Suppose
μ
does not depend on
x
. Then
μ
x
= 0. Therefore,
μ
(
N
x

M
y
) =
μ
y
M
. Using the
assumption that (
N
x

M
y
)
/M
=
Q
(
y
), we can find an integrating factor
μ
by choosing
μ
which satisfies
μ
y
/μ
=
Q
. We conclude that
μ
(
y
) = exp
R
Q
(
y
)
dy
is an integrating factor of
the differential equation.
24. Suppose
μ
is an integrating factor which will make the equation exact. Then multiplying
the equation by
μ
, we have
μMdx
+
μNdy
= 0. Then we need (
μM
)
y
= (
μN
)
x
. That is, we
need
μ
y
M
+
μM
y
=
μ
x
N
+
μN
x
. Then we rewrite the equation as
μ
(
N
x

M
y
) =
μ
y
M

μ
x
N
.
By the given assumption, we need
μ
to satisfy
μR
(
xM

yN
) =
μ
y
M

μ
x
N
. This equation
is satisfied if
μ
y
= (
μx
)
R
and
μ
x
= (
μy
)
R
. Consider
μ
=
μ
(
xy
). Then
μ
x
=
μ
0
y
and
μ
y
=
μ
0
x
2.6.
EXACT EQUATIONS AND INTEGRATING FACTORS
73
where
0
=
d/dz
for
z
=
xy
. Therefore, we need to choose
μ
to satisfy
μ
0
=
μR
. This equation
is separable with solution
μ
= exp(
R
R
(
z
)
dz
).
25.(a) Since (
M
y

N
x
)
/N
= 3 is a function of
x
only, we know that
μ
=
e
3
x
is an integrating
factor for this equation. Multiplying the equation by
μ
, we have
e
3
x
(3
x
2
y
+ 2
xy
+
y
3
)
dx
+
e
3
x
(
x
2
+
y
2
)
dy
= 0
.
Then
M
y
=
e
3
x
(3
x
2
+ 2
x
+ 3
y
2
) =
N
x
. Therefore, this new equation is exact. Integrating
M
with respect to
x
, we conclude that
ψ
= (
x
2
y
+
y
3
/
3)
e
3
x
+
h
(
y
). Then
ψ
y
= (
x
2
+
y
2
)
e
3
x
+
h
0
(
y
) =
N
=
e
3
x
(
x
2
+
y
2
). Therefore,
h
0
(
y
) = 0 and we conclude that the solution is given
implicitly by (3
x
2
y
+
y
3
)
e
3
x
=
c
.
(b)
26.(a) Since (
M
y

N
x
)
/N
=

1 is a function of
x
only, we know that
μ
=
e

x
is an
integrating factor for this equation. Multiplying the equation by
μ
, we have
(
e

x

e
x

ye

x
)
dx
+
e

x
dy
= 0
.
Then
M
y
=

e

x
=
N
x
. Therefore, this new equation is exact. Integrating
M
with respect
to
x
, we conclude that
ψ
=

e

x

e
x
+
ye

x
+
h
(
y
). Then
ψ
y
=
e

x
+
h
0
(
y
) =
N
=
e

x
.
Therefore,
h
0
(
y
) = 0 and we conclude that the solution is given implicitly by

e

x

e
x
+
ye

x
=
c
.
(b)
74
CHAPTER 2.
FIRST ORDER DIFFERENTIAL EQUATIONS
27.(a) Since (
N
x

M
y
)
/M
= 1
/y
is a function of
y
only, we know that
μ
(
y
) =
e
R
1
/y dy
=
y
is an integrating factor for this equation. Multiplying the equation by
μ
, we have
ydx
+ (
x

y
sin
y
)
dy
= 0
.
Then for this equation,
M
y
= 1 =
N
x
. Therefore, this new equation is exact. Integrating
M
with respect to
x
, we conclude that
ψ
=
xy
+
h
(
y
). Then
ψ
y
=
x
+
h
0
(
y
) =
N
=
x

y
sin
y
.
Therefore,
h
0
(
y
) =

y
sin
y
which implies that
h
(
y
) =

sin
y
+
y
cos
y
, and we conclude that
the solution is given implicitly by
xy

sin
y
+
y
cos
y
=
c
.
(b)
28.(a) Since (
N
x

M
y
)
/M
= (2
y

1)
/y
is a function of
y
only, we know that
μ
(
y
) =
e
R
2

1
/y dy
=
e
2
y
/y
is an integrating factor for this equation. Multiplying the equation by
μ
,
we have
e
2
y
dx
+ (2
xe
2
y

1
/y
)
dy
= 0
.
Then for this equation,
M
y
=
N
x
. Therefore, this new equation is exact. Integrating
M
with
respect to
x
, we conclude that
ψ
=
xe
2
y
+
h
(
y
). Then
ψ
y
= 2
xe
2
y
+
h
0
(
y
) =
N
= 2
xe
2
y

1
/y
.
Therefore,
h
0
(
y
) =

1
/y
which implies that
h
(
y
) =

ln
y
, and we conclude that the solution
is given implicitly by
xe
2
y

ln
y
=
c
or
y
=
e
2
x
+
ce
x
+ 1.
You've reached the end of your free preview.
Want to read all 70 pages?
 Spring '13
 YOYL
 initial condition, First Order Equations, τ dτ