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Differential Equation Help

  1. Mar 16, 2005 #1
    I don't know why, but I am stuck on this seemingly easy question. Here's the question and the work I've done.

    ---------------------
    A certain model for spread of rumors states that [tex]\frac{dy}{dt} = 3y(3-2y)[/tex] , where [tex]y[/tex] is the proportion of the population that has heard the rumor at time [tex]t[/tex]. What proportion of the population has heard the rumor when it is spreading the fastest?

    --------

    Ok. You are given the derivative of the proportion function, so setting it equal to 0 will give you when it is changing the fastest/slowest. Solving the equation [tex]3y(3-2y) = 0[/tex] you get 0 and 1.5....

    Next part is to find the original equation and evaluate it at 1.5. So I will need to separate the variables, and when I do I get:

    [tex]\frac{1}{3y(3-2y)}dy = dt[/tex]

    This integral (I did it on my calculator) is [tex]\frac{-\ln{\frac{\mid2x-3\mid}{\mid{x}\mid}}}{9}[/tex]

    When I evaulate 1.5 I get [tex]\infty[/tex]

    Help me please.
    Jameson
     
    Last edited: Mar 16, 2005
  2. jcsd
  3. Mar 16, 2005 #2
    Your procedure isn't quite right. Can you explain why the time when the proportion is changing fastest is when [tex]y^\prime = 0[/tex]?
     
  4. Mar 16, 2005 #3
    Also you need an initial condition.
     
  5. Mar 16, 2005 #4
    You need to take the derivative one more time before you set it equal to zero. You are maximizing dy/dt, not y.

    edit: sorry that wasn't very clear--I should say take the derivative with respect to y, since they have given you dy/dt as a function of y.
     
    Last edited: Mar 16, 2005
  6. Mar 16, 2005 #5
    Way to give away the answer to my question! :p
     
  7. Mar 16, 2005 #6
    Alright then. So [tex]\frac{d^{2}y}{dx^{2}} = 9 - 12y[/tex] and setting it equal to zero you get [tex] y = \frac{3}{4}[/tex]

    Is my integral correct from my first post correct? So now I can just plug .75 in for y?
     
  8. Mar 16, 2005 #7
    Can anyone give their thoughts to this question?
     
  9. Mar 16, 2005 #8
    What is the initial condition?
     
  10. Mar 17, 2005 #9

    saltydog

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    Well, its been a day so I'd like to write this one up if you've moved on (it took me this long to figure it out so I couldn't help at the start). Can someone check it. I don't want to make mistakes.

    The equation modeling rumor spread is:

    [tex]\frac{dy}{dt} = 3y(3-2y)[/tex]

    Separating variables and integrating from [itex]y_0[/itex] to y yields:

    [tex]ln|{\frac{2y}{3-2y}}|=9t+k[/tex]

    or:

    [tex]y(t)=\frac{1}{2}[\frac{3e^{9t+k}}{1+e^{9t+k}}][/tex]

    where:

    [tex] k=ln|\frac{2y_0}{3-2y_0}|[/tex]

    Since y is a proportion: [itex] 0<y\leq1[/itex], (assume y>0 since if no one knows the rumor at time 0 then it won't spread) the logarithm quantity is always positive and thus I can omit the absolute values.

    A plot (for y(0)=0.1) is attached. Looking at the plot, one can see that the rate of y is fastest at the point of inflection, that is, when the second derivative is zero. But we know what the first derivative is:

    [tex]y^{'}=9y-6y^2[/tex]

    Thus:

    [tex]y^{''}=9y^{'}-12yy^{'}[/tex]

    or substituting in the first derivative:

    [tex]y^{''}=72y^3-162y^2+81y=0[/tex]

    Solving this cubic equation, yields the roots:

    0, 3/4, and 3/2.

    Since y is a proportion between 0 and 1, we take the root 3/4 and conclude the rumor is spreading fastest when 75% of the group knows about it. Using the logarithm version of the solution, we can plug in 0.75 and determine, for a specific initial condition, the time when this occurs.
     

    Attached Files:

    Last edited: Mar 17, 2005
  11. Mar 18, 2005 #10

    saltydog

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    Ok, thanks to Daniel (from another post) I now understand why y cannot be 0 or 3/2: separating variables, one assumes that y can't be this since that would be dividing by zero. This is the reason we neglect absolute values and also why we must choose 3/4 as the root to the cubic. Yea, I know it's basic and in every Calculus text; I don't claim to be a wiz.
     
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