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Math Help - Differential Equations and exponential growth

  1. #1
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    Differential Equations and exponential growth

    The volume of water in a solution during a chemical process varies with time and satisfies the equation: dx/dt=-3x/(1 plus t)^2 (sorry no plus sign on this laptop :P )

    Anyway, initially at rest, t=0, x= 1000. Show that at time T the volume is given by: x=1000e^(-3t/(1 plus t))


    And then prove that the volume of water tends to a limit as t tends to infinite and find that limit to the nearest litre.


    Any help would be VERY appreciated coz I am so stumped. Thank you!!!!
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  2. #2
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    Quote Originally Posted by Yehia View Post
    The volume of water in a solution during a chemical process varies with time and satisfies the equation: dx/dt=-3x/(1 plus t)^2 (sorry no plus sign on this laptop :P )

    Anyway, initially at rest, t=0, x= 1000. Show that at time T the volume is given by: x=1000e^(-3t/(1 plus t))


    And then prove that the volume of water tends to a limit as t tends to infinite and find that limit to the nearest litre.


    Any help would be VERY appreciated coz I am so stumped. Thank you!!!!
    \frac{dx}{dt} = \frac{-3x}{(1+t)^2} <br />

    \frac{dx}{x} = \frac{-3}{(1+t)^2} \, dt

    \ln|x| = \frac{3}{1+t} + C

    x = Ae^{\frac{3}{1+t}}

    1000 = Ae^{3}

    A = 1000e^{-3}

    x = 1000e^{-3} \cdot e^{\frac{3}{1+t}}

    x = 1000e^{-3 + \frac{3}{1+t}}<br />

    x = 1000e^{\frac{-3t}{1+t}}
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  3. #3
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    Quote Originally Posted by Yehia View Post
    The volume of water in a solution during a chemical process varies with time and satisfies the equation: dx/dt=-3x/(1 plus t)^2 (sorry no plus sign on this laptop :P )

    Anyway, initially at rest, t=0, x= 1000. Show that at time T the volume is given by: x=1000e^(-3t/(1 plus t))


    And then prove that the volume of water tends to a limit as t tends to infinite and find that limit to the nearest litre.


    Any help would be VERY appreciated coz I am so stumped. Thank you!!!!
    \frac{dx}{dt} = \frac{3x}{(1+t)^2}

    I'd go with separation of variables here

    \frac{dx}{3x} = \frac{dt}{(1+t)^2}

    Integrating both sides

    \frac{1}{3}\ln|x| = -\frac{1}{1+t} + C

    ln|x| = -\frac{3}{1+t} + C

    x = e^{-\frac{3}{1+t} + C} = C'e^{-\frac{3}{1+t}}

    where  C' = e^C
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