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

`C(t)=(r)/(k)-((r-kC_o)/(k))e^(-kt)`

Step-by-step explanation:

The differential equation is given as:

`(dC)/(dt) = r- kC`

`(dC)/(r- kC) = dt`

Taking integral of both sides; we have:

`\int\limits (dC)/(r- kC) = \int\limits dt`

`-(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=(r)/(k)-(A)/(k)e^(-kt)`

`C(t)=(r)/(k)-(A)/(k)e^(-kt)` ------- equation (1)

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

C(0)=
`(r)/(k)-(A)/(k)e^0` (where; A is an integration constant)

`C_o = (r)/(k)- (A)/(k)`

`C_o=(r-A)/(k)`

`kC_o=r-A`

`A=r-kC_o`

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

`C(t)=(r)/(k)-((r-kC_o)/(k))e^(-kt)`