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Math Help - differencial equations

  1. #1
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    differencial equations

    1.A fruit fly population of 182 flies is in a closed container. The number of flies grows exponentially, reaching 340 in 18 days. Find the doubling time (time for the population to double) to the nearest tenth of a day.

    19.9 days
    20.0 days
    18.0 days
    2.0 days

    2.One thousand dollars is invested at 5% continuous annual interest. This means the value of the investment will grow exponentially, with k equaling the decimal rate of interest. What will the value of the investment be after 7 1/2 years?
    $1,375.00
    $375.00
    $1,454.99
    $454.99

    3.A population grows exponentially, doubling in 4 hours. How long, to the nearest tenth of an hour, will it take the population to triple?
    8.0 hours
    10.3 hours
    6.3 hours
    6.6 hours


    4.The half-life of a newly discovered radioactive element is 30 seconds. To the nearest tenth of a second, how long will it take for a sample of 9 grams to decay to 0.72 grams?
    109.3 seconds
    42.5 seconds
    30.8 seconds
    99.7 seconds


    5.
    Find the particular solution to y dx = x dy given that y(8) = 2.

    x = 4y
    y = 4y
    ln|x|=ln|y|+ln6
    ln|x|=ln|y|+ln4

    6.Which of the following is a first-order homogeneous differential equation?
    dy/dx = (3x^3-2x^2y)/xy^2
    dy/dx=(x^2y^3 - y^5)/(2xy^5)
    dy/dx=(5xy-5y-5x)/y
    none
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by harry View Post
    1.A fruit fly population of 182 flies is in a closed container. The number of flies grows exponentially, reaching 340 in 18 days. Find the doubling time (time for the population to double) to the nearest tenth of a day.

    19.9 days
    20.0 days
    18.0 days
    2.0 days
    The population t later is:

    p(t) = p(0) e^{lambda t}

    so:

    340 = 182 e^{lambda 18}

    18 lambda = ln(340/182) ~= 0.624939

    lambda = 0.03471883.

    For the doubling time t_2, we have:

    364 = 182 e^{0.03471883 t_2}

    so:

    t_2 = ln(2)/0.03471883 ~=19.9646

    So to the nearest 1/10th of a day the doubling time is 20 days.

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by harry View Post

    4.The half-life of a newly discovered radioactive element is 30 seconds. To the nearest tenth of a second, how long will it take for a sample of 9 grams to decay to 0.72 grams?
    109.3 seconds
    42.5 seconds
    30.8 seconds
    99.7 seconds

    Mass remaining after t seconds is:

    M = M_0 2^(-t/30),

    so for this problem we have:

    0.72 = 9 2^(-t/30)

    t = (-30) log_2(0.72/9) = (-1/30) ln(0.72/9)/ln(2) ~= 109.3

    RonL
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by harry View Post

    5.
    Find the particular solution to y dx = x dy given that y(8) = 2.

    x = 4y
    y = 4y
    ln|x|=ln|y|+ln6
    ln|x|=ln|y|+ln4
    The DE may be rewritten:

    (1/x) dx = (1/y) dy

    and so integrated immeadiatly to:

    ln(|x|) = ln(|y|) + C,

    Now as y(8) = 2, we have:

    ln(8) = ln(2) + C,

    so

    C = ln(8) - ln(2) = ln(8/2) = ln(4)

    hence:

    ln(|x|) = ln(|y|) + ln(4).

    Except as the DE has a singularity in dy/dx at x=0 and we cannot integrate
    through a singularity this solution is only valid for x>0 (as the initial condition
    is in this segment of the real line

    RonL
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