# Diff Eqs

1. Jun 25, 2013

### Zondrina

1. The problem statement, all variables and given/known data

A certain drug is being administered intravenously to a hospital patient. Fluid containing
5 mg/cm3 of the drug enters the patient’s bloodstream at a rate of 100 cm3 /h.

The drug is absorbed by body tissues or otherwise leaves the bloodstream at a rate proportional to the amount present, with a rate constant of 0.4 (h)-1.

(a) Assuming that the drug is always uniformly distributed throughout the bloodstream,
write a differential equation for the amount of the drug that is present in the bloodstream
at any time.

(b) How much of the drug is present in the bloodstream after a long time?

2. Relevant equations

3. The attempt at a solution

(a) Since the distribution of the drug is uniform, we know that there will be :

$(5 \frac{mg}{cm^3}) (100 \frac{cm^3}{h}) = 500 \frac{mg}{h}$

of fluid distributed per hour regardless.

The drug is being absorbed or leaving the body at a rate proportional to the amount present with a constant rate of 0.4 (h)-1. So when the drug is entering the body, mg > 0, when the drug leaves the body, mg < 0. Hence we must account for how much fluid enters and leaves the body as well. Thus we get the equation :

$\frac{d(mg)}{dt} = 500 - 0.4mg$

(b) I believe this amounts to solving this equation first so :

$\frac{d(mg)}{dt} = 500 - 0.4mg$
$\int \frac{1}{500 - 0.4mg} d(mg) = \int dt$

Cleaning this up a bit I got :

$mg(t) = ke^{-0.4t} + 1250$

Now we can observe the nature of the position with respect to time rather than the velocity. So as $t → ∞$ we see $mg(t) → 1250$.

So after 'forever' has happened there will be 1250 mg of the drug in the blood stream.

Does this look okay? I'm not very experienced with diff eqs word questions, but they really caught my eye today.

Last edited: Jun 25, 2013
2. Jun 25, 2013

### pasmith

The working is correct, but it would be better to have a single-letter name for the amount of drug in the bloodstream.

3. Jun 25, 2013

### Staff: Mentor

Looks fine to me. The only change I would make is to use a different variable, say A (for amount) rather than mg, which is a unit. mg might also be confused as the product of m and g, which appear in problems involving masses that are falling.