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    Consider a total population N(t) containing n(t) diseased individuals. Suppose thaton average the members of the diseased population n(t) infect undiseased individualsat a rate α per diseased individual, per unit time, per undiseased individual, and thatthey die off at a rate one per time τ per diseased individual (τ is usually known asthe lifetime).
    a) If the population is initially all infected (noting that this means n(t) = N(t)since there is no recovery from this disease), write a differential equation for thesubsequent evolution of N(t) and solve it, given that the initial population is N0 attime t = 0.
    b) If the initial total population is N0, and the initial number of diseased individualsis n0 < N0, find a differential equation for the total population N(t) of this system.[Hints: The final answer should be of the form dN/dt = f(N, t) i.e. not involvingn(t) on the RHS. Find equations for the rate of change of N and for the rate ofchange of the number of healthy individuals nhealthy = N − n. Then note that youcan use the chain rule to find an equation for dnhealthy/dN which you can solve torelate nhealthy to N. Also note dA/dt + dB/dt = d(A + B)/dt.]

    Can anyone help here? I literally have no idea how to begin with either of these questions and feel I am missing something really obvious. I appreciate the help :-)
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    (Original post by Shadow11)
    ...
    For the first part you can forget about people becoming infected, as everyone is. So you just have a population and a die-off rate.

    Can you get an ODE from that?
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    (Original post by ghostwalker)
    For the first part you can forget about people becoming infected, as everyone is. So you just have a population and a die-off rate.

    Can you get an ODE from that?
    Well I have two different perspective answers depending on the wording of the question: if one dies by time tau then N(t)=N0-t/tau or if they all die by time tau in which case its the same except t/tau is multiplied by N0. But honestly I'm not sure, I've done tonnes of these styles of questions in the past and for some reason this one doesn't click with me.
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    (Original post by Shadow11)
    Well I have two different perspective answers depending on the wording of the question: if one dies by time tau then N(t)=N0-t/tau or if they all die by time tau in which case its the same except t/tau is multiplied by N0. But honestly I'm not sure, I've done tonnes of these styles of questions in the past and for some reason this one doesn't click with me.
    OK.

    You're initially looking set up an ODE, and looking for a formula for the rate of change of N(t).

    "die off at a rate one per time τ per diseased individual"

    I think you're missing the bit I've put in bold in your thinking. The death rate is not just dependent on time, but also on the number of currently diseased individuals - which in this case equals the number of currently living individuals, as they're all diseased.
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    (Original post by ghostwalker)
    OK.

    You're initially looking set up an ODE, and looking for a formula for the rate of change of N(t).

    "die off at a rate one per time τ per diseased individual"

    I think you're missing the bit I've put in bold in your thinking. The death rate is not just dependent on time, but also on the number of currently diseased individuals - which in this case equals the number of currently living individuals, as they're all diseased.
    Using what you have suggested I've think I've come up with something that to me looks like the kind of equation I was hoping to get. It makes more sense to my real, world understanding:
    N=N0e^-t/tau. Thank you for how helpful you've been so far.
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    (Original post by Shadow11)
    Using what you have suggested I've think I've come up with something that to me looks like the kind of equation I was hoping to get. It makes more sense to my real, world understanding:
    N=N0e^-t/tau. Thank you for how helpful you've been so far.
    For the second part, there's quite a bit of advice in the question itself. Have a go with that and see what you can do. If it's not coming out, post some working and say where the problem is.
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    (Original post by ghostwalker)
    For the second part, there's quite a bit of advice in the question itself. Have a go with that and see what you can do. If it's not coming out, post some working and say where the problem is.
    I've managed to get an answer for part b) but while the working I used to get there makes sense, I feel the answer I get does not, so I hope you will be able to spot my mistake, or if there isn't a mistake please explain how I am able to answer the following question :
    c)Check that your equation makes sense i) when n0 = 0 and ii) when n0 = N0 and iii) when α->0
    Provide solutions to support you statements where possible.


    Working:
    dN/dt = -n/τ , dn/dt = αn(N-n) – n/τ
    Therefore as dA/dt + dB/dt = d(A + B)/dt, so dA/dt – dB/dt= d(A - B)/dt
    d(N-n)/dt = dN/dt – dn/dt
    d(N-n)/dt = -n/τ –( αn(N-n) – n/τ)
    d(N-n)/dt = -αn(N-n)
    N-n = nhealthy
    d(N-n)/dt x dt/dN = dnhealthy/dN
    = -αn(N-n) x –τ/n
    = ατ(N-n)
    = ατnhealthy
    Int(1/nhealthy dnhealthy) = ατInt(dN)
    ln(nhealthy) = ατN + ln(k)
    ke^(ατN) = nhealthy
    N- nhealthy = n
    = N - ke^(ατN)
    Therefore dN/dt = (-N+ke^(ατN))/τ

    I don't feel that this answer is right due to the positive power of e and if it is right please suggest how I go about doing c).

    Thanks again









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    (Original post by Shadow11)
    I've managed to get an answer for part b) but while the working I used to get there makes sense, I feel the answer I get does not, so I hope you will be able to spot my mistake, or if there isn't a mistake please explain how I am able to answer the following question :
    1. You *really* need to use Latex.

    2. For what it's worth, I had a quick go at this last night and got the same result as you:

    \frac{dh}{dN} = \alpha \tau h

    and integrated this to get:

    h(t)=(N_0-n_0)e^{\alpha \tau (N(t)-N_0)}

    where h is the no of healthy members but I went no further. I guess it's pretty straightforward from there though.
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    (Original post by atsruser)
    1. You *really* need to use Latex.

    2. For what it's worth, I had a quick go at this last night and got the same result as you:

    \frac{dh}{dN} = \alpha \tau h

    and integrated this to get:

    h(t)=(N_0-n_0)e^{\alpha \tau (N(t)-N_0)}

    where h is the no of healthy members but I went no further. I guess it's pretty straightforward from there though.
    Why do you get N - N_0 in the power?
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    (Original post by Shadow11)
    Why do you get N - N_0 in the power?
    That's the way it seemed to come out - I'll put up my working later when I've checked it.
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    (Original post by Shadow11)

    Int(1/nhealthy dnhealthy) = ατInt(dN)
    ln(nhealthy) = ατN + ln(k)
    At this point I would have put it as:

    \ln (N(t)-n(t))=\alpha \tau N(t)+c for some constant c

    And then substitute in your initial condition, N_0,n_0, for t=0 to determine c.
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    (Original post by ghostwalker)
    At this point I would have put it as:

    \ln (N(t)-n(t))=\alpha \tau N(t)+c for some constant c

    And then substitute in your initial condition, N_0,n_0, for t=0 to determine c.
    Why is c needed when i can just figure the k using log rules to be N_0 - n_0. By logic, this must be the starting number of healthy people surely?:confused:
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    (Original post by Shadow11)
    Why is c needed when i can just figure the k using log rules to be N_0 - n_0. By logic, this must be the starting number of healthy people surely?:confused:
    As long as you can figure k out correctly, that's fine.

    It's just clearer, to me anyway, where the N_0 - n_0 in the exponent comes from, which you were querying.
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    (Original post by ghostwalker)
    As long as you can figure k out correctly, that's fine.

    It's just clearer, to me anyway, where the N_0 - n_0 in the exponent comes from, which you were querying.
    Wait are we talking about N_0 - n_0 for the coefficient of e or the power N(t)-N_0 because i dont get where the latter comes from, but do understand the former. Is there some extra constant or something that i am missing?
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    (Original post by Shadow11)
    Wait are we talking about N_0 - n_0 for the coefficient of e or the power N(t)-N_0 because i dont get where the latter comes from, but do understand the former. Is there some extra constant or something that i am missing?
    Sorry, yes, it's the N(t)-N_0 in the exponent.

    Using the +c constant and substituting makes it clear, IMO.

    Edit: Just going out for the night now.
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    (Original post by ghostwalker)
    Sorry, yes, it's the N(t)-N_0 in the exponent.

    Using the +c constant and substituting makes it clear, IMO.

    Edit: Just going out for the night now.
    So can we confirm the answer to this is
    \frac{dN}{dt} = \frac{(N_0-n_0)e^{\alpha \tau (N(t)-N_0)} - N}{\tau} and if so how do i go about rearranging it for question d)
    d) The Lambert function W(x) is the solution to x = W eW . It can be numericallytabulated in much the same way as L(x) = log(x) which is the solution to x = eL.You will almost certainly not have seen it at A-level but no matter; given that such afunction exists, show using your differential equation that the population approachesa constant value
    Nconst = −(ατ )^−1 W (-ατ (N0-n0)e^-ατN0)
    I assume you put the differential equation equal to zero but i can't get a single Nconst on its own.
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    (Original post by Shadow11)
    ...
    I've already chucked my working, so I'll have to work it through again from scratch. Not got time right now.
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    (Original post by Shadow11)
    So can we confirm the answer to this is
    \frac{dN}{dt} = \frac{(N_0-n_0)e^{\alpha \tau (N(t)-N_0)} - N}{\tau}
    Yes (one word - took an hour's work)

    and if so how do i go about rearranging it for question d)
    d) The Lambert function W(x) is the solution to x = W eW . It can be numericallytabulated in much the same way as L(x) = log(x) which is the solution to x = eL.You will almost certainly not have seen it at A-level but no matter; given that such afunction exists, show using your differential equation that the population approachesa constant value
    Nconst = −(ατ )^−1 W (-ατ (N0-n0)e^-ατN0)
    I assume you put the differential equation equal to zero but i can't get a single Nconst on its own.
    Nope, can't see how to do that bit. The minus N at the end is causing problems, which makes me wonder if it's correct. But assuming it approaches a constant then setting dN/dt=0 would be my approach.
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    (Original post by ghostwalker)
    Yes (one word - took an hour's work)



    Nope, can't see how to do that bit. The minus N at the end is causing problems, which makes me wonder if it's correct. But assuming it approaches a constant then setting dN/dt=0 would be my approach.
    Thank you very much for all your help and while i agree the final question leads to questioning of the answer, it seems to make sense as far as the initial equations are concerned and makes sense it terms of the test conditions in question c). There might be some rearrangement I'm missing for part d) but overall a and b are the important questions.
    Again thank you, it is very much appreciated
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    (Original post by Shadow11)
    Again thank you, it is very much appreciated
    OK, just "seen" the solution to finding the constant.

    Setting dN/dt= 0, cancel tau and move the N over to get:

    N=(N_0-n_0)e^{\alpha\tau(N-N_0)}


    Multiply both sides by -\alpha\tau e^{-\alpha\tau N}

    This gets all references to N over to the left, and the correct form for the Lambert function.
 
 
 
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