Thread: Differential equation - involving 2 rates, very confused

1. 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.

2. Differential equation

Hello mitch_nufc
Originally Posted by mitch_nufc
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

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

4. Differential equation

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