Exercises 2.5
equation has an integrating factor depending on
y
alone. Also, since
∂M/∂y
−
∂N/∂x
N
=
(6
y
2
+4
y
)
−
(3
y
2
+2
y
)
3
y
2
x
xy
=
3
y
2
y
x
(3
y
2
y
)
=
1
x
,
the equation has an integrating factor depending on
x
.
Writing the equation in the form
dx
dy
=
−
3
y
2
x
xy
2
y
3
+
y
2
=
−
xy
(3
y
+2)
2
y
2
(
y
+1)
=
−
y
(3
y
2
y
2
(
y
x
we conclude that it is separable and linear with
x
as the dependent variable.
3.
This equation is not separable because of the factor (
y
2
xy
). It is not linear because of the
factor
y
2
. To see if it is exact, we compute
M
y
(
x, y
)and
N
x
(
x, y
), and see that
M
y
(
x, y
)2
y
x
6
=
−
2
x
=
N
x
(
x, y
)
.
Therefore, the equation is not exact. To see if we can Fnd an integrating factor of the form
µ
(
x
), we compute
∂M
∂y
−
∂N
∂x
N
=
2
y
x
−
x
2
,
which is not a function of
x
alone. To see if we can Fnd an integrating factor of the form
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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