2.4
Exact Equations
The left side is the total differential of
x
2
+
y
2
and the right side is the total differential of
x
+
c
. Thus
x
2
+
y
2
=
x
+
c
is a solution of the differential equation.
44.
To see that the statement is true, write the separable equation as
−
g
(
x
)
dx
+
dy/h
(
y
) = 0. Identifying
M
=
−
g
(
x
)
and
N
= 1
/h
(
y
), we see that
M
y
= 0 =
N
x
, so the differential equation is exact.
45. (a)
In differential form we have (
v
2
−
32
x
)
dx
+
xv dv
= 0. This is not an exact form, but
µ
(
x
) =
x
is an
integrating factor. Multiplying by
x
we get (
xv
2
−
32
x
2
)
dx
+
x
2
v dv
= 0. This form is the total differential
of
u
=
1
2
x
2
v
2
−
32
3
x
3
, so an implicit solution is
1
2
x
2
v
2
−
32
3
x
3
=
c
. Letting
x
= 3 and
v
= 0 we find
c
=
−
288.
Solving for
v
we get
v
= 8
x
3
−
9
x
2
.
(b)
The chain leaves the platform when
x
= 8, so the velocity at this time is
v
(8) = 8
8
3
−
9
64
≈
12
.
7 ft/s
.
46. (a)
Letting
M
(
x, y
) =
2
xy
(
x
2
+
y
2
)
2
and
N
(
x, y
) = 1 +
y
2
−
x
2
(
x
2
+
y
2
)
2
we compute
M
y
=
2
x
3
−
8
xy
2
(
x
2
+
y
2
)
3
=
N
x
,
so the differential equation is exact. Then we have
∂f
∂x
=
M
(
x, y
) =
2
xy
(
x
2
+
y
2
)
2
= 2
xy
(
x
2
+
y
2
)
−
2
f
(
x, y
) =
−
y
(
x
2
+
y
2
)
−
1
+
g
(
y
) =
−
y
x
2
+
y
2
+
g
(
y
)
∂f
∂y
=
y
2
−
x
2
(
x
2
+
y
2
)
2
+
g
(
y
) =
N
(
x, y
) = 1 +
y
2
−
x
2
(
x
2
+
y
2
)
2
.
Thus,
g
(
y
) = 1 and
g
(
y
) =
y
. The solution is
y
−
y
x
2
+
y
2
=
c
. When
c
= 0 the solution is
x
2
+
y
2
= 1.
(b)
The first graph below is obtained in
Mathematica
using
f
(
x, y
) =
y
−
y/
(
x
2
+
y
2
) and
ContourPlot[f[x, y],
{
x, -3, 3
}
,
{
y, -3, 3
}
,
Axes
−
>
True, AxesOrigin
−
>
{
0, 0
}
, AxesLabel
−
>
{
x, y
}
,
Frame
−
>
False, PlotPoints
−
>
100, ContourShading
−
>
False,
Contours
−
>
{
0, -0.2, 0.2, -0.4, 0.4, -0.6, 0.6, -0.8, 0.8
}
]
The second graph uses
x
=
−
y
3
−
cy
2
−
y
c
−
y
and
x
=
y
3
−
cy
2
−
y
c
−
y
.
In this case the
x
-axis is vertical and the
y
-axis is horizontal.
To obtain the third graph, we solve
y
−
y/
(
x
2
+
y
2
) =
c
for
y
in a CAS. This appears to give one real and two complex solutions. When
graphed in
Mathematica
however, all three solutions contribute to the graph. This is because the solutions
involve the square root of expressions containing
c
. For some values of
c
the expression is negative, causing
an apparent complex solution to actually be real.