1
Exact Differentials and Integrating Factors
J. R. Rice, AM 105b, 6 February 2008
Consider a first order ode which is given in the form
dy
dx
=
±
M
(
x
,
y
)
N
(
x
,
y
)
.
To solve such an ode is equivalent, after multiplying by
dx
, and rearranging, to finding functions
y
=
y
(
x
)
satisfying
the differential form
M
(
x
,
y
)
dx
+
N
(
x
,
y
)
dy
=
0
As a special case. Let us suppose that the form
M
(
x
,
y
)
dx
+
N
(
x
,
y
)
dy
is a
perfect
differential
(sometimes called an
exact
differential).
That means that a function
f
(
x
,
y
)
exists so that
M
(
x
,
y
)
dx
+
N
(
x
,
y
)
dy
=
df
(
x
,
y
)
,
where
df
(
x
,
y
)
=
±
f
(
x
,
y
)
±
x
dx
+
±
f
(
x
,
y
)
±
y
dy
.
Thus, if such an
f
(
x
,
y
)
does exists, which means that
Mdx
+
Ndy
=
df
for arbitrary ratios of
dx
to
dy,
then it must be the case that
M
(
x
,
y
)
=
±
f
(
x
,
y
)
±
x
and
N
(
x
,
y
)
=
±
f
(
x
,
y
)
±
y
, implying that
±
M
(
x
,
y
)
±
y
=
±
2
f
(
x
,
y
)
±
y
±
x
=
±
2
f
(
x
,
y
)
±
x
±
y
=
±
N
(
x
,
y
)
±
x
To turn that around, if
M
and
N
satisfy
±
M
(
x
,
y
)
±
y
=
±
N
(
x
,
y
)
±
x
, then it can be shown (please
show it) that the differential form
M
(
x
,
y
)
dx
+
N
(
x
,
y
)
dy
is perfect, i.e., that a function
f
(
x
,
y
)
exists (and is unique to within an additive constant) such that
M
(
x
,
y
)
dx
+
N
(
x
,
y
)
dy
=
df
(
x
,
y
)
.
In such case it is trivial to solve
dy
dx
=
±
M
(
x
,
y
)
N
(
x
,
y
)
, and the solutions are just
f
(
x
,
y
)
=
C
(an
arbitrarily chosen constant). If we want the specific solution satisfying
y
=
b
when
x
=
a
, we just
choose the constant
C
=
f
(
a
,
b
)
.
Of course, in general if we are given two functions
M
(
x
,
y
) and
N
(
x
,
y
)
, it must be
anticipated that
±
M
(
x
,
y
)
±
y
²
±
N
(
x
,
y
)
±
x
, and hence that
M
(
x
,
y
)
dx
+
N
(
x
,
y
)
dy
is
not
a
perfect