Question

Intravenous infusion of glucose into the blood-stream of a patient is an important medical technique. To study this process, let G(t) be the amount of glucose in the patient’s blood stream t minutes after the process begins. Assume that glucose is infused into the bloodstream at a constant rate of k(in g/min). Also assume that at the same time, the glucose is converted and removed from the bloodstream at a rate proportional to the amount of glucose still present, with the proportionality constant r.(a) Write a first-order differential equation describing this process that should be satisfied by G(t).(b) Suppose that glucose is not present initially in the patient’s bloodstream; solve the differential equation in (a) above which satisfies the initial condition G(0) = 0.
(c) Find
lim t→∞G(t).

Answer 1

Answer on Question #79346 - Math - Calculus

(a)

$G^{\prime}(t)=k-rG(t).$

(b) Consider the homogeneous equation

$G^{\prime}(t)=-rG(t).$

The general solution of this equation is $G(t)=Ce^{-rt}$ where $C$ is an arbitrary constant. It easy to see that a particular solution of the equation $G^{\prime}(t)=k-rG(t)$ is $G(t)=\frac{k}{r}$. Thus the general solution of the equation $G^{\prime}(t)=k-rG(t)$ is $G(t)=\frac{k}{r}+Ce^{-rt}$. From the initial condition $G(0)=0$ we have $C=-\frac{k}{r}$. Finally $G(t)=\frac{k}{r}-\frac{k}{r}e^{-rt}$.

(c) From the general solution formula of the equation $G^{\prime}(t)=k-rG(t)$ it is easy to see that $\lim\limits_{t\rightarrow\infty}G(t)=\frac{k}{r}$ for any initial condition.

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