Results 1 to 4 of 4

Math Help - Differential equations + exponenetial models

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    16

    Differential equations + exponenetial models

    7a

    A differential equation is given by \frac{dx}{dt}=-kte^\frac{x}{2}, where k is a positive constant. Note: The power to e is x/2, tried making it bigger but still suck at using latex.

    (i) Solve the differential equation.
    (ii) Hence, given that x=6 when t=0, show that x=-2\ln(\frac{kt^2}{4}+\frac{1}{e^3}).

    7b The population of a colony of insects is decreasing according to the model \frac{dx}{dt}=-kte^\frac{x}{2}, where x thousands is the number of insects in the colony after time t minutes. Initially, there were 6000 insects in the colony.

    Given that k=0.004 , find:

    (i) the population of the colony after 10 minutes, giving your answer to the nearest hundred.

    (ii) the time after which there will be no insects left in the colony, giving your answer to the nearest 0.1 of a minute.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by gamboo View Post
    7a

    A differential equation is given by \frac{dx}{dt}=-kte^\frac{x}{2}, where k is a positive constant. Note: The power to e is x/2, tried making it bigger but still suck at using latex.

    (i) Solve the differential equation.
    (ii) Hence, given that x=6 when t=0, show that x=-2\ln(\frac{kt^2}{4}+\frac{1}{e^3}).

    7b The population of a colony of insects is decreasing according to the model \frac{dx}{dt}=-kte^\frac{x}{2}, where x thousands is the number of insects in the colony after time t minutes. Initially, there were 6000 insects in the colony.

    Given that k=0.004 , find:

    (i) the population of the colony after 10 minutes, giving your answer to the nearest hundred.

    (ii) the time after which there will be no insects left in the colony, giving your answer to the nearest 0.1 of a minute.
    Separate the variables:

    \frac{dx}{dt}=-kte^\frac{x}{2}

    \frac{dx}{e^\frac{x}{2}} = -kt dt

    Note that \frac{1}{e^\frac{x}{2}} = e^{-\frac{x}{2}}

    -2e^{-\frac{x}{2}} = -\frac{k}{2}t^2 +C where C is the constant of integration

    Now we need to use a data point to solve for C. In this case we have x=6 and t=0

    -2e^{-\frac{6}{2}} = -2e^{-3}= C which makes the equation:

    -2e^{-\frac{x}{2}} = -\frac{k}{2}t^2 - 2e^{-3}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2008
    Posts
    16
    Thanks, matches mine.

    for b I got (i) 3800 insects
    (ii) 30.8 minutes

    can anyone confirm this for me
    Follow Math Help Forum on Facebook and Google+

  4. #4
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by gamboo View Post
    Thanks, matches mine.

    for b I got (i) 3800 insects
    (ii) 30.8 minutes

    can anyone confirm this for me
    I got those two answers too
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. non linear differential eqns population models
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: June 7th 2010, 10:56 PM
  2. Differential Equation Models
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: March 31st 2010, 03:10 PM
  3. Math models and equations
    Posted in the Algebra Forum
    Replies: 3
    Last Post: January 14th 2009, 05:48 PM
  4. Replies: 2
    Last Post: July 17th 2007, 05:11 AM
  5. Replies: 3
    Last Post: February 19th 2006, 10:20 AM

Search Tags


/mathhelpforum @mathhelpforum