Linear Differential
Equations
The first special case of first order differential equations
that we will look at is the linear first order differential equation. In this
case, unlike most of the first order cases that we will look at, we can
actually derive a formula for the general solution. The general solution is
derived below. However, I would suggest
that you do not memorize the formula itself. Instead of memorizing the formula
you should memorize and understand the process that I'm going to use to derive
the formula. Most problems are actually easier to work by using the process
instead of using the formula.
So, let's see how to solve a linear first order differential
equation. Remember as we go through this process that the goal is to arrive at
a solution that is in the form . It's sometimes easy to lose sight of the goal
as we go through this process for the first time.
In order to solve a linear first order differential equation
we MUST start with the differential equation in the form shown below. If the
differential equation is not in this form then the process we’re going to use
will not work.
Where both p(t)
and g(t) are continuous
functions. Recall that a quick and dirty
definition of a continuous function is that a function will be continuous
provided you can draw the graph from left to right without ever picking up your
pencil/pen. In other words, a function
is continuous if there are no holes or breaks in it.
Now, we are going to assume that there is some magical
function somewhere out there in the world, ,
called an integrating factor. Do
not, at this point, worry about what this function is or where it came from. We
will figure out what is once we have the formula for
the general solution in hand.
So, now that we have assumed the existence of multiply everything in (1) by .
This will give.
Now, this is where the magic of comes into play. We are going to assume that
whatever is, it will satisfy the following.
Again do not worry about how we can find a that will satisfy (3). As
we will see, provided p(t) is
continuous we can find it. So substituting (3)
into (2)
we now arrive at.
At this point we need to recognize that the left side of (4)
is nothing more than the following product rule.
So we can replace the left side of (4)
with this product rule. Upon doing this (4)
becomes
Now, recall that we are after y(t). We can now do
something about that. All we need to do
is integrate both sides then use a little algebra and we'll have the solution.
So, integrate both sides of (5) to get.


(6)

Note the constant of integration, c, from the left side integration is included
here. It is vitally important that this be included. If it is left out you will
get the wrong answer every time.
The final step is then some algebra to solve for the
solution, y(t).
Now, from a notational standpoint we know that the constant
of integration, c,
is an unknown constant and so to make our life easier we will absorb the minus
sign in front of it into the constant and use a plus instead. This will NOT
affect the final answer for the solution. So with this change we have.
Again, changing the sign on the constant will not affect our
answer. If you choose to keep the minus
sign you will get the same value of c
as I do except it will have the opposite sign.
Upon plugging in c we will get
exactly the same answer.
There is a lot of playing fast and loose with constants of
integration in this section, so you will need to get used to it. When we do this we will always to try to
make it very clear what is going on and try to justify why we did what we did.
So, now that we’ve got a general solution to (1)
we need to go back and determine just what this magical function is. This is actually an easier process than you
might think. We’ll start with (3).
Divide both sides by ,
Now, hopefully you will recognize the left side of this from
your Calculus I class as nothing more than the following derivative.
As with the process above all we need to do is integrate
both sides to get.
You will notice that the constant of integration from the
left side, k, had been moved to the
right side and had the minus sign absorbed into it again as we did
earlier. Also note that we’re using k here because we’ve already used c and in a little bit we’ll have both of
them in the same equation. So, to avoid
confusion we used different letters to represent the fact that they will, in
all probability, have different values.
Exponentiate both sides to get out
of the natural logarithm.
Now, it’s time to play fast and loose with constants
again. It is inconvenient to have the k in the exponent so we’re going to get
it out of the exponent in the following way.
Now, let’s make use of the fact that k is an unknown constant. If
k is an unknown constant then so is so we might as well just rename it k and make our life easier. This will give us the following.
So, we now have a formula for the general solution, (7),
and a formula for the integrating factor, (8). We do have a problem however. We’ve got two unknown constants and the more
unknown constants we have the more trouble we’ll have later on. Therefore, it would be nice if we could find
a way to eliminate one of them (we’ll not be able to eliminate both….).
This is actually quite easy to do. First, substitute (8)
into (7)
and rearrange the constants.
So, (7) can be written in such a
way that the only place the two unknown constants show up is a ratio of the
two. Then since both c and k are unknown constants so is the ratio of the two constants. Therefore we’ll just call the ratio c and then drop k out of (8)
since it will just get absorbed into c
eventually.
The solution to a linear first order differential equation
is then
where,
Now, the reality is that (9) is
not as useful as it may seem. It is
often easier to just run through the process that got us to (9)
rather than using the formula. We will
not use this formula in any of my examples.
We will need to use (10) regularly, as that formula
is easier to use than the process to derive it.
Solution Process
The solution process for a first order linear differential
equation is as follows.
 Put
the differential equation in the correct initial form, (1).
 Find
the integrating factor, ,
using (10).
 Multiply
everything in the differential equation by and verify that the left side becomes the
product rule and write it as such.
 Integrate
both sides, make sure you properly deal with the constant of integration.
 Solve
for the solution y(t).
Let’s work a couple of examples. Let’s start by solving the differential
equation that we derived back in the Direction
Field section.
Example 1 Find
the solution to the following differential equation.
Solution
First we need to get the differential equation in the
correct form.
From this we can see that p(t)=0.196 and so is then.
Note that officially there should be a constant of
integration in the exponent from the integration. However, we can drop that for exactly the
same reason that we dropped the k
from (8).
Now multiply all the terms in the differential equation by
the integrating factor and do some simplification.
Integrate both sides and don't forget the constants of
integration that will arise from both integrals.
Okay. It’s time to
play with constants again. We can
subtract k from both sides to get.
Both c and k are unknown constants and so the
difference is also an unknown constant.
We will therefore write the difference as c. So, we now have
From this point on we will only put one constant of
integration down when we integrate both sides knowing that if we had written
down one for each integral, as we should, the two would just end up getting
absorbed into each other.
The final step in the solution process is then to divide
both sides by or to multiply both sides by . Either will work, but I usually prefer the multiplication
route. Doing this gives the general
solution to the differential equation.

From the solution to this example we can now see why the
constant of integration is so important in this process. Without it, in this
case, we would get a single, constant solution, v(t)=50. With the constant of
integration we get infinitely many solutions, one for each value of c.
Back in the direction
field section where we first derived the differential equation used in the
last example we used the direction field to help us sketch some solutions.
Let's see if we got them correct. To sketch some solutions all we need to do is
to pick different values of c
to get a solution. Several of these are shown in the graph below.
So, it looks like we did pretty good sketching the graphs
back in the direction field section.
Now, recall from the Definitions
section that the Initial Condition(s) will
allow us to zero in on a particular solution.
Solutions to first order differential equations (not just linear as we
will see) will have a single unknown constant in them and so we will need
exactly one initial condition to find the value of that constant and hence find
the solution that we were after. The
initial condition for first order differential equations will be of the form
Recall as well that a differential equation along with a
sufficient number of initial conditions is called an Initial Value Problem (IVP).
Example 2 Solve
the following IVP.
Solution
To find the solution to an IVP we must first find the
general solution to the differential equation and then use the initial
condition to identify the exact solution that we are after. So, since this is
the same differential equation as we looked at in Example 1, we already have its general solution.
Now, to find the solution we are after we need to identify
the value of c
that will give us the solution we are after. To do this we simply plug in the
initial condition which will give us an equation we can solve for c. So let's do this
So, the actual solution to the IVP is.
A graph of this solution can be seen in the figure above.

Let’s do a couple of examples that are a little more
involved.
Example 3 Solve
the following IVP.
Solution :
Rewrite the differential equation to get the coefficient
of the derivative a one.
Now find the integrating factor.
Can you do the integral?
If not rewrite tangent back into sines and cosines and then use a
simple substitution. Note that we
could drop the absolute value bars on the secant because of the limits on x.
In fact, this is the reason for the limits on x.
Also note that we made use of the following fact.
(11)
This is an important fact that you should always remember
for these problems. We will want to
simplify the integrating factor as much as possible in all cases and this
fact will help with that simplification.
Now back to the example.
Multiply the integrating factor through the differential equation and
verify the left side is a product rule.
Note as well that we multiply the integrating factor through the
rewritten differential equation and NOT the original differential
equation. Make sure that you do
this. If you multiply the integrating
factor through the original differential equation you will get the wrong
solution!
Integrate both sides.
Note the use of the trig formula that made the integral easier. Next, solve for the solution.
Finally, apply the initial condition to find the value of c.
The solution is then.
Below is a plot of the solution.

Example 4 Find the
solution to the following IVP.
Solution
First,
divide through by the t to get the differential equation into the
correct form.
Now let’s
get the integrating factor, .
Now, we need to simplify . However, we can’t use (11)
yet as that requires a coefficient of one in front of the logarithm. So, recall that
and rewrite the integrating factor in a form that will
allow us to simplify it.
We were able to drop the absolute value bars here because
we were squaring the t, but often
they can’t be dropped so be careful with them and don’t drop them unless you
know that you can. Often the absolute
value bars must remain.
Now, multiply the rewritten differential equation
(remember we can’t use the original differential equation here…) by the
integrating factor.
Integrate both sides and solve for the solution.
Finally, apply the initial condition to get the value of c.
The solution is then,
Here is a plot of the solution.

Example 5 Find
the solution to the following IVP.
Solution
First, divide through by t to get the differential equation in the
correct form.
Now that we have done this we can find the integrating
factor, .
Do not forget that the “” is part of p(t). Forgetting this
minus sign can take a problem that is very easy to do and turn it into a very
difficult, if not impossible problem so be careful!
Now, we just need to simplify this as we did in the
previous example.
Again, we can drop the absolute value bars since we are
squaring the term.
Now multiply the differential equation by the integrating factor
(again, make sure it’s the rewritten one and not the original differential
equation).
Integrate both sides and solve for the solution.
Apply the initial condition to find the value of c.
The solution is then
Below is a plot of the solution.

Let’s work one final example that looks more at interpreting
a solution rather than finding a solution.
Example 6 Find
the solution to the following IVP and determine all possible behaviors of the
solution as . If this behavior depends on the value of y_{0} give this dependence.
Solution
First, divide through by a 2 to
get the differential equation in the correct form.
Now find .
Multiply through the differential equation and
rewrite the left side as a product rule.
Integrate both sides and solve for the solution.
Apply the initial condition to find the value of c and note that it will contain y_{0} as we don’t have a value
for that.
So the solution is
Now that we have the solution, let’s look at the long term
behavior (i.e. ) of the solution. The first two terms of the solution will
remain finite for all values of t. It is the last term that will determine the
behavior of the solution. The
exponential will always go to infinity as ,
however depending on the sign of the coefficient c (yes we’ve already found it, but for ease of this discussion
we’ll continue to call it c). The following table gives the long term
behavior of the solution for all values of c.
Range of c

Behavior of solution as

c < 0


c = 0

remains finite

c > 0


This behavior can also be seen in the following graph of
several of the solutions.
Now, because we know how c relates to y_{0}
we can relate the behavior of the solution to y_{0}. The
following table give the behavior of the solution in terms of y_{0} instead of c.
Note that for the solution will remain finite. That will not always happen.

Investigating the long term behavior of solutions is
sometimes more important than the solution itself. Suppose that the solution above gave the
temperature in a bar of metal. In this
case we would want the solution(s) that remains finite in the long term. With this investigation we would now have the
value of the initial condition that will give us that solution and more
importantly values of the initial condition that we would need to avoid so that
we didn’t melt the bar.