as a refresher our teacher gave us a little assignment (not for credit so asking this wouldn't be cheating)that had many calc and DE's refresher questions. But its been a few years since I took the DE's class and one of the questions is screwing me up... anyone know if I am doing this right?

    The question: dx/dt + 3x = 0, x(0)=2, x(t)=?

    Okay, if memory serves, the first thing to do is get the dx and dt on separate sides and then integrate.

    dx/dt=-3x

    dt=(-1/3)dx

    Integrating... t=(-1/3) ln(x) + C

    Plugging in the values above to solve for C... 2=(-1/3) ln (0) + C

    But the natural log of zero is undefined, so I know that I am doing something wrong... any help?

  2. #2
    Toaster Zach L.'s Avatar
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    x(0)=2 -> x=2 when t=0
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    Cheesy Poofs! PJYelton's Avatar
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    Ahhh... so then

    -3t + C = ln(x)

    e^(-3t) + C = x

    e^(0) + C = 2

    C=1

    x(t) = e^(-3t) + 1

    Thanks!

  4. #4
    Registered User sean345's Avatar
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    >dt=(-1/3)dx
    >Integrating... t=(-1/3) ln(x) + C
    Wouldn't it become:
    t=(-1/3)x + C
    -> X=-3t + C

    <edit>
    I see now. That first line I quoted should have an X with 3.

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    Cheesy Poofs! PJYelton's Avatar
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    Oops, you're right, tis a typo.

    Change

    dt=(-1/3)dx

    to

    dt=(-1/3x)dx

  6. #6
    Toaster Zach L.'s Avatar
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    Originally posted by PJYelton
    Ahhh... so then

    -3t + C = ln(x)

    e^(-3t) + C = x

    e^(0) + C = 2

    C=1

    x(t) = e^(-3t) + 1

    Thanks!
    Err... not quite. Its not a solution to dx/dt=-3x. Here (note, I'm sloppy with the 'C' because I'm not looking for a particular form of it).

    dx/dt = -3x

    (-1/3)(dx/x) = dt

    dx/x = -3dt

    ln x = -3t + C

    x = e^(-3t+C)

    x = Ce^(-3t) [The sloppiness I mentioned]

    2 = Ce^0

    C = 2

    x(t) = 2e^(-3t)
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  7. #7
    Cheesy Poofs! PJYelton's Avatar
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    Jeez I feel like an idiot For some reason I thought:

    e^(-3t+C)=e^(-3t)+e^(C) instead of e^(-3t)*e^(C)

    which of course is wrong... Can you tell its been a few years since I've done anything like this?

    You're right, the answer should be 2e^(-3t), thanks!

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