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Math Help - Problem with first order separabel differential equation

  1. #1
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    Problem with first order separabel differential equation

    Hi!

    First of all I am Norwegian, and my math-english might not be totally correct.

    Okay, I have a problem with a first order separabel differential equation:

    where A is a real constant and 0 < y < A.

    I seperate:



    Splitting the fraction and differentiate it:



    I now have to put the y on the one side of the equation, but I'm unsure of how to do this. This is as far as i have gotten:



    Is there something I am doing wrong in the steps above?

    Any help would be greatly appreciated!

    Thank you in advance,
    Thomas
    Last edited by luckytommy; December 9th 2009 at 03:59 AM. Reason: Edited y and A > 0
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  2. #2
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    Quote Originally Posted by luckytommy View Post
    Hi!

    First of all I am Norwegian, and my math-english might not be totally correct.

    Okay, I have a problem with a first order separabel differential equation:



    I seperate:



    Splitting the fraction and differentiate it:

    Mr F says: This should be {\color{red} \frac{1}{A \lambda} \ln \left(\frac{y}{A {\color{blue}-} y}\right) = t + C}

    I now have to put the y on the one side of the equation, but I'm unsure of how to do this. This is as far as i have gotten:



    Is there something I am doing wrong in the steps above?

    Any help would be greatly appreciated!

    Thank you in advance,
    Thomas
    \frac{1}{A \lambda} \ln \frac{y}{A - y} = t + C

    \Rightarrow \ln \frac{y}{A - y} = A \lambda (t + C) = A \lambda t + K

    \Rightarrow \frac{y}{A - y} = e^{A \lambda t + K} = D e^{A \lambda t}

    and now your job is to make y the subject.
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  3. #3
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    Thank you mr F.

    I did some more manipulation, and came to this answer:

    y=Ae^{A\lambda t+C}-ye^{A\lambda t+C}

    y(1+e^{A\lambda t+C})=Ae^{A\lambda t+C}

    y=\frac{Ae^{A\lambda t+C}}{1+e^{A\lambda t+C}}

    Correctamente?

    I see from your post that you created a new constant D. That is a better way to do it... You really are fantastic aren't you?

    Much obliged
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  4. #4
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    Quote Originally Posted by luckytommy View Post
    Thank you mr F.

    I did some more manipulation, and came to this answer:

    y=Ae^{A\lambda t+C}-ye^{A\lambda t+C}

    y(1+e^{A\lambda t+C})=Ae^{A\lambda t+C}

    y=\frac{Ae^{A\lambda t+C}}{1+e^{A\lambda t+C}}

    Correctamente?

    I see from your post that you created a new constant D. That is a better way to do it... You really are fantastic aren't you?

    Much obliged
    Correctamente!
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