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Math Help - Differential equation - involving 2 rates, very confused

  1. #1
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    Thumbs down Differential equation - involving 2 rates, very confused

    In this model, let the number of those not yet ill, but susceptibles, be S(t); let the number currently infected be I(t). The rate at which the susceptibles are infected is given by:

    S'(t) = dS/dt = -alpha(SI)

    (Where alpha > 0 is a constant). The rate at which the number of those infected can change depends on both the susceptibles getting ill and others recovering; this process is described by:

    I'(t) = dI/dt = alpha(SI) - beta(I)

    (Where beta > 0 is a constant). Form the differential equation I(S) and, given that I = 1 when S = N, find the solution for I(S).

    The part that really throws me is the stuff in blue :S how do I get the function I(S)? I tried dividing I'(t)/S'(t) to give dI/dS then integrating but this seemed totally wrong... Am I missing some sort of elementary trick here? Help very appreciated.
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  2. #2
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    Differential equation

    Hello mitch_nufc
    Quote Originally Posted by mitch_nufc View Post
    In this model, let the number of those not yet ill, but susceptibles, be S(t); let the number currently infected be I(t). The rate at which the susceptibles are infected is given by:

    S'(t) = dS/dt = -alpha(SI)

    (Where alpha > 0 is a constant). The rate at which the number of those infected can change depends on both the susceptibles getting ill and others recovering; this process is described by:

    I'(t) = dI/dt = alpha(SI) - beta(I)

    (Where beta > 0 is a constant). Form the differential equation I(S) and, given that I = 1 when S = N, find the solution for I(S).

    The part that really throws me is the stuff in blue :S how do I get the function I(S)? I tried dividing I'(t)/S'(t) to give dI/dS then integrating but this seemed totally wrong... Am I missing some sort of elementary trick here? Help very appreciated.
    I think that your method is perfectly OK. You just get:

    \frac{dI}{dS} = \frac{dI}{dt}\div \frac{dS}{dt}= \frac{(\alpha S - \beta)I}{-\alpha SI}

    = \frac{\beta - \alpha S}{\alpha S}

    \Rightarrow \int dI= \int\left(\frac{\beta}{\alpha S }-1 \right)dS

    which integrates OK.

    Plug in I=1 when S=N, and you're there.

    Grandad
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  3. #3
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    Thanks , I got I=c*ln(S/N)-S+1+N seems right
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  4. #4
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    Differential equation

    Quote Originally Posted by mitch_nufc View Post
    Thanks , I got I=c*ln(S/N)-S+1+N seems right
    So did I - with c = \frac{\beta}{\alpha}
    .
    Grandad
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