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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?

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 { }

\]

Differential Equations

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Numerade Educator

Harvey Mudd College

University of Michigan - Ann Arbor

Boston College

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.

California Baptist University

Differential Equations