# ODE Word Problem

• Apr 16th 2010, 01:46 PM
cdlegendary
ODE Word Problem
Drug Metabolism The rate at which a drug is absorbed into the bloodstream is modeled by the first-order differential equation:

http://homework.math.ucsb.edu/webwor...a12d8e9041.png

where $a$ and $b$ are positive constants and $C(t)$ denotes the concentration of the drug in the bloodstream at time . Assume no drug is initially present in the bloodstream.

Find a formula for $C(t)$. You may need to use $a$ and $b$ in your answer.

I think it may be the constants a and b, but something's throwing me off on how to do this. So far I've tried:

$dC/dt = a-bC(t)$
$dC/dt +C(t)= a$

And then I'm guessing I need to use an integrating factor? Or am I just completely wrong here. Any help is greatly appreciated!
• Apr 16th 2010, 02:27 PM
Prove It
Quote:

Originally Posted by cdlegendary
Drug Metabolism The rate at which a drug is absorbed into the bloodstream is modeled by the first-order differential equation:

http://homework.math.ucsb.edu/webwor...a12d8e9041.png

where $a$ and $b$ are positive constants and $C(t)$ denotes the concentration of the drug in the bloodstream at time . Assume no drug is initially present in the bloodstream.

Find a formula for $C(t)$. You may need to use $a$ and $b$ in your answer.

I think it may be the constants a and b, but something's throwing me off on how to do this. So far I've tried:

$dC/dt = a-bC(t)$
$dC/dt -C(t)= a$

And then I'm guessing I need to use an integrating factor? Or am I just completely wrong here. Any help is greatly appreciated!

$\frac{dC}{dt} = a - b\,C$

$\frac{dC}{dt} + b\,C = a$.

This is first order linear, so use the integrating factor $e^{\int{b\,dt}} = e^{b\,t}$.

Multiplying through gives

$e^{b\,t}\frac{dC}{dt} + b\,e^{b\,t}C = a\,e^{b\,t}$

$\frac{d}{dt}(C\,e^{b\,t}) = a\,e^{b\,t}$

$C\,e^{b\,t} = \int{a\,e^{b\,t}\,dt}$

$C\,e^{b\,t} = \frac{a\,e^{b\,t}}{b} + d$

$C = \frac{a}{b} + d\,e^{-b\,t}$.

When $t = 0, C = 0$.

So $0 = \frac{a}{b} + d\,e^{0}$

$0 = \frac{a}{b} + d$

$d = -\frac{a}{b}$.

Therefore

$C = \frac{a}{b} - \frac{a}{b}\,e^{-b\,t}$.
• Apr 23rd 2010, 05:43 PM
student150150
How could you find the limiting concentration as t approaches infinity? and at what time does it reach its half life?

I can't do this problem!!! =[