Question

Drug Dissemination A mathematical model for the rate at which a drug disseminates into the bloodstream is given by

$\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{r}\mathbf{-}\mathit{k}\mathit{x}$

where r and k are positive constants. The function x(t) describes the concentration of the drug in the bloodstream at time t.

1. Since the DE is autonomous, use the phase portrait concept of Section 2.1 to find the limiting value of x(t) as t : '.
2. Solve the DE subject to$\mathit{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$. Sketch the graph of x(t) and verify your prediction in part (a). At what time is the concentration one-half this limiting value?

Short answer

(a) As we can see from the graph, when the limiting value of x equals r/k, t goes to infinity.

$$\begin{gathered} b)x=\frac{r}{k} \\ c)x=\frac{r}{k}-\frac{r}{k}{e}^{-kt} \end{gathered}$$

Step-by-step solution

Given Information.

The given value is:

$\frac{dx}{dt}=r-kx$

Graphing the phase portrait and solution curve

This is the autonomous differential equation's phase portrait and solution curve.

$\frac{dx}{dt}=r-kx$

where x has a limiting value of $\frac{r}{k}$and $t=\infty$.

At time t, the drug's concentration in the bloodstream

We have a differential equation that describes medication absorption into the bloodstream.

$\frac{dx}{dt}=r-kx$

with the predicament

$x\left(0\right)=0$

If this differential equation is of first order and separable, we can solve it as follows:

$\begin{array}{rcl}\frac{dr}{r-kx}& =& dt\\ \int \frac{1}{r-kx}dx& =& \int dt\\ -\frac{1}{k}\int \frac{-k}{r-kx}dx& =& \int dt\end{array}$

$\begin{array}{rcl}\ln (r-kx)& =& -kt+c_1\\ e^{\ln (r-kx)}& =& e^{-kt+c_1}\\ r-kx& =& e^{c_1}{e}^{-kt}\\ kx& =& r-c{e}^{-kt}\end{array}$

Then there's

$x\left(t\right)=\frac{r}{k}-n{e}^{-kt}$ (1)

After that, we must apply the point of condition (x,t)=(0,0) into equation(1) to obtain the value of constant n.

$0=\frac{r}{k}-n{e}^0$

Then there's

$n=\frac{r}{k}$

then, in equation (1), substitute the value of constant n, and we have

$x=\frac{r}{k}-\frac{r}{k}{e}^{-kt}$

is the drug's concentration in the blood at time t.

Conclusion

(a) As we can see from the graph, when the limiting value of x equals r/k, t goes to infinity.

$$\begin{gathered} b)x=\frac{r}{k} \\ c)x=\frac{r}{k}-\frac{r}{k}{e}^{-kt} \end{gathered}$$