khan (fak337) – HW  9.1, 9.2, 9.3 – nibert – (52800)
1
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printout
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have
20
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before answering.
001
5.0points
If
y
1
satisfies the equations
y
′
+ 4
y
2
ln
x
= 0
,
y
(1) =
1
6
for
x >
1
2
, determine the value of
y
1
(
e
).
1.
y
1
(
e
) =
1
11
2.
y
1
(
e
) =
1
9
3.
y
1
(
e
) =
1
10
correct
4.
y
1
(
e
) =
1
13
5.
y
1
(
e
) =
1
12
Explanation:
After separating variables and integrating,
the given differential equation becomes
−
integraldisplay
1
y
2
dy
=
1
y
=
integraldisplay
4 ln
x dx.
To integrate the right hand side we integrate
by parts. Thus
1
y
= 4(
x
ln
x
−
x
) +
C
with
C
an arbitrary constant. For
y
1
the value
of
C
is determined by the condition
y
(1) =
1
6
since
y
(1) =
1
6
=
⇒
C
= 10
.
Consequently,
y
1
(
x
) =
1
4(
x
ln
x
−
x
) + 10
.
At
x
=
e
, therefore,
y
1
(
e
) =
1
10
.
002
5.0points
Solve
dy
dx
=
8
x
15
y
1 +
x
16
for
y.
1.
arcsin(1 +
x
16
) +
C
2.
arctan(1 +
x
16
) +
C
3.
None of these
correct
4.
arccos(1 +
x
16
) +
C
Explanation:
y
′
=
8
x
15
y
1 +
x
16
y
′
y
=
8
x
15
1 +
x
16
integraldisplay
y
′
y
dx
=
integraldisplay
8
x
15
dx
1 +
x
16
ln

y

=
1
2
integraldisplay
16
x
15
dx
1 +
x
16
=
1
2
ln

1 +
x
16

+
C
0
y
=
e
1
2
ln

1+
x
16

+
C
0
=
e
ln
√
1+
x
16
e
C
0
y
=
C
radicalbig
1 +
x
16
003
5.0points
Use the direction field of the differential
equation
y
′
=
y
3
to sketch a solution curve
that passes through the point (0
,
1).
khan (fak337) – HW  9.1, 9.2, 9.3 – nibert – (52800)
2
1.
−
2
2
−
2
2
x
y
2.
−
2
2
−
2
2
x
y
3.
−
2
2
−
2
2
x
y
4.
−
2
2
−
2
2
x
y
correct
5.
−
2
2
−
2
2
x
y
6.
−
2
2
−
2
2
x
y
Explanation:
The vector field for
y
′
=
y
3
with a sample
curve through our initial condition
y
(0) = 1
looks like the following
−
2
2
−
2
2
x
y
004(part1of3)5.0points
For the differential equation
dy
dx
= 6
x
2
y,
khan (fak337) – HW  9.1, 9.2, 9.3 – nibert – (52800)
3
(i) first find its general solution.
007
5.0points
khan (fak337) – HW  9.1, 9.2, 9.3 – nibert – (52800)
4
If
y
0
is the particular solution of
dy
dx
= 4
x
3
y,
y
(0) = 4
,
find the value of
y
0
(2).
1.
y
0
(2) = 4
e
16
correct
2.
y
0
(2) = 4
e
−
16
3.
y
0
(2) = 4
e
−
8
4.
y
0
(2) = 4
e
5.
y
0
(2) = 4
e
8
Explanation:
The differential equation becomes
integraldisplay
dy
y
=
integraldisplay
4
x
3
dx
after separating
variables and
integrating.
Thus its general solution is
ln
y
=
x
4
+
C,
which can also be written as
y
=
e
C
e
x
4
=
Ae
x
4
with
A
an arbitrary constant. The value of
A