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Math Help - Differential equations with initial conditions

  1. #1
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    Differential equations with initial conditions

    I have a practice problem that confuses me.

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

    I worked this out as

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

    then

     e^{2y} = 2t^2 -2t +2C<br />
    <br />
          2y= ln(2t^2 -2t + 2C)<br />
    <br /> <br />
          y =  \frac{1}{2} ln(2t^2 -2t + 2C)<br />

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

    <br />
3 = y(0) =\frac{1}{2} ln(2t^2 -2t + 2C)<br />

    <br />
     3 = \frac{1}{2} ln(2(0)^2 -2(0) + 2C)<br />
    <br />
     3 = \frac{1}{2} ln 2C<br />
    <br />
     e^6 = 2C<br />
    <br />
     \frac{e^6}{2}<br />

    Which would result in a final equation looking like:
    <br /> <br />
y = \frac{1}{2} ln(2t^2 -2t + \frac{e^6}{2})<br />
    But my worksheet declares the correct answer to be:
    <br />
y = \frac{1}{2} ln(2t^2 -2t + e^6)<br />

    How did I end up with the extra  \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.

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

    I worked this out as

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

    then

     e^{2y} = 2t^2 -2t +2C<br />
    <br />
          2y= ln(2t^2 -2t + 2C)<br />
    <br /> <br />
          y =  \frac{1}{2} ln(2t^2 -2t + 2C)<br />

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

    <br />
3 = y(0) =\frac{1}{2} ln(2t^2 -2t + 2C)<br />

    <br />
     3 = \frac{1}{2} ln(2(0)^2 -2(0) + 2C)<br />
    <br />
     3 = \frac{1}{2} ln 2C<br />
    <br />
     e^6 = 2C<br />
    <br />
     \frac{e^6}{2}<br />

    Which would result in a final equation looking like:
    <br /> <br />
y = \frac{1}{2} ln(2t^2 -2t + \frac{e^6}{2})<br />
    But my worksheet declares the correct answer to be:
    <br />
y = \frac{1}{2} ln(2t^2 -2t + e^6)<br />

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


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


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

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

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

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

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

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

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

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


    Now using the initial condition:

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

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

    6 = \ln{C}

    C = e^6.


    Therefore:

    y = \frac{1}{2}\ln{(2t^2 - 2t + e^6)}.
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  3. #3
    Junior Member
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    Florida
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    e^{2y} = 2t^2 - 2t + C, where 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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