# Differential equation modeling glucose in a patient's body

#### ForceBoy

Problem Statement
"A hospital patient is fed glucose intravenously (directly to the bloodstream) at a rate of r units per minute. The body removes glucose from the bloodstream at a rate proportional to the amount Q(t) present in the bloodstream at time t. " (Finney; Weir; Giordano, 452)

A) write the differential eq.
B) Solve the diff. eq $Q(0) = Q_0$
C) find the limit as t goes to infinity
Relevant Equations
The chapter this problem is found in is one on separable differential equations
The rate at which glucose enters the bloodstream is $r$ units per minute so:

$\frac{dI}{dt} = r$

The rate at which it leaves the body is:

$\frac {dE}{dt} = -k Q(t)$

Then the rate at which the glucose in the body changes is:

A) $Q'(t) = \frac{dI}{dt} + \frac {dE}{dt} = r - k Q(t)$

I don't see how this is a separable differential equation. I still try to solve it.

$\frac{dQ}{dt} + k Q = r$

$Q e^{kt} = \int r e^{kt} dt$

$Q = \frac{r}{e^{kt}}\frac{e^{kx}}{k}$

B) $Q(t) = \frac{r}{k}$

This tells me that the glucose in the bloodstream at any point in time will be a constant. I know this is wrong. It would be appreciated if someone could point me onto the right path to solve this diff. eq. Thank you.

Related Calculus and Beyond Homework News on Phys.org

#### Orodruin

Staff Emeritus
Homework Helper
Gold Member
2018 Award
You missed the integration constant which you will need to adapt the solution to the initial condition.

I don't see how this is a separable differential equation.
What do you know about separable ODEs? The point of a separable ODE is that you should be able to write it on the form
$$y'(t) f(y(t)) = g(t).$$
This is possible in this situation.

#### ForceBoy

What do you know about separable ODEs? The point of a separable ODE is that you should be able to write it on the form

y′(t)f(y(t))=g(t).​
Thank you. I put my equation in the form you gave and solved just fine:

$\frac{dQ}{dt} = Q'(t)$
$r - kQ(t) = g(Q(t))$
________________________

$Q'(t) = g(Q(t))$
$\frac{Q'(t)}{g(Q(t))} =1$

$Q'(t) f(Q(t)) = 1$

$\frac{dQ}{dt} \frac{1}{r-kQ} = 1$

$\frac{dQ}{r-kQ} = dt$

$\int\frac{dQ}{r-kQ} = \int dt$

$\ln|r-kQ | = t +C$ (Can't forget the C now, thanks)

$r-kQ = Ae^{t}$

$Q = \frac{r - A e^{t}}{k}$

If the above is correct, then I can solve the rest of the problem.

#### Mark44

Mentor
Thank you. I put my equation in the form you gave and solved just fine:

$\frac{dQ}{dt} = Q'(t)$
$r - kQ(t) = g(Q(t))$
________________________

$Q'(t) = g(Q(t))$
$\frac{Q'(t)}{g(Q(t))} =1$

$Q'(t) f(Q(t)) = 1$

$\frac{dQ}{dt} \frac{1}{r-kQ} = 1$

$\frac{dQ}{r-kQ} = dt$

$\int\frac{dQ}{r-kQ} = \int dt$
Things are OK but a bit verbose up to the line above.
For example, you can go directly from $\frac{dQ}{dt} = r - kQ$ and separate the equation to $\frac{dQ}{r - kQ} = dt$, and skip several of the lines you wrote.
ForceBoy said:
$\ln|r-kQ | = t +C$ (Can't forget the C now, thanks)
Mistake in the line above. What is your substitution when you do the integration?
ForceBoy said:
$r-kQ = Ae^{t}$

$Q = \frac{r - A e^{t}}{k}$

If the above is correct, then I can solve the rest of the problem.

#### hilbert2

Gold Member

$\displaystyle\int_{0}^{t}dt' = \int_{Q(0)}^{Q(t)}\frac{dQ}{r-kQ} = -\frac{1}{k}\int_{Q(0)}^{Q(t)}\frac{-kdQ}{r-kQ}$

Then you already have the integration constant in terms of the initial condition $Q(0)$.

#### ForceBoy

Mistake in the line above. What is your substitution when you do the integration?
Oh, I hadn't caught that! Thanks alot.

Here is the correct version: $Q(t) = \frac{r-Ae^{-kt}}{k}$

$\displaystyle\int_{0}^{t}dt' = \int_{Q(0)}^{Q(t)}\frac{dQ}{r-kQ} = -\frac{1}{k}\int_{Q(0)}^{Q(t)}\frac{-kdQ}{r-kQ}$

Then you already have the integration constant in terms of the initial condition $Q(0)$.
This is a great tip. Thanks alot this will save me
work.

$\displaystyle\frac{-1}{k}\int_{Q(0)}^{Q(t)} \frac{-kdQ}{r- kQ} = \int_{0}^{t} dt'$

$\displaystyle\ln(r-kQ)_{Q(0)}^{Q(t)} = -kt$

$\displaystyle\frac{r-kQ(t)}{r-kQ(0)} = e^{-kt}$

$\displaystyle r- kQ(t) = (r-kQ(0))e^{-kt}$

$\displaystyle Q(t) = \frac{r-(r-kQ(0))e^{-kt}}{k}$

$\displaystyle Q(t) = \frac{r-re^{-kt}+kQ(0)e^{-kt}}{k}$

$\displaystyle Q(t) = \frac{r-(r-kQ_{0})e^{-kt}}{k}$

So this last equation must be the answer. Thank you all for your time

#### Ray Vickson

Homework Helper
Dearly Missed
The rate at which glucose enters the bloodstream is $r$ units per minute so:

$\frac{dI}{dt} = r$

The rate at which it leaves the body is:

$\frac {dE}{dt} = -k Q(t)$

Then the rate at which the glucose in the body changes is:

A) $Q'(t) = \frac{dI}{dt} + \frac {dE}{dt} = r - k Q(t)$

I don't see how this is a separable differential equation. I still try to solve it.

$\frac{dQ}{dt} + k Q = r$

$Q e^{kt} = \int r e^{kt} dt$

$Q = \frac{r}{e^{kt}}\frac{e^{kx}}{k}$

B) $Q(t) = \frac{r}{k}$

This tells me that the glucose in the bloodstream at any point in time will be a constant. I know this is wrong. It would be appreciated if someone could point me onto the right path to solve this diff. eq. Thank you.
You can render the equation separable by changing to $P = Q- r/k$, so that $dP/dt +kP = 0$ (because, of course, $dP/dt = dQ/dt$).

#### Orodruin

Staff Emeritus
Homework Helper
Gold Member
2018 Award
You can render the equation separable by changing to $P = Q- r/k$, so that $dP/dt +kP = 0$ (because, of course, $dP/dt = dQ/dt$).
The equation was already separable. As demonstrated in #3 by the OP.

"Differential equation modeling glucose in a patient's body"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving