Graph the direction field of the differential equation below: (b) y-y(y -2)2

Graph the direction field of the differential equation below: (b) y-y(y -2)2

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  1. simplest reason is that Infinity is not a number, it is an idea.

    So 1∞ is a bit like saying 1beauty or 1tall.

    Maybe we could say that 1∞= 0, ... but that is a problem too, because if we divide 1 into infinite pieces and they end up 0 each, what happened to the 1?

    In fact 1∞ is known to be undefined.

    But We Can Approach It!

    So instead of trying to work it out for infinity (because we can't get a sensible answer), let's try larger and larger values of x:

    graph 1/x

    x 1x

    11.00000

    20.50000

    40.25000

    100.10000

    1000.01000

    1,0000.00100

    10,0000.00010

    Now we can see that as x gets larger, 1x tends towards 0

    We are now faced with an interesting situation:

    Step-by-step explanation:

  2. The answer is shown below

    Step-by-step explanation:

    Let y(t) be the fraction of the population that has heard the rumor at time t and assume that the rate at which the rumor spreads is proportional to the product of the fraction y of the population that has heard the rumor and the fraction 1−y that has not yet heard the rumor.

    a)

    [tex]\frac{dy}{dt}\ \alpha\ y(1-y)[/tex]

    [tex]\frac{dy}{dt}=ky(1-y)[/tex]

    where k is the constant of proportionality, dy/dt =  rate at which the rumor spreads

    b)

    [tex]\frac{dy}{dt}=ky(1-y)\\\frac{dy}{y(1-y)}=kdt\\\int\limits {\frac{dy}{y(1-y)}} \, =\int\limit {kdt}\\\int\limits {\frac{dy}{y}} +\int\limits {\frac{dy}{1-y}} =\int\limit {kdt}\\\\ln(y)-ln(1-y)=kt+c\\ln(\frac{y}{1-y}) =kt+c\\taking \ exponential \ of\ both \ sides\\\frac{y}{1-y} =e^{kt+c}\\\frac{y}{1-y} =e^{kt}e^c\\let\ A=e^c\\\frac{y}{1-y} =Ae^{kt}\\y=(1-y)Ae^{kt}\\y=\frac{Ae^{kt}}{1+Ae^{kt}} \\at \ t=0,y=10\%\\0.1=\frac{Ae^{k*0}}{1+Ae^{k*0}} \\0.1=\frac{A}{1+A} \\A=\frac{1}{9} \\[/tex]

    [tex]y=\frac{\frac{1}{9} e^{kt}}{1+\frac{1}{9} e^{kt}}\\y=\frac{1}{1+9e^{-kt}}[/tex]

    At t = 2, y = 40% = 0.4

    c) At y = 75% = 0.75

    [tex]y=\frac{1}{1+9e^{-0.8959t}}\\0.75=\frac{1}{1+9e^{-0.8959t}}\\t=3.68\ days[/tex]

  3. The slopes shows that the direction of the field is from -2 to +2, with three point charges, q₁, q₂ and q₃ at -2, 0 and +2 respectively.

    Explanation:

    Given;

    The slope, dy/dx = 2x(y-6) - 4

    2x(y-6) - 4 = 2xy - 12x - 4, divide through by 'x'

    dy/dx = 2y -12 - 4/x

    The slopes of the linear elements on the lines, x =0, y = 5, y = 6, y = 7.

    At x = 0, and y = 5

    dy/dx = 2y -12 - 4/x

    dy/dx = 2(5) - 12 = -2

    At x = 0, and y = 6

    dy/dx = 2y -12 - 4/x

    dy/dx = 2(6) - 12 = 0

    At x = 0, and y = 7

    = 2y -12 - 4/x

    dy/dx = 2(7) - 12 = 2

    Therefore, the slopes shows that the direction of the field is from -2 to +2, with three point charges, q₁, q₂ and q₃ at -2, 0 and +2 respectively.

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