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Math Help - Urgent Differential Equation Modelling

  1. #1
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    Urgent Differential Equation Modelling

    Hi everyone,
    I have this problem that I really need somehelp with.

    the rate of spread of a rumour is proportional to the product of the fraction y of the population L that has heard the rumour and the fraction who have not.

    (1) Find the differential equation that is satisfied by y.

    This is my answer

    dy/dt = k * [y(L-y) / (L^2)]

    (2) Show that the solution with initial condition y(0) = y0 is given by

    y = (Ly0e^(kLt)) / ((L-y0)+y0e^(kLt))

    I am lost on this.

    (3) A school has 100 students. At 9am, 20 students have heard a rumour. At 11am 50 students have heard the rumour. At what time will 80 students will have heard the rumour.


    I dont know what to do here either, however, im guessing these are the initial conditions. i.e.

    y(0) = 20
    y(2) = 50

    Please help me
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by arguabsysi View Post
    Hi everyone,
    I have this problem that I really need somehelp with.

    the rate of spread of a rumour is proportional to the product of the fraction y of the population L that has heard the rumour and the fraction who have not.

    (1) Find the differential equation that is satisfied by y.

    This is my answer

    dy/dt = k * [y(L-y) / (L^2)]

    (2) Show that the solution with initial condition y(0) = y0 is given by

    y = (Ly0e^(kLt)) / ((L-y0)+y0e^(kLt))

    I am lost on this.

    (3) A school has 100 students. At 9am, 20 students have heard a rumour. At 11am 50 students have heard the rumour. At what time will 80 students will have heard the rumour.


    I dont know what to do here either, however, im guessing these are the initial conditions. i.e.

    y(0) = 20
    y(2) = 50

    Please help me
    I tried this differential equation out, and it worked:

    \frac{dy}{dt}=ky(L-y)

    This is not separable. However, if I write it as follows:

    \frac{dy}{dt}=kLy-ky^2 \implies \color{red}\boxed{\frac{dy}{dt}-kLy=-ky^2}.

    This equation has the form of a Bernoulli's Equation (I cover this in my Differential Equations Tutorial):

    \frac{dy}{dt}+P(t)y=Q(t)y^n (general form of Bernoulli Equation)

    If we make a substitution v=y^{1-n} the Bernoulli Eqn will become a linear equation we know how to solve.

    v=y^{1-2}=y^{-1} \implies y=v^{-1}

    We also need a substitution for \frac{dy}{dt}.

    We can get this by applying a definition of the chain rule: \frac{dy}{dt}=\frac{dy}{dv}\cdot \frac{dv}{dt} \implies \frac{dy}{dt}=-\frac{1}{v^2}\frac{dv}{dt}.

    Substituting this into the DE, we get:

    -\frac{1}{v^2}\frac{dv}{dt}-kL\frac{1}{v}=-k\frac{1}{v^2}

    Multiply both sides by -v^2 to get a linear DE:

    \frac{dv}{dt}+kLv=k

    Apply Integrating factor:

    \rho(t)=e^{\int P(t)\,dt}=e^{\int kL \,dt}=e^{kLt}

    It should be known by now: when we multiply through by the int. factor, we get on the left side of the equation the derivative of \rho(t)\cdot v

    Thus, our DE becomes:

    \left[e^{kLt}v\right]^{/}=ke^{kLt}

    Solving the DE, we get:

    e^{kLt}v=\frac{1}{L}e^{kLt}+C \implies v=\frac{1}{L}+Ce^{-kLt}

    We don't want v. We want y. So sub back in v=y^{-1}

    \therefore \frac{1}{y}=\frac{1}{L}+Ce^{-kLt}\implies y=\frac{1}{\frac{1}{L}+Ce^{-kLt}}=\frac{e^{kLt}}{\frac{e^{kLt}}{L}+C}

    When y(0)=y_0, we have:

    y_0=\frac{1}{\frac{1}{L}+C}\implies y_0\left(\frac{1}{L}+C\right)=1\implies \frac{1}{L}+C=\frac{1}{y_0} \implies C=\frac{1}{L}-\frac{1}{y_0}=\frac{L-y_0}{Ly_0}

    \therefore y=\frac{e^{kLt}}{\frac{e^{kLt}}{L}+\frac{L-y_0}{Ly_0}}\implies y=\frac{e^{kLt}}{\frac{y_0e^{kLt}+(L-y_0)}{Ly_0}} \implies \color{red}\boxed{y=\frac{Ly_0e^{kLt}}{(L-y_0)+y_0e^{kLt}}}

    I believe in this case y(0)=y_0=20 and L=100

    Thus,

    y=\frac{2000e^{100kt}}{(80)+20e^{100kt}}

    Since y(2)=50, we see that

    50=\frac{2000e^{200k}}{(80)+20e^{200k}}

    My calculator tells me that k=\frac{\ln(2)}{100}

    Thus, y=\frac{2000e^{\ln(2)t}}{(80)+20e^{\ln(2)t}}.

    Now find t when y(t)=80

    80=\frac{2000e^{\ln(2)t}}{(80)+20e^{\ln(2)t}} \implies \color{red}\boxed{t=4}.

    So at 1 PM, 80 students will have heard the rumor.

    (I hope my math was right! It does make sense!)

    Hope this answered your question!
    Last edited by Chris L T521; May 20th 2008 at 09:58 PM.
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by Chris L T521 View Post
    This is not separable. However, if I write it as follows:

    \frac{dy}{dt}=kLy-ky^2 \implies \color{red}\boxed{\frac{dy}{dt}-kLy=-ky^2}.

    Hope this answered your question!
    I hate to say this but it is separable...

     <br />
\frac{dy}{dt}=kLy-ky^2 \iff \frac{dy}{dt}=k(L-y)y<br />

    \frac{dy}{(L-y)y}=kdx

    You can use partial fractions on the left hand side or my favorite trick

    \frac{1}{(L-y)y}=\frac{1}{L}\cdot\frac{L}{(L-y)y}=\frac{1}{L}\cdot\frac{(L-y)+y}{(L-y)y}=\frac{1}{L}\left( \frac{(L-y)}{(L-y)y}+\frac{y}{(L-y)y}\right)=

    \frac{1}{L}\frac{1}{y}+\frac{1}{L}\frac{1}{(L-y)}

    Rock on!!!
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    I hate to say this but it is separable...

     <br />
\frac{dy}{dt}=kLy-ky^2 \iff \frac{dy}{dt}=k(L-y)y<br />

    \frac{dy}{(L-y)y}=kdx

    You can use partial fractions on the left hand side or my favorite trick

    \frac{1}{(L-y)y}=\frac{1}{L}\cdot\frac{L}{(L-y)y}=\frac{1}{L}\cdot\frac{(L-y)+y}{(L-y)y}=\frac{1}{L}\left( \frac{(L-y)}{(L-y)y}+\frac{y}{(L-y)y}\right)=

    \frac{1}{L}\frac{1}{y}+\frac{1}{L}\frac{1}{(L-y)}

    Rock on!!!
    I keep forgetting that...

    Thanks for catching that!

    Both ways are acceptable, but I think your way is easier.
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