which of the following equation is an exact differential equation?
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Section 2-3 : Exact 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.
2
x
y
−
9
x
2
+
(
2
y
+
x
2
+
1
)
d
y
d
x
=
0
Show Solution
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.
M
(
x
,
y
)
+
N
(
x
,
y
)
d
y
d
x
=
0
(2)
Note that it’s important that it must 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,
Ψ
x
=
M
(
x
,
y
)
and
Ψ
y
=
N
(
x
,
y
)
then we call the differential equation exact. In these cases we can write the differential equation as
Ψ
x
+
Ψ
y
d
y
d
x
=
0
(3)
Then using the chain rule from your Multivariable Calculus class we can further reduce the differential equation to the following derivative,
d
d
x
(
Ψ
(
x
,
y
(
x
)
)
)
=
0
The (implicit) solution to an exact differential equation is then
Ψ
(
x
,
y
)
=
c
(4)
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 its exact we know that somewhere out there is a function
Ψ
(
x
,
y
)
that satisfies
Ψ
x
=
M
Ψ
y
=
N
Now, provided
Ψ
(
x
,
y
)
is continuous and its first order derivatives are also continuous we know that
Ψ
x
y
=
Ψ
y
x
However, we also have the following.
Ψ
x
y
=
(
Ψ
x
)
y
=
(
M
)
y
=
M
y
Ψ
y
x
=
(
Ψ
y
)
x
=
(
N
)
x
=
N
x
Therefore, if a differential equation is exact and
Ψ
(
x
,
y
)
meets all of its continuity conditions we must have.
M
y
=
N
x
(5)
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.
2
x
y
−
9
x
2
+
(
2
y
+
x
2
+
1
)
d
y
d
x
=
0
,
y
(
0
)
=
−
3
Show 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.
2
x
y
2
+
4
=
2
(
3
−
x
2
y
)
y
′
y
(
−
1
)
=
8
Show Solution
Example 4 Find the solution and interval of validity to the following IVP.
2
t
y
t
2
+
1
−
2
t
−
(
2
−
ln
(
t
2
+
1
)
)
y
′
=
0
y
(
5
)
=
0
Show Solution
Example 5 Find the solution and interval of validity for the following IVP.
3
y
3
e
3
x
y
−
1
+
(
2
y
e
3
x
y
+
3
x
y
2
e
3
x
y
)
y
′
=
0
y
(
0
)
=
1
Show Solution
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