Download the App!

Get 24/7 study help with the Numerade app for iOS and Android! Enter your email for an invite.

Differential equations have been used extensively in the study of drug dissolution for patients given oral medication. One such equation is the Weibull equation for the concentration c(t) of the drug:

$ \frac {dc}{dt} = \frac {k}{t^b} (c_a - c) $where $ k $ and $ c_a $ are positive constants and $ 0 < b < 1. $ Verify that

$ c(t) = c_a (1 - e^{-at^{1-b}}) $is a solution of the Weibull equation for $ 1 > 0, $ where $ a = k/(1 - b). $ What does the differential equation say about how drug dissolution occurs?

Get the answer to your homework problem.

Try Numerade free for 7 days

Like

Report

If $c(t)=c_{s}\left(1-e^{-\alpha t^{1-b}}\right)=c_{s}-c_{s} e^{-\alpha t^{2-b}}$ for $t>0,$ where $k>0, c_{s}>0,0<b<1,$ and $\alpha=k /(1-b),$ then\[\frac{d c}{d t}=c_{s}\left[0-e^{-\alpha t^{1-b}} \cdot \frac{d}{d t}\left(-a t^{1-b}\right)\right]=-c_{s} e^{-\alpha t^{2-b}} \cdot(-\alpha)(1-b) t^{-b}=\frac{\alpha(1-b)}{t^{b}} c_{s} e^{-\alpha t^{1-b}}=\frac{k}{t^{b}}\left(c_{s}-c\right) . \text { }\]

03:15

Vishal Parmar

Calculus 2 / BC

Chapter 9

Differential Equations

Section 1

Modeling with Differential Equations

Campbell University

Baylor University

University of Nottingham

Lectures

13:37

A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself and its derivatives of various orders. An ordinary differential equation (ODE) is a differential equation containing one or more derivatives of a function and their rates of change with respect to the function itself; it can be used to model a wide variety of phenomena. Differential equations can be used to describe many phenomena in physics, including sound, heat, electrostatics, electrodynamics, fluid dynamics, elasticity, quantum mechanics, and general relativity.

33:32

02:54

13-15 Drug dissolution Dif…

01:12

The Noyes-Whitney equation…

05:10

Applications of first orde…

11:09

Solving the differential e…

03:45

02:03

Use separation of variable…

So what we're going to start off with, um is this equation where we have that d C D t is equal to K some constant over t to be times he as minus c. So with this in mind, we can divide um, both sides by this portion right here. And then we can put the DT over and replacement. So we'll do that. This will become DT little have this, Then what we can do is take the integral of both sides. So will integrate this from zero to see, and we will integrate this. Okay, deputy, what we end up getting as a result, looking at this is going to be the natural log of C s minus C. Over. Um, we want to keep this within. I'm seeing so CS minus C over CS and then on this side. What we end up seeing as a result is going to be a negative k times t to the one minus B and that will be divided by one minus beef. With this, we can now raise both sides to the power of E. The reason we'll do that is so we can get rid of this natural log and ultimately, this right here. We'll have e to the power of this whole thing. Once we've done that, we can simplify this a little further. To be one minus c oversee base s, and we can simplify this to ultimately be e to the negative. Okay, It's a negative U t the one minus b where we assume, uh, that you is k over one minus b country kindness. Um, we can add the one over and then multiplied by C over s. So for our final answer, we end up getting exactly what we wanted to show, which is that C is equal to C s times one plus e to the negative u t to the one minus b, and that will be our final result.

View More Answers From This Book

Find Another Textbook

Numerade Educator