Exact Differential Equations
Definition: Let F(x,y) be a function of two real variables such that F has continuous first
partial derivatives in a domain D in the xy-plane.
The total differential of F, denoted by dF, is given by the expression:
!F
!F
dF(x, y) =
(x, y)!dx +
(x, y)!dy,!for!all!(x, y) "D
!x
!y
Example:
Given F(x,y) = x3sin(y) + y2x,
then its partial derivatives are:
!F
!F
= 3x 2 sin(y)!!and!!
= x 3 cos(y) + 2yx
!x
!y
they are continuous functions in the whole xy-plane.
Therefore,
dF(x,y) = (3x2 sin(y)) dx + (x3 cos(y) + 2yx) dy
Definition: The expression M(x,y) dx + N(x,y) dy is called an exact differential form or
a conservative differential from in a domain D if there exists a function F(x,y), called
the potential function, defined on D such that dF(x,y) = M(x,y) dx + N(x,y) dy for all
(x,y) in D.
Remark:
If M(x,y) dx + N(x,y) dy is an exact differential form and F(x,y) is a potential function
!F
!F
(x, y) = M(x, y)!!and!! (x, y) = N(x, y) .
then
!x
!y
Definition: If M(x,y) dx + N(x,y) dy is an exact differential form, then the equation
M(x,y) dx + N(x,y) dy = 0
is called an exact differential equation.
Example:
The form 2xy dx + x2 dy is an exact differential form, since it is the total differential of
the function F(x,y) = x2y.
Then the equation 2xy dx + x2 dy = 0 is an exact differential equation.
We need a test to determine whether or not a given first order differential equation is
exact.
!M
!N
,!and!
continuous functions in a domain D.
!y
!x
The first order differential equation M(x,y) dx +N(x,y) dy = 0 is an exact equation in D if
!M !N
=
and only if
for all (x,y) in D.
!y
!x
Proof:
Theorem: Let M(x,y), N(x,y),
1) Assume M(x,y) dx +N(x,y) dy = 0 is exact in D,
then, M(x,y) dx +N(x,y) dy is an exact differential form in D,
then, there exists a function F(x,y) defined on D, such that dF = M(x,y) dx +N(x,y) dy,
!F
!F
(x, y) = M(x, y)!!and!! (x, y) = N(x, y) in D,
then,
!x
!y
!2 F !M
!2 F !N
=
!!and!
=
then,
in D.
!y!x !y
!x!y !x
!M
!N
,!and!
Since
are continuous functions in a domain D, the mixed partial
!y
!x
!M !N
=
derivatives are continuous, therefore they are equal which implies
for all (x,y)
!y
!x
in D.
!M !N
=
for all (x,y) in D, we must show that M(x,y) dx + N(x,y) dy = 0 is
!y
!x
exact. This means we must find a function F(x,y) defined in D such that
!F
!F
(x, y) = M(x, y)!!and!! (x, y) = N(x, y)
dF = M(x,y) dx + N(x,y) dy or
!x
!y
!F
Let F(x,y) be a function such that
= M(x, y) . If we do partial integration with respect
!x
to x, holding the variable y as a constant and integrating with respect to x. Instead of
getting an arbitrary constant we get an arbitrary function of the variable y.
F(x, y) = ! M(x, y)dx + B(y)
2) Assume
Now, we have to determine the function B(y), to do so, we take the partial derivative of
F(x,y) with respect to y.
!F !
dB
=
M(x,
y)dx
+
= N(x, y)
!y !y "
dy
dB
"
= N(x, y) !
M(x, y)dx
so
dy
"y #
then
and
$
"
'
B(y) = # & N(x, y) !
M(x, y)dx ) dy
#
"y
%
(
$
#
'
F(x, y) = ! M(x, y)dx + ! & N(x, y) "
M(x, y)dx ) dy
!
#y
%
(
Remark: We use the assumption that
!M !N
=
to show that
!y
!x
dB
"
= N(x, y) !
M(x, y)dx is only function of y. Look at the text-book for those
dy
"y #
details.
Remark: We can also obtain F(x,y) assuming F(x,y) be a function such that
!F
= N(x, y) . So, we do partial integration with respect to y, holding the variable x as a
!y
constant and integrating with respect to y. Instead of getting an arbitrary constant we get
an arbitrary function of the variable x.
F(x, y) = ! N(x, y)dy + Q(x)
Remark: this proof contains the germ of a method for obtaining a set of solutions for the
exact equation M(x,y) dx + N(x,y) dy = 0.
If M(x,y) dx + N(x,y) dy = 0 is exact,
then dF = M(x,y) dx + N(x,y) dy = 0,
then dF = 0 = d(c),
then F = c is the family of solutions.
Example:
Find the solution of the equation:
1) 3x(xy – 2) dx + (x3 + 2y) dy = 0
Solution:
Since M(x,y) = 3x(xy – 2) and N(x,y) = x3 + 2y
!M
!N
= 3x 2 !!and!!
= 3x 2
we have
!y
!x
then, it is an exact equation.
F(x, y) = ! M(x, y)dx = ! 3x 2 y + 6x dx = x 3y " 3x 2 + B(y) .
(
)
!F
= N(x, y) ,
!y
Therefore, x3 + B’(y) = N(x,y) = x3 + 2y
then B’(y) = 2y
then B(y) = ! B'(y)dy = y2 + c, where c is an arbitrary constant
To determine B(y) we use the fact
and F(x,y) = x3y – 3x2 + y2 + c.
The one-parameter family solution of the equation is F(x,y) = x3y – 3x2 + y2 + c = k
or F(x,y) = x3y – 3x2 + y2 = s, where s is an arbitrary constant.
2) (2x3 – xy2 – 2y + 3) dx – (x2y + 2x) dy = 0
Solution:
Since M(x,y) = 2x3 – xy2 – 2y + 3 and N(x,y) = – (x2y + 2x)
!M
!N
= "2xy " 2 =
then
, the equation is exact.
!y
!x
This time we get F(x,y) by
!F
x2y2
F(x, y) = "
dy = " N(x, y)dy = " #x 2 y # 2x dy = #
# 2xy + Q(x) .
!y
2
(
)
Let’s determine Q(x),
&
!F(x, y) ! # x 2 y 2
=
"
"
2xy
+
Q(x)
= "xy 2 " 2y + Q)(x) = M(x, y) = 2x 3 " xy 2 " 2y + 3
%
(
!x
!x $
2
'
then Q’(x) = 2x3 + 3,
! ( 2x
)
1 4
x + 3x + c , where c is an arbitrary constant
2
therefore F(x,y) = -½ x2y2 – 2xy + ½ x4 + 3x + c
and the one-parameter family of solutions of the equation is
F(x,y) = -½ x2y2 – 2xy + ½ x4 + 3x = k
then Q(x) = ! Q '(x)dx =
3
+ 3 dx =
3) Solve the I.V.P.
(cos(y) + y cos(x)) dx + (sin(x) – x sin(y)) dy = 0
y(0) = -1
Solution:
Since M(x,y) = cos(y) + y cos(x) and N(x,y) = sin(x) – x sin(y),
then, F(x, y) = ! M(x, y)dx = ! ( cos(y) + y cos(x)) dx = x cos(y) + y sin(x) + B(y)
then,
!F(x, y) !
= ( x cos(y) + y sin(x) + B(y)) = "x sin(y) + sin(x) + B'(x) = N(x, y) = sin x " x sin(y)
!y
!y
then B’(y) = 0,
then B(y) = c, where c is an arbitrary constant.
Thus, F(x,y) = x cos(y) + y sin(x) + c.
Hence, a one-parameter family of solutions is
F(x,y) = x cos(y) + y sin(x) = k
Applying the initial condition y = -1 when x = 0, we have
F(0,1) = 0 cos(-1) – 1 sin(0) = 0 = c
then c = 0, and the unique solution of the I.V.P. is
x cos(y) + y sin(x) = 0.
Remark: The equation (cos(y) + y cos(x)) dx + (sin(x) – x sin(y)) dy = 0 can be solved
by grouping:
(cos(y) dx – x sin(y) dy) + (y cos(x) dx + sin(x) dy) = 0
d(x cos(y)) + d(y sin(x)) = 0 = d(c)
x cos(y) + y sin(x) = c