Mathematical Modelling question

[Boogzy]

Though this question is about medicine, the actual question has little to do with medicine and more to do with modelling ..

Homework Statement

A patient is put on an intravenous drip at time t=0, the drip supplies a drug into the patients bloodstream at a constant rate λ. At the same time (t=0) the patient is given M grams of the same drug orally which immediately starts dissolving at a rate directly proportional to the mass of the drug in the stomach (co-efficient of proportionalility α) The drug in the blood stream is eliminated from the blood stream at rate directly proportional to the mass of the drug in the blood (co-efficient of proportionalility β)

Find a model for the mass of the drug in the patients stomach and bloodstream at time t in terms of λ, α, β, M and t.

Homework Equations

Let S = S(t) = Mass of drug in stomach at time t.
Let B = B(t) = Mass of drug in bloodstream at time t.

The Attempt at a Solution

Stomach :

\frac{dS}{dt} = -α.S ... then using separation of variables
\frac{dS}{α.S}= -dt ... integrate both sides to get

\frac{ln(α.S)}{α} = -t + C ... (where C is arbitrary constant)
ln(α.S) = -αt + C ... raising both sides to e, we get

α.S = e^{-αt+C}
α.S = e^{-αt}.e^{C} ... (e^{C} is an arbitrary constant)

S = \frac{C.e^{-αt}}{α}

Using: at t=0, S=M, we can find that C = M.α

S(t) = \frac{M.a.e^{-αt}}{α}
S(t) = M.e^{-αt}

I think its right up to here, but I'm stuggling with the bloodstream part..

Here's what I tried ...

In flow = λ+α.S(t)
Out flow = β.B(t)

\frac{dB}{dt}= λ + α.S(t) - β.B(t)

but this now has 2 dependant variables so I'm not too sure where to go.

Maybe substituting S(t) = M.e^{-αt} to get

\frac{dB}{dt}= λ + α.M.e^{-αt} - β.B(t)

but then I wouldn't know how to solve this differential equation

 

[gneill]

Your S(t) yields how much mass remains in the stomach. Ergo, the rate at which the mass goes into the bloodstream from the stomach is -dS(t)/dt.

You might find that you can solve the resulting differential equation fairly easily using Laplace transforms.