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Exercises 2.5
Solving these equations simultaneously, we obtain
n
=1and
m
=1
. So
,
µ
(
x, y
)=
xy.
With these choices for
n
and
m
we obtain the exact equation
(2
xy
3
−
6
x
2
y
2
)
dx
+(3
x
2
y
2
−
4
x
3
y
)
dy
=0
.
Solving this equation, we have
F
(
x, y
)=
Z
(2
xy
3
−
6
x
2
y
2
)
dx
=
x
2
y
3
−
2
x
3
y
2
+
g
(
y
)
⇒
F
y
(
x, y
)=3
x
2
y
2
−
4
x
3
y
+
g
0
(
y
)=
N
(
x, y
)=3
x
2
y
2
−
4
x
3
y.
Therefore,
g
0
(
y
) = 0. Since the constant of integration can be incorporated into the constant
C
of the solution, we can pick
g
(
y
)
≡
0. Thus, we have
F
(
x, y
)=
x
2
y
3
−
2
x
3
y
2
and the solution becomes
x
2
y
3
−
2
x
3
y
2
=
C.
Since we have multiplied the original equation by
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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