Question

A glucose solution is administered intravenously into the bloodstream at a constant rate r. As the glucose is added, it is converted into other substances and removed from the bloodstream at a rate that is proportional to the concentration at that time. Thus a model for the concentration C = C(t) of the glucose solution in the bloodstream is dC dt = r − k C where k is a positive constant. (a) Suppose that the concentration at time t = 0 is C0. Determine the concentration at any time t by solving the differential equation. C(t)

Answer 1

Answer:

$C(t)=\frac{r}{k}-(\frac{r-kC_o}{k})e^{-kt}$

Explanation:

The differential equation is given as:

$\frac{dC}{dt} = r- kC$

$\frac{dC}{r- kC} = dt$

Taking integral of both sides; we have:

$\int\limits \frac{dC}{r- kC} = \int\limits dt$

$-\frac{1}{k} In(r-kC) = t+D\\In(r-kC)=-kt-kD$

$r-kC=e^{-kt-kD}$

$r-kC=e^{-kt}e^{-kD}$

$r-kC=Ae^{-kt}$

$kC=r-Ae^{-kt}$

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

$C(t)=\frac{r}{k}-\frac{A}{k}e^{-kt}$     ------- equation (1)

If C(0)= $C_o$ ; we have:

C(0)= $\frac{r}{k}-\frac{A}{k}e^0$         (where; A is an integration constant)

$C_o = \frac{r}{k}- \frac{A}{k}$

$C_o=\frac{r-A}{k}$

$kC_o=r-A$

$A=r-kC_o$

Substituting $A=r-kC_o$ into equation (1); we have;

$C(t)=\frac{r}{k}-(\frac{r-kC_o}{k})e^{-kt}$