Exact Differential Equations
Solving an Exact DE
Making a DE Exact
MATH 312
Section 2.4: Exact Differential Equations
Prof. Jonathan Duncan
Walla Walla University
Spring Quarter, 2008
Conclusion
Exact Differential Equations
Solving an Exact DE
Outline
1
Exact Differential Equations
2
Solving an Exact DE
3
Making a DE Exact
4
Conclusion
Making a DE Exact
Conclusion
Exact Differential Equations
Solving an Exact DE
Making a DE Exact
A Motivating Example
Our tools so far allow us to solve first-order differential equations
which are separable and/or linear.
Example
Is the following differential equation separable or linear?
(tan x − sin x sin y )dx + (cos x cos y )dy = 0
After rewriting as shown, what do you notice?
dy
sin x sin y − tan x
=
dx
cos x cos y
The equation is not separable.
The equation is not linear.
We need a new solution method for this DE!
Conclusion
Exact Differential Equations
Solving an Exact DE
Making a DE Exact
Conclusion
Working Backwards
We develop our method using Calculus notation.
Differentials
Recall that if f (x, y ) has continuous first partials on some region of the
xy -plane, then with z = f (x, y ) the differential is:
dz =
∂f
∂f
dx +
dy
∂x
∂y
Why is this of use? Recall our motivating example.
Example
Now, to solve
(tan x − sin x sin y )dx + (cos x cos y )dy = 0
∂f
we find an f (x, y ) for which ∂x
= (tan x − sin x sin y ) and
∂f
=
(cos
x
cos
y
),
and
set
f
(x,
y
) = c for any constant c so that dz = 0.
∂y
Exact Differential Equations
Solving an Exact DE
Making a DE Exact
Conclusion
Exact Differentials and Equations
We now formalize this type of solution with several definitions.
Definition 2.3
A differential expression of the form M(x, y ) dx + N(x, y ) dy is an exact
differential in a region R of the xy -plane if it corresponds to the
differential of some function f (x, y ) defined on R.
Definition 2.3, Part II
A first order differential equation of the form
M(x, y ) dx + N(x, y ) dy = 0
is an exact equation if the left side is an exact differential.
Solving an Exact Equation
If the differential of f (x, y ) is M(x, y ) dx + N(x, y ) dy , then f (x, y ) = c
is an implicit solution to the DE M(x, y ) dx + N(x, y ) dy = 0
Exact Differential Equations
Solving an Exact DE
Making a DE Exact
Conclusion
Verifying Exactness
We now consider how to tell if a DE is exact.
Example
Is the differential equation below exact?
(2x − 1) dx + (3y + 7) dy = 0
Theorem 2.1
Let M(x, y ) and N(x, y ) be continuous with continuous first partial
derivatives on a rectangular region R of the xy -plane. Then, a necessary
and sufficient condition that M(x, y ) dx + N(x, y ) dy be an exact
∂N
differential is that ∂M
∂y = ∂x .
Step 1:
∂M
∂y
=
∂N
∂x
implies exactness.
Step 2: exactness implies
∂M
∂y
=
∂N
∂x .
Exact Differential Equations
Solving an Exact DE
Making a DE Exact
Conclusion
Solution Method
To solve an exact differential equation, we will follow the procedure
from Step 2 of the theorem proof.
Solution Method
To solve an exact differential equation, follow these steps.
∂f
= M(x, y )
∂x
Z
f (x, y ) =
∂f
∂
=
∂y
∂y
M(x, y ) dx + g (y )
Z
M(x, y ) dx
+ g 0 (y ) = N(x, y )
Exact Differential Equations
Solving an Exact DE
Making a DE Exact
Examples
Use this procedure to find an implicit solution to each of the
following exact differential equations.
Example
Solve
(2xy 2 − 3) dx + (2x 2 y + 4) dy = 0
x 2 y 2 − 3x + 4y = c
Example
Solve
(tan x − sin x sin y ) dy + (cos x cos y ) dy = 0
cos x sin y − ln | cos x| = c
Conclusion
Exact Differential Equations
Solving an Exact DE
Making a DE Exact
Conclusion
Not Everything is Exact
Unfortunately, not every differential equation which has the form
M(x, y ) dx + N(x, y ) dy = 0 is exact.
Example
Show that the differential equation below is not exact.
(2y 2 + 3x) dx + 2xy dy = 0
Sometimes we can find an integrating factor µ(x, y ) so that the
equation obtained by multiplying by µ(x, y ) (shown below) is exact.
µ(x, y )M(x, y ) dx + µ(x, y )N(x, y ) dy = 0
Exact Differential Equations
Solving an Exact DE
Making a DE Exact
Using an Integrating Factor
In order for our integrating factor to work, we need the following
to be exact.
µ(x, y )M(x, y ) dx + µ(x, y )N(x, y ) dy = 0
Applying the exactness theorem, this means:
∂
∂
(µM) =
(µN)
∂y
∂x
∂M
∂µ
∂N
∂µ
M+
µ=
N+
µ
∂y
∂y
∂x
∂x
∂µ
∂µ
∂M
∂N
N−
M=µ
−
∂x
∂y
∂y
∂x
Conclusion
Exact Differential Equations
Solving an Exact DE
Making a DE Exact
Conclusion
Finding an Integrating Factor
To continue, we must assume that µ depends only on x or y .
Suppose µ = µ(x).
∂µ
∂µ
∂N
∂M
N−
M=µ
−
∂x
∂y
∂y
∂x
∂µ
∂N
∂M
N=µ
−
∂x
∂y
∂x
dµ
My − N x
=
µ
dx
N
Finding µ
M −N
N −M
If y N x depends only on x or if x M y depends only on y , then we can
My −Nx
Nx −My
solve the separable DE dµ
µ or dµ
µ respectively to
dx =
N
dy =
M
find the integrating factor µ.
Exact Differential Equations
Solving an Exact DE
Making a DE Exact
Completing our Example
Recall our problem example from a few slides previous.
Example
Solve
(2y 2 + 3x) dx + 2xy dy = 0
My − N x
4y − 2y
=
N
2xy
dµ 1
− µ=0
dx
x
N x − My
2y − 4y
= 2
M
2y + 3x
and
is linear, so
µ = e−
R
− x1 dx
x(2y 2 + 3x) dx + x(2xy ) dy = 0 is exact
x 2y 2 + x 3 = c
=x
Conclusion
Exact Differential Equations
Solving an Exact DE
Making a DE Exact
Important Concepts
Things to Remember from Section 2.4
1
Identifying exact differential equations
2
Solution method for exact differential equations
3
Using an integrating factor to make a DE exact
Conclusion