Modeling with First
Order Differential Equations
We now move into one of the main applications of
differential equations both in this class and in general. Modeling is the process of writing a
differential equation to describe a physical situation. Almost all of the differential equations that
you will use in your job (for the engineers out there in the audience) are
there because somebody, at some time, modeled a situation to come up with the
differential equation that you are using.
This section is not intended to completely teach you how to
go about modeling all physical situations.
A whole course could be devoted to the subject of modeling and still not
cover everything! This section is
designed to introduce you to the process of modeling and show you what is
involved in modeling. We will look at
three different situations in this section : Mixing Problems, Population
Problems, and Falling Bodies.
In all of these situations we will be forced to make
assumptions that do not accurately depict reality in most cases, but without
them the problems would be very difficult and beyond the scope of this
discussion (and the course in most cases to be honest).
So let’s get started.
Mixing Problems
In these problems we will start with a substance that is
dissolved in a liquid. Liquid will be
entering and leaving a holding tank. The
liquid entering the tank may or may not contain more of the substance dissolved
in it. Liquid leaving the tank will of
course contain the substance dissolved in it.
If Q(t) gives the amount of
the substance dissolved in the liquid in the tank at any time t we want to develop a differential
equation that, when solved, will give us an expression for Q(t). Note as well that in
many situations we can think of air as a liquid for the purposes of these kinds
of discussions and so we don’t actually need to have an actual liquid, but
could instead use air as the “liquid”.
The main assumption that we’ll be using here is that the
concentration of the substance in the liquid is uniform throughout the
tank. Clearly this will not be the case,
but if we allow the concentration to vary depending on the location in the tank
the problem becomes very difficult and will involve partial differential
equations, which is not the focus of this course.
The main “equation” that we’ll be using to model this
situation is :
Rate of
change of
Q(t)

=

Rate at
which Q(t)
enters the
tank


Rate at
which Q(t)
exits the
tank

where,
Rate of change of Q(t)
=

Rate at which Q(t) enters
the tank = (flow rate of liquid entering) x
(concentration of substance in liquid entering)

Rate at which Q(t) exits
the tank = (flow rate of liquid exiting) x
(concentration of substance in liquid exiting)

Let’s take a look at the first problem.
Example 1 A
1500 gallon tank initially contains 600 gallons of water with 5 lbs of salt
dissolved in it. Water enters the tank
at a rate of 9 gal/hr and the water entering the tank has a salt
concentration of lbs/gal.
If a well mixed solution leaves the tank at a rate of 6 gal/hr, how
much salt is in the tank when it overflows?
Solution
First off, let’s address the “well mixed solution”
bit. This is the assumption that was
mentioned earlier. We are going to
assume that the instant the water enters the tank it somehow instantly
disperses evenly throughout the tank to give a uniform concentration of salt
in the tank at every point. Again,
this will clearly not be the case in reality, but it will allow us to do the
problem.
Now, to set up the IVP that we’ll need to solve to get Q(t) we’ll need the flow rate of the
water entering (we’ve got that), the concentration of the salt in the water
entering (we’ve got that), the flow rate of the water leaving (we’ve got
that) and the concentration of the salt in the water exiting (we don’t have
this yet).
So, we first need to determine the concentration of the
salt in the water exiting the tank.
Since we are assuming a uniform concentration of salt in the tank the
concentration at any point in the tank and hence in the water exiting is
given by,
The amount at any time t
is easy it’s just Q(t). The volume is also pretty easy. We start with 600 gallons and every hour 9
gallons enters and 6 gallons leave.
So, if we use t in hours,
every hour 3 gallons enters the tank, or at any time t there is 600 + 3t
gallons of water in the tank.
So, the IVP for this situation is,
This is a linear differential equation and it isn’t too
difficult to solve (hopefully). We
will show most of the details, but leave the description of the solution
process out. If you need a refresher
on solving linear first order differential equations go back and take a look
at that section.
So, here’s the general solution. Now, apply the initial condition to get the
value of the constant, c.
So, the amount of salt in the tank at any time t is.
Now, the tank will overflow at t = 300 hrs. The amount of
salt in the tank at that time is.
Here’s a graph of the salt in the tank before it
overflows.
Note that the whole graph should have small oscillations
in it as you can see in the range from 200 to 250. The scale of the oscillations however was
small enough that the program used to generate the image had trouble showing
all of them.

The work was a little messy with that one, but they will
often be that way so don’t get excited about it. This first example also assumed that nothing
would change throughout the life of the process. That, of course will usually not be the
case. Let’s take a look at an example
where something changes in the process.
Example 2 A
1000 gallon holding tank that catches runoff from some chemical process
initially has 800 gallons of water with 2 ounces of pollution dissolved in
it. Polluted water flows into the tank
at a rate of 3 gal/hr and contains 5 ounces/gal of pollution in it. A well
mixed solution leaves the tank at 3 gal/hr as well. When the amount of pollution in the holding
tank reaches 500 ounces the inflow of polluted water is cut off and fresh
water will enter the tank at a decreased rate of 2 gal/hr while the outflow
is increased to 4 gal/hr. Determine
the amount of pollution in the tank at any time t.
Solution
Okay, so clearly the pollution in the tank will increase
as time passes. If the amount of
pollution ever reaches the maximum allowed there will be a change in the
situation. This will necessitate a
change in the differential equation describing the process as well. In other words, we’ll need two IVP’s for
this problem. One will describe the
initial situation when polluted runoff is entering the tank and one for after
the maximum allowed pollution is reached and fresh water is entering the
tank.
Here are the two IVP’s for this problem.
The first one is fairly straight forward and will be valid
until the maximum amount of pollution is reached. We’ll call that time t_{m}. Also, the
volume in the tank remains constant during this time so we don’t need to do
anything fancy with that this time in the second term as we did in the previous
example.
We’ll need a little explanation for the second one. First notice that we don’t “start over” at t = 0.
We start this one at t_{m},
the time at which the new process starts.
Next, fresh water is flowing into the tank and so the concentration of
pollution in the incoming water is zero.
This will drop out the first term, and that’s okay so don’t worry
about that.
Now, notice that the volume at any time looks a little
funny. During this time frame we are
losing two gallons of water every hour of the process so we need the “2” in
there to account for that. However, we
can’t just use t as we did in the
previous example. When this new
process starts up there needs to be 800 gallons of water in the tank and if
we just use t there we won’t have
the required 800 gallons that we need in the equation. So, to make sure that we have the proper
volume we need to put in the difference in times. In this way once we are one hour into the
new process (i.e t  t_{m} =
1) we will have 798 gallons in the tank as required.
Finally, the second process can’t continue forever as
eventually the tank will empty. This
is denoted in the time restrictions as t_{e}. We can also note that t_{e }= t_{m}
+ 400 since the tank will empty 400 hours after this new process starts up. Well, it will end provided something
doesn’t come along and start changing the situation again.
Okay, now that we’ve got all the explanations taken care
of here’s the simplified version of the IVP’s that we’ll be solving.
The first IVP is a fairly simple linear differential
equation so we’ll leave the details of the solution to you to check. Upon solving you get.
Now, we need to find t_{m}. This isn’t too bad all we need to do is
determine when the amount of pollution reaches 500. So we need to solve.
So, the second process will pick up at 35.475 hours. For completeness sake here is the IVP with
this information inserted.
This differential equation is both linear and separable
and again isn’t terribly difficult to solve so I’ll leave the details to you
again to check that we should get.
So, a solution that encompasses the complete running time
of the process is
Here is a graph of the amount of pollution in the tank at
any time t.

As you can surely see, these problems can get quite
complicated if you want them to. Take
the last example. A more realistic
situation would be that once the pollution dropped below some predetermined
point the polluted runoff would, in all likelihood, be allowed to flow back in
and then the whole process would repeat itself.
So, realistically, there should be at least one more IVP in the process.
Let’s move on to another type of problem now.
Population
These are somewhat easier than the mixing problems although,
in some ways, they are very similar to mixing problems. So, if P(t)
represents a population in a given region at any time t the basic equation that we’ll use is identical to the one that we
used for mixing. Namely,
Rate of
change of
P(t)

=

Rate at
which P(t)
enters the
region



Rate at
which P(t)
exits the
region






Here the rate of change of P(t) is still the derivative.
What’s different this time is the rate at which the population enters
and exits the region. For population
problems all the ways for a population to enter the region are included in the
entering rate. Birth rate and migration
into the region are examples of terms that would go into the rate at which the
population enters the region. Likewise,
all the ways for a population to leave an area will be included in the exiting
rate. Therefore things like death rate,
migration out and predation are examples of terms that would go into the rate
at which the population exits the area.
Here’s an example.
Example 3 A
population of insects in a region will grow at a rate that is proportional to
their current population. In the
absence of any outside factors the population will triple in two weeks
time. On any given day there is a net
migration into the area of 15 insects and 16 are eaten by the local bird
population and 7 die of natural causes.
If there are initially 100 insects in the area will the population
survive? If not, when do they die out?
Solution
Let’s start out by looking at the birth rate. We are told that the insects will be born
at a rate that is proportional to the current population. This means that the birth rate can be written
as
where r is a
positive constant that will need to be determined. Now, let’s take everything into account and
get the IVP for this problem.
Note that in the first line we used parenthesis to note
which terms went into which part of the differential equation. Also note that we don’t make use of the
fact that the population will triple in two weeks time in the absence of
outside factors here. In the absence
of outside factors means that the ONLY thing that we can consider is birth
rate. Nothing else can enter into the
picture and clearly we have other influences in the differential equation.
So, just how does this tripling come into play? Well, we should also note that without
knowing r we will have a difficult
time solving the IVP completely. We
will use the fact that the population triples in two week time to help us
find r. In the absence of outside factors the
differential equation would become.
Note that since we used days as the time frame in the
actual IVP I needed to convert the two weeks to 14 days. We could have just as easily converted the
original IVP to weeks as the time frame, in which case there would have been
a net change of 56
per week instead of the 8
per day that we are currently using in the original differential equation.
Okay back to the differential equation that ignores all
the outside factors. This differential
equation is separable and linear and is a simple differential equation to
solve. I’ll leave the detail to you to
get the general solution.
Applying the initial condition gives c = 100. Now apply the
second condition.
We need to solve this for r. First divide both sides
by 100, then take the natural log of both sides.
We made use of the fact that here to simplify the problem. Now, that we have r we can go back and solve the original differential
equation. We’ll rewrite it a little
for the solution process.
This is a fairly simple linear differential equation, but
that coefficient of P always get
people bent out of shape, so we’ll go through at least some of the details
here.
Now, don’t get excited about the integrating factor
here. It’s just like only this time the constant is a little more
complicated than just a 2, but it is a constant! Now, solve the differential equation.
Again, do not get excited about doing the right hand
integral, it’s just like integrating ! Applying the initial condition gives the
following.
Now, the exponential has a positive exponent and so will
go to plus infinity as t
increases. Its coefficient, however,
is negative and so the whole population will go negative eventually. Clearly, population can’t be negative, but
in order for the population to go negative it must pass through zero. In other words, eventually all the insects
must die. So, they don’t survive and
we can solve the following to determine when they die out.
So, the insects will survive for around 7.2 weeks. Here is a graph of the population during
the time in which they survive.

As with the mixing problems, we could make the population
problems more complicated by changing the circumstances at some point in
time. For instance, if at some point in
time the local bird population saw a decrease due to disease they wouldn’t eat
as much after that point and a second differential equation to govern the time
after this point.
Let’s now take a look at the final type of problem that
we’ll be modeling in this section.
Falling Body
This will not be the first time that we’ve looked into
falling bodies. If you recall, we looked
at one of these when we were looking at Direction
Fields. In that section we saw that
the basic equation that we’ll use is Newton’s
Second Law of Motion.
The two forces that we’ll be looking at here are gravity and
air resistance. The main issue with
these problems is to correctly define conventions and then remember to keep
those conventions. By this we mean
define which direction will be termed the positive direction and then make sure
that all your forces match that convention.
This is especially important for air resistance as this is usually
dependent on the velocity and so the “sign” of the velocity can and does affect
the “sign” of the air resistance force.
Let’s take a look at an example.
Example 4 A
50 kg mass is shot from a cannon straight up with an initial velocity of
10m/s off a bridge that is 100 meters above the ground. If air resistance is given by 5v determine the velocity of the mass
when it hits the ground.
Solution
First, notice that when we say straight up, we really mean
straight up, but in such a way that it will miss the bridge on the way back
down. Here is a sketch of the
situation.
Notice the conventions that we set up for this
problem. Since the vast majority of
the motion will be in the downward direction we decided to assume that
everything acting in the downward direction should be positive. Note that we also defined the “zero
position” as the bridge, which makes the ground have a “position” of 100.
Okay, if you think about it we actually have two
situations here. The initial phase in
which the mass is rising in the air and the second phase when the mass is on
its way down. We will need to examine
both situations and set up an IVP for each.
We will do this simultaneously.
Here are the forces that are acting on the object on the way up and on
the way down.
Notice that the air resistance force needs a negative in
both cases in order to get the correct “sign” or direction on the force. When the mass is moving upwards the
velocity (and hence v) is negative,
yet the force must be acting in a downward direction. Therefore, the “” must be part of the force
to make sure that, overall, the force is positive and hence acting in the
downward direction. Likewise, when the
mass is moving downward the velocity (and so v) is positive. Therefore,
the air resistance must also have a “” in order to make sure that it’s
negative and hence acting in the upward direction.
So, the IVP for each of these situations are.
In the second IVP, the t_{0}
is the time when the object is at the highest point and is ready to start on
the way down. Note that at this time
the velocity would be zero. Also note
that the initial condition of the first differential equation will have to be
negative since the initial velocity is upward.
In this case, the differential equation for both of the
situations is identical. This won’t
always happen, but in those cases where it does, we can ignore the second IVP
and just let the first govern the whole process.
So, let’s actually plug in for the mass and gravity (we’ll
be using g = 9.8 m/s^{2}
here). We’ll go ahead and divide out
the mass while we’re at it since we’ll need to do that eventually anyway.
This is a simple linear differential equation to solve so
we’ll leave the details to you. Upon
solving we arrive at the following equation for the velocity of the object at
any time t.
Okay, we want the velocity of the ball when it hits the
ground. Of course we need to know when it hits the ground before we can
ask this. In order to find this we
will need to find the position function.
This is easy enough to do.
We can now use the fact that I took the convention that s(0) = 0 to find that c = 1080. The position at any time is then.
To determine when the mass hits the ground we just need to
solve.
We’ve got two solutions here, but since we are starting
things at t = 0, the negative is
clearly the incorrect value. Therefore
the mass hits the ground at t =
5.98147. The velocity of the object
upon hitting the ground is then.

This last example gave us an example of a situation where
the two differential equations needed for the problem ended up being identical
and so we didn’t need the second one after all.
Be careful however to not always expect this. We could very easily change this problem so
that it required two different differential equations. For instance we could have had a parachute on
the mass open at the top of its arc changing its air resistance. This would have completely changed the second
differential equation and forced us to use it as well. Or, we could have put a river under the
bridge so that before it actually hit the ground it would have first had to go
through some water which would have a different “air” resistance for that phase
necessitating a new differential equation for that portion.
Or, we could be really crazy and have both the parachute and
the river which would then require three IVP’s to be solved before we
determined the velocity of the mass before it actually hits the solid ground.
Before leaving this section let’s work a couple examples
illustrating the importance of remembering the conventions that you set up for
the positive direction in these problems.
Awhile back I gave my students a problem in which a sky
diver jumps out of a plane. Most of my
students are engineering majors and following the standard convention from most
of their engineering classes they defined the positive direction as upward,
despite the fact that all the motion in the problem was downward. There is nothing wrong with this assumption,
however, because they forgot the convention that up was positive they did not
correctly deal with the air resistance which caused them to get the incorrect
answer.
So, let’s take a look at the problem and set up the IVP that
will give the sky diver's velocity at any time t.
Example 5 Set
up the IVP that will give the velocity of a 60 kg sky diver that jumps out of
a plane with no initial velocity and an air resistance of . For this example assume that the positive
direction is upward.
Solution
Here are the forces that are acting on the sky diver
Because of the conventions the force due to gravity is
negative and the force due to air resistance is positive. As set up, these forces have the correct
sign and so the IVP is
The problem arises when you go to remove the absolute
value bars. In order to do the problem
they do need to be removed. This is
where most of the students made their mistake. Because they had forgotten about the
convention and the direction of motion they just dropped the absolute value
bars to get.
So, why is this incorrect?
Well remember that the convention is that positive is upward. However in this case the object is moving
downward and so v is negative! Upon dropping the absolute value bars the
air resistance became a negative force and hence was acting in the downward
direction!
To get the correct IVP recall that because v is negative then v = v. Using this, the air
resistance becomes F_{A} = 0.8v
and despite appearances this is a positive force since the “” cancels out
against the velocity (which is negative) to get a positive force.
The correct IVP is then
Plugging in the mass gives
For the sake of completeness the velocity of the sky
diver, at least until the parachute opens, which I didn’t include in this
problem is.

This mistake was made in part because the students were in a
hurry and weren’t paying attention, but also because they simply forgot about
their convention and the direction of motion!
Don’t fall into this mistake.
Always pay attention to your conventions and what is happening in the
problems.
Just to show you the difference here is the problem worked
by assuming that down is positive.
Example 6 Set
up the IVP that will give the velocity of a 60 kg sky diver that jumps out of
a plane with no initial velocity and an air resistance of . For this example assume that the positive
direction is downward.
Solution
Here are the forces that are acting on the sky diver
In this case the force due to gravity is positive since
it’s a downward force and air resistance is an upward force and so needs to
be negative. In this case since the
motion is downward the velocity is positive so v = v. The air resistance is then F_{A }=
0.8v. The IVP for this case is
Plugging in the mass gives
Solving this gives
This is the same solution as the previous example, except
that it’s got the opposite sign. This
is to be expected since the conventions have been switched between the two
examples.
