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Math Help - Another differential equation

  1. #1
    Tau
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    Another differential equation

    Hello

    We got a container holding 100 liters of saltwater (concentration: 1%).

    We pour 5 liters/min saltwater (2% conc.) and remove 5 liters/min of the liquid in the tank.

    The concentration of salt is the same in all of the container.

    When is the concentration 1.5%?
    ----

    Ok I have nearly solved this problem, some details that don't fit in though.

    My (near) solution:

    Some trivial facts:
    y(0) = 1
    y(infinity) = 2

    where y(t) is the amount of salt at a specific time.

    y' = \frac{1}{10} - \frac{5y(t)}{100}

    We get this equation:

    y' + \frac{5}{100}y = \frac{1}{10}

    Solve it...

    y(t)\cdot e^{\frac{5x}{100}} = \frac{1}{10} \cdot \frac{100}{5}e^{\frac{5x}{100}} + C

    y(t) = \frac{1}{2} + Ce^{-\frac{x}{20}}

    Well I think that I got it wrong somewhere, the above equation doesn't seem to fit.

    Thanks for your replies.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Tau View Post
    Hello

    We got a container holding 100 liters of saltwater (concentration: 1%).

    We pour 5 liters/min saltwater (2% conc.) and remove 5 liters/min of the liquid in the tank.

    The concentration of salt is the same in all of the container.

    When is the concentration 1.5%?
    ----

    Ok I have nearly solved this problem, some details that don't fit in though.

    My (near) solution:

    Some trivial facts:
    y(0) = 1
    y(infinity) = 2

    where y(t) is the amount of salt at a specific time.

    y' = \frac{1}{10} - \frac{5y(t)}{100}

    We get this equation:

    y' + \frac{5}{100}y = \frac{1}{10}

    Solve it...

    y(t)\cdot e^{\frac{5x}{100}} = \frac{1}{10} \cdot \frac{100}{5}e^{\frac{5x}{100}} + C

    y(t) = \frac{1}{2} + Ce^{-\frac{x}{20}}

    Well I think that I got it wrong somewhere, the above equation doesn't seem to fit.

    Thanks for your replies.

     <br />
\frac{1}{10} \cdot \frac{100}{5}=2<br />

    CB
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