A drug is administered intravenously at a constant rate of r mg/hour and is excreted at a rate proportional to the quantity present, with constant of proportionality k > 0.
(a) Set up a differential equation for the quantity, Q, in milligrams, of the drug in the body at time t hours. (Your answer will contain the unknown constants r and k.)
(b) Solve this differential equation, assuming there is no drug in the body initially. (Your answer will contain r and k.)
(c) What is the limiting long-run value of Q?

Answer:

a.)Q' = -kQ + r

b.)Q = [tex]\frac{r}{k} [ 1 - e^{-kt} ][/tex]

c.) Limiting long run value of Q = [tex]\frac{r}{k}[/tex]

Step-by-step explanation:

a.)

The rate of change is directly proportional to the quantity, Q

⇒[tex]\frac{dQ}{dt}[/tex] ∝ Q

⇒[tex]\frac{dQ}{dt}[/tex] = -kQ ( because it is decreasing )

Also given, quantity is increasing with a constant rate r

⇒[tex]\frac{dQ}{dt}[/tex] = -kQ + r

⇒Q' = -kQ + r

b.)

As we have

[tex]\frac{dQ}{dt}[/tex] = -kQ + r

⇒[tex]\frac{dQ}{-kQ + r}[/tex] = dt

⇒∫[tex]\frac{dQ}{-kQ + r}[/tex] = ∫dt

⇒-[tex]\frac{1}{k}log(-kQ + r) = t + C[/tex]

⇒log(-kQ + r) = -kt -kC

Taking exponential both side, we get

⇒-kQ + r = [tex]e^{-kt +A}[/tex]

⇒-kQ = -r + [tex]e^{-kt +A}[/tex]

⇒Q = [tex]\frac{r}{k}[/tex] - [tex]\frac{1}{k}[/tex][tex]e^{-kt +A}[/tex]

⇒Q = [tex]\frac{r}{k}[/tex] - [tex]\frac{1}{k}[/tex][tex]e^{-kt}.e^{A}[/tex]

⇒Q = [tex]\frac{r}{k}[/tex] - [tex]\frac{1}{k}[/tex][tex]Be^{-kt}[/tex]      .......(1)

Now,

At t = 0, Q = 0

0 = [tex]\frac{r}{k}[/tex] - [tex]\frac{1}{k}[/tex]B

⇒[tex]\frac{1}{k}B = \frac{r}{k}[/tex]

⇒B = r

∴ equation (1) becomes

Q = [tex]\frac{r}{k}[/tex] - [tex]\frac{r}{k}[/tex][tex]e^{-kt}[/tex]

⇒Q = [tex]\frac{r}{k} [ 1 - e^{-kt} ][/tex]

c.)

for limiting long run value of Q

[tex]\lim_{n \to \infty} Q = \lim_{n \to \infty}[/tex]   [tex]\frac{r}{k} [ 1 - e^{-kt} ][/tex]

               =  [tex]\frac{r}{k}[/tex][tex]\lim_{n \to \infty} [ 1 - e^{-kt} ]= \frac{r}{k} [ 1 - e^{\infty} ] = \frac{r}{k}[ 1-0][/tex]

               = [tex]\frac{r}{k}[/tex]

⇒Limiting long run value of Q = [tex]\frac{r}{k}[/tex]

Q' = -kQ + r

Q = r/k (1- [tex]\rm e^{-kt}[/tex])

The limiting long-run value of Q = [tex]\rm \frac{r}{k}[/tex]

A drug is administered intravenously at a constant rate of r mg/hour and is excreted at a rate proportional to the quantity present, with a constant of proportionality k > 0.

The differential equation is an equation that contains the derivative of the unknown function.

a) It is given that the rate of change is directly proportional to the quantity Q

[tex]\rm \frac{dQ}{dt}[/tex] ∝ Q

[tex]\rm \frac{dQ}{dt}[/tex] = -kQ

So, Q' = -kQ + r

where quantity Q is increasing with constant rate r and k is unknown constant.

b) Q' = -kQ + r

[tex]\rm \frac{dQ}{dt}[/tex] = -kQ + r

[tex]\rm \frac{dQ}{-kQ+r}=dt\\[/tex]

[tex]\int\limits {\rm \frac{dQ}{-kQ+r} = \int\limits dt[/tex]

log(-kQ + r) = -kt -kC

by taking exponential both side, we get

-kQ + r = [tex]e^{-kt+A}[/tex]

-kQ = -r + [tex]e^{-kt+A}[/tex]

-Q /k= -r/k + [tex]e^{-kt+A}[/tex]/k

Q = -r/k -[tex]\rm \frac{1}{k}Be^{-kt}[/tex]

At t = 0, Q = 0

[tex]\rm \frac{1}{k} B=\frac{r}{k}[/tex]

B = r

by substituting the value in the above equation

Q = r/k (1- [tex]\rm e^{-kt}[/tex])

c) The limiting long-run value of Q

[tex]\rm \lim_{n \to \infty} Q= \lim_{n \to \infty} \frac{r}{k} [1-e^{-kt} ]\\\rm= \frac{r}{k}\lim_{n \to \infty} [1-e^{-kt} ]\\\rm =\frac{r}{k}[1-0]\\\rm=\frac{r}{k}[/tex]

The value Q =[tex]\rm \frac{r}{k}[/tex]

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