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Math Help - Help with a simple first order differential equation that seems to contradict itself.

  1. #1
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    Help with a simple first order differential equation that seems to contradict itself.

    So given a differential equation that models the mass of a substance (m) produced during a chemical reaction after time (t).

    equation: dm/dt = (50-m)/500 (given m=0 when t=0)


    after i solved it, i get: m = (50) + (500/(t-10))

    but this looks problematic.
    firstly it implies that the substance keeps getting smaller mass, shouldn't m be getting bigger as t does?!?!?!?

    secondly, during the first 10 seconds the mass is negative!

    and also t can never equal 10 then!!! (denominator will be 0).

    and lastly, i'm asked to show that the mass will never exceed 50...but wait isn't it ALWAYS going to be bigger than 50?!?! it never gets below 50!.




    please help or show me what i did wrong, so appreciated thanks!!!!
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  2. #2
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    e^(i*pi)'s Avatar
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    You might have got a couple of sign errors in there because I get an answer different in sign. I have put my working in spoiler

    Spoiler:
    \dfrac{dm}{(50-m)^2} = \dfrac{dt}{500}

    u = 50-m \rightarrow du = -dm \implies dm = -du


    Giving -\dfrac{du}{u^2} = \dfrac{dt}{500}

    \dfrac{1}{u} = \dfrac{1}{50-m} = \dfrac{t}{500} + C

    Using your initial condition  C = \dfrac{10}{500}


    1 = (50-m) \left(\dfrac{t+10}{500}\right)

    50-m = \dfrac{500}{t+10}


    m = 50 - \dfrac{500}{t+10}


    What do you know about t+10 and it's relationship to 0?
    Last edited by e^(i*pi); February 23rd 2011 at 01:04 PM. Reason: questions end with a question mark, not a full stop
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  3. #3
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    Quote Originally Posted by e^(i*pi) View Post
    You might have got a couple of sign errors in there because I get an answer different in sign. I have put my working in spoiler

    Spoiler:
    \dfrac{dm}{(50-m)^2} = \dfrac{dt}{500}

    u = 50-m \rightarrow du = -dm \implies dm = -du


    Giving -\dfrac{du}{u^2} = \dfrac{dt}{500}

    \dfrac{1}{u} = \dfrac{1}{50-m} = \dfrac{t}{500} + C

    Using your initial condition  C = \dfrac{10}{500}


    1 = (50-m) \left(\dfrac{t+10}{500}\right)

    50-m = \dfrac{500}{t+10}


    m = 50 - \dfrac{500}{t+10}


    What do you know about t+10 and it's relationship to 0?
    *sigh*

    Thank you so much, i mistakenly forgot to multiply the (-1/50-m) by -1, so as you said, i got my signs mixed up.
    that clarifies everything and clearly t can never be -10 seconds, so the denominator will never be 0.
    and clearly as t tends towards infinite, m tends towards 50 (but never reaches it) which explains my quiery.

    i really want to kick myself for spending ages on something that was so obvious!
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