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Thread: Differential equations with initial conditions

  1. #1
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    Differential equations with initial conditions

    I have a practice problem that confuses me.

    $\displaystyle y'= 2te^{-2y} -e^{-2y} $ y(0)=3

    I worked this out as

    $\displaystyle \int e^{2y} \frac {dy}{dt} dt= \int 2t -1 dt $

    then

    $\displaystyle e^{2y} = 2t^2 -2t +2C
    $
    $\displaystyle
    2y= ln(2t^2 -2t + 2C)
    $
    $\displaystyle

    y = \frac{1}{2} ln(2t^2 -2t + 2C)
    $

    when y(0)=3 is used, the worksheet and I disagree on the outcome.

    $\displaystyle
    3 = y(0) =\frac{1}{2} ln(2t^2 -2t + 2C)
    $

    $\displaystyle
    3 = \frac{1}{2} ln(2(0)^2 -2(0) + 2C)
    $
    $\displaystyle
    3 = \frac{1}{2} ln 2C
    $
    $\displaystyle
    e^6 = 2C
    $
    $\displaystyle
    \frac{e^6}{2}
    $

    Which would result in a final equation looking like:
    $\displaystyle

    y = \frac{1}{2} ln(2t^2 -2t + \frac{e^6}{2})
    $
    But my worksheet declares the correct answer to be:
    $\displaystyle
    y = \frac{1}{2} ln(2t^2 -2t + e^6)
    $

    How did I end up with the extra $\displaystyle \frac{1}{2} $ ?


    I'm quite confused, grateful for any explanation, thanks.
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  2. #2
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    Quote Originally Posted by Quixotic View Post
    I have a practice problem that confuses me.

    $\displaystyle y'= 2te^{-2y} -e^{-2y} $ y(0)=3

    I worked this out as

    $\displaystyle \int e^{2y} \frac {dy}{dt} dt= \int 2t -1 dt $

    then

    $\displaystyle e^{2y} = 2t^2 -2t +2C
    $
    $\displaystyle
    2y= ln(2t^2 -2t + 2C)
    $
    $\displaystyle

    y = \frac{1}{2} ln(2t^2 -2t + 2C)
    $

    when y(0)=3 is used, the worksheet and I disagree on the outcome.

    $\displaystyle
    3 = y(0) =\frac{1}{2} ln(2t^2 -2t + 2C)
    $

    $\displaystyle
    3 = \frac{1}{2} ln(2(0)^2 -2(0) + 2C)
    $
    $\displaystyle
    3 = \frac{1}{2} ln 2C
    $
    $\displaystyle
    e^6 = 2C
    $
    $\displaystyle
    \frac{e^6}{2}
    $

    Which would result in a final equation looking like:
    $\displaystyle

    y = \frac{1}{2} ln(2t^2 -2t + \frac{e^6}{2})
    $
    But my worksheet declares the correct answer to be:
    $\displaystyle
    y = \frac{1}{2} ln(2t^2 -2t + e^6)
    $

    How did I end up with the extra $\displaystyle \frac{1}{2} $ ?


    I'm quite confused, grateful for any explanation, thanks.
    $\displaystyle \frac{dy}{dt} = 2t\,e^{-2y} - e^{-2y}$ with $\displaystyle y(0) = 3$.


    $\displaystyle \frac{dy}{dt} = e^{-2y}(2t - 1)$

    $\displaystyle e^{2y}\,\frac{dy}{dt} = 2t - 1$

    $\displaystyle \int{e^{2y}\,\frac{dy}{dt}\,dt} = \int{2t - 1\,dt}$

    $\displaystyle \int{e^{2y}\,dy} = t^2 - t + C_1$

    $\displaystyle \frac{1}{2}e^{2y} + C_2 = t^2 - t + C_1$

    $\displaystyle e^{2y} = 2t^2 - 2t + C$, where $\displaystyle C = 2(C_1 - C_2)$

    $\displaystyle 2y = \ln{(2t^2 - 2t + C)}$

    $\displaystyle y = \frac{1}{2}\ln{(2t^2 - 2t + C)}$.


    Now using the initial condition:

    $\displaystyle 3 = \frac{1}{2}\ln{[2(0)^2 - 2(0) + C]}$

    $\displaystyle 3 = \frac{1}{2}\ln{C}$

    $\displaystyle 6 = \ln{C}$

    $\displaystyle C = e^6$.


    Therefore:

    $\displaystyle y = \frac{1}{2}\ln{(2t^2 - 2t + e^6)}$.
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  3. #3
    Junior Member
    Joined
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    Florida
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    $\displaystyle e^{2y} = 2t^2 - 2t + C$, where $\displaystyle C = 2(C_1 - C_2)$
    Ah. I get it now. You have my eternal gratitude! It would have taken me -days- to work that out.
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