Differential Equations (Part 3. Exact Differential Equations)
Posted by Suhaivi Hamdan on 6:48 PM with No comments
Exact Differential Equations
The next type of first order differential equations that
we’ll be looking at is exact differential equations. Before we get into the full details behind
solving exact differential equations it’s probably best to work an example that
will help to show us just what an exact differential equation is. It will also show some of the behind the
scenes details that we usually don’t bother with in the solution process.
The vast majority of the following example will not be done
in any of the remaining examples and the work that we will put into the
remaining examples will not be shown in this example. The whole point behind this example is to
show you just what an exact differential equation is, how we use this fact to
arrive at a solution and why the process works as it does. The majority of the actual solution details
will be shown in a later example.
Example 1 Solve
the following differential equation.
Solution
Let’s start off by supposing that somewhere out there in
the world is a function Ψ(x,y) that we can find.
For this example the function that we need is
Do not worry at this point about where this function came
from and how we found it. Finding the
function, Ψ(x,y), that is needed for any particular differential equation is
where the vast majority of the work for these problems lies. As stated earlier however, the point of
this example is to show you why the solution process works rather than
showing you the actual solution process.
We will see how to find this function in the next example, so at this
point do not worry about how to find it, simply accept that it can be found
and that we’ve done that for this particular differential equation.
Now, take some partial derivatives of the function.
Now, compare these partial derivatives to the differential
equation and you’ll notice that with these we can now write the differential
equation as.
Now, recall from your multivariable calculus class
(probably Calculus III) that (1) is nothing more than
the following derivative (you’ll need the multivariable chain rule for
this…).
So, the differential equation can now be written as
Now, if the ordinary (not partial…) derivative of
something is zero, that something must have been a constant to start
with. In other words, we’ve got to
have
This then is an implicit solution for our differential
equation! If we had an initial
condition we could solve for c. We
could also find an explicit solution if we wanted to, but we’ll hold off on
that until the next example.

Okay, so what did we learn from the last example? Let’s look at things a little more
generally. Suppose that we have the
following differential equation.

Note that it’s important that it be in this form! There must be an “= 0” on one side and the
sign separating the two terms must be a “+”.
Now, if there is a function somewhere out there in the world, Ψ(x,y),
so that,


then we call the differential equation exact. In these cases we can
write the differential equation as

(3)

Then using the chain rule from Calculus III we can further
reduce the differential equation to the following derivative,


The (implicit) solution to an exact differential equation is
then

Well, it’s the solution provided we can find Ψ(x,y)
anyway. Therefore, once we have the
function we can always just jump straight to (4) to
get an implicit solution to our differential equation.
Finding the function Ψ(x,y) is clearly the central task in
determining if a differential equation is exact and in finding its solution.
As we will see, finding Ψ(x,y) can be a somewhat lengthy process
in which there is the chance of mistakes. Therefore, it would be nice if there
was some simple test that we could use before even starting to see if a
differential equation is exact or not. This will be especially useful if it turns
out that the differential equation is not exact, since in this case Ψ(x,y) will
not exist. It would be a waste of time to try and find a nonexistent function!
So, let's see if we can find a test for exact differential
equations. Let's start with (2) and assume that the
differential equation is in fact exact.
Since it’s exact we know that somewhere out there is a function Ψ(x,y)
that satisfies


Now, provided Ψ(x,y) is continuous and its first order
derivatives are also continuous we know that


However, we also have the following.


Therefore, if a differential equation is exact and Ψ(x,y)
meets all of its continuity conditions we must have.

Likewise if (5) is not true there is no
way for the differential equation to be exact.
Therefore, we will use (5) as
a test for exact differential equations.
If (5)
is true we will assume that the differential equation is exact and that Ψ(x,y)
meets all of its continuity conditions and proceed with finding it. Note that for all the examples here the
continuity conditions will be met and so this won’t be an issue.
Okay, let’s go back and rework the first example. This time we will use the example to show how
to find Ψ(x,y). We’ll also add in an initial condition to the
problem.
Example 2 Solve
the following IVP and find the interval of validity for the solution.
Solution
First identify M
and N and check that the
differential equation is exact.
So, the differential equation is exact according to the
test. However, we already knew that as
we have given you Ψ(x,y).
It’s not a bad thing to verify it however and to run through the test
at least once however.
Now, how do we actually find Ψ(x,y)?
Well recall that
We can use either of these to get a start on finding Ψ(x,y) by integrating as follows.
However, we will need to be careful as this won’t give us
the exact function that we need. Often it doesn’t matter which one you choose
to work with while in other problems one will be significantly easier than
the other. In this case it doesn’t
matter which one we use as either will be just as easy.
So, I’ll use the first one.
Note that in this case the “constant” of integration is
not really a constant at all, but instead it will be a function of the
remaining variable(s), y in this
case.
Recall that in integration we are asking what function we
differentiated to get the function we are integrating. Since we are working with two variables
here and talking about partial differentiation with respect to x, this means that any term that
contained only constants or y’s
would have differentiated away to zero, therefore we need to acknowledge that
fact by adding on a function of y
instead of the standard c.
Okay, we’ve got most of Ψ(x,y) we just need to determine h(y) and we’ll be done. This is actually easy to do. We used
From this we can see that
Note that at this stage h(y) must be only a function of y and so if there are any x’s
in the equation at this stage we have made a mistake somewhere and it’s time
to go look for it.
We can now find h(y)
by integrating.
You’ll note that we included the constant of integration, k, here. It will turn out however that this will end
up getting absorbed into another constant so we can drop it in general.
So, we can now write down Ψ(x,y).
With the exception of the k this is identical to the function that we used in the first
example. We can now go straight to the
implicit solution using (4).
We’ll now take care of the k. Since both k and c are unknown constants all we need to do is subtract one from
both sides and combine and we still have an unknown constant.
Therefore, we’ll not include the k in anymore problems.
This is where we left off in the first example. Let’s now apply the initial condition to
find c.
The implicit solution is then.
Now, as we saw in the separable differential equation
section, this is quadratic in y and
so we can solve for y(x) by using
the quadratic formula.
Now, reapply the initial condition to figure out which of
the two signs in the
So, it looks like the “” is the one that we need. The explicit solution is then.
Now, for the interval of validity. It looks like we might well have problems
with square roots of negative numbers.
So, we need to solve
Upon solving this equation is zero at x =
So, it looks like there are two intervals where the
polynomial will be positive.
However, recall that intervals of validity need to be
continuous intervals and contain the value of x that is used in the initial condition. Therefore the interval of validity must be.
Here is a quick graph of the solution.

That was a long example, but mostly because of the initial
explanation of how to find Ψ(x,y).
The remaining examples will not be as long.
Example 3 Find
the solution and interval of validity for the following IVP.
Solution
Here, we first need to put the differential equation into
proper form before proceeding. Recall
that it needs to be “= 0” and the sign separating the two terms must be a
plus!
So we have the following
and so the differential equation is exact. We can either integrate M with respect to x or integrate N with
respect to y. In this case either would be just as easy
so we’ll integrate N this time so
we can say that we’ve got an example of both down here.
This time, as opposed to the previous example, our
“constant” of integration must be a function of x since we integrated with respect to y. Now differentiate with
respect to x and compare this to M.
So, it looks like
Again, we’ll drop the constant of integration that
technically should be present in h(x)
since it will just get absorbed into the constant we pick up in the next
step. Also note that, h(x) should only involve x’s at this point. If there are any y’s left at this point a mistake has been made so go back and
look for it.
Writing everything down gives us the following for Ψ(x,y).
So, the implicit solution to the differential equation is
Applying the initial condition gives,
The solution is then
Using the quadratic formula gives us
Reapplying the initial condition shows that this time we
need the “+” (we’ll leave those details to you to check). Therefore, the explicit solution is
Now let’s find the interval of validity. We’ll need to avoid x = 0 so we don’t get division by zero. We’ll also have to watch out for square
roots of negative numbers so solve the following equation.
The only real solution here is x = 3.217361577. Below is
a graph of the polynomial.
So, it looks like the polynomial will be positive, and
hence okay under the square root on
Now, this interval can’t be the interval of validity
because it contains x = 0 and we
need to avoid that point. Therefore,
this interval actually breaks up into two different possible intervals of
validity.
The first one contains x
= 1, the x value from the initial
condition. Therefore, the interval of
validity for this problem is
Here is a graph of the solution.

Example 4 Find
the solution and interval of validity to the following IVP.
Solution
So, first deal with that minus sign separating the two
terms.
Now, find M and N and check that it’s exact.
So, it’s exact.
We’ll integrate the first one in this case.
Differentiate with respect to y and compare to N.
So, it looks like we’ve got.
This gives us
The implicit solution is then,
Applying the initial condition gives,
The implicit solution is now,
This solution is much easier to solve than the previous
ones. No quadratic formula is needed
this time, all we need to do is solve for y. Here’s what we get for an explicit
solution.
Alright, let’s get the interval of validity. The term in the logarithm is always
positive so we don’t need to worry about negative numbers in that. We do need to worry about division by zero
however. We will need to avoid the
following point(s).
We now have three possible intervals of validity.
The last one contains t
= 5 and so is the interval of validity for this problem is

Example 5 Find
the solution and interval of validity for the following IVP.
Solution
Let’s identify M
and N and check that it’s exact.
So, it’s exact.
With the proper simplification integrating the second one isn’t too
bad.
However, the first is already set up for easy integration
so let’s do that one.
Differentiate with respect to y and compare to N.
So, it looks like we’ve got
Recall that actually h(y)
= k, but we drop the k because
it will get absorbed in the next step.
That gives us h(y) = 0. Therefore, we get.
The implicit solution is then
Apply the initial condition.
The implicit solution is then
This is as far as we can go. There is no way to solve this for y and get an explicit solution.

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