pacheco (jnp926) – Homework 6 – staron – (52840)
1
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printout
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have
13
questions.
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before answering.
001(part1of3)10.0points
For the differential equation
dy
dx
+ 3
y
= 8
e
3
x
,
(i) first find its general solution.
1.
y
=
4
3
e
3
x
+
C e
−
3
x
correct
2.
y
=
4
3
e
3
x
+
Ce
3
x
3.
y
=
1
6
e
3
x
+
Ce
−
3
x
4.
y
=
4
3
e
−
3
x
+
Ce
−
3
x
5.
y
=
4
3
e
−
3
x
+
Ce
3
x
Explanation:
The integrating factor for the first order
differential equation
dy
dx
+ 3
y
= 8
e
3
x
is
μ
=
e
integraltext
3
dx
=
e
3
x
.
Thus after multiplying both sides by
e
3
x
we
can rewrite it as
d
dx
parenleftBig
y e
3
x
parenrightBig
= 8
e
6
x
.
Consequently, its general solution is given by
y e
3
x
=
integraldisplay
e
6
x
dx
=
4
3
e
6
x
+
C
where
C
is an arbitrary constant, so
y
=
4
3
e
3
x
+
C e
−
3
x
with
C
an arbitrary constant.
002(part2of3)10.0points
(ii) Then find the particular solution
y
0
such
that
y
0
(0) = 8.
1.
y
0
=
4
3
e
3
x
+
20
3
e
−
3
x
correct
2.
y
0
=

4
3
e
3
x
+
20
3
e
−
3
x
3.
y
0
=
4
3
e
3
x

20
3
e
−
3
x
4.
y
0
=
1
6
e
3
x
+
20
3
e
−
3
x
5.
y
0
=
4
3
e
−
3
x
+
20
3
e
3
x
Explanation:
For the particular solution
y
0
the value of
C
is determined by the condition
y
(0) = 8 since
y
(0) = 8
=
⇒
8 =
4
3
+
C.
Consequently,
y
0
=
4
3
e
3
x
+
20
3
e
−
3
x
.
003(part3of3)10.0points
(iii) For the particular solution
y
0
in (ii), deter
mine the value of
y
0
(1).
1.
4
3
e
3
+
20
3
e
−
3
correct
2.
1
6
e
3

20
3
e
−
3
3.
1
6
e
3
+
20
3
e
−
3
4.
4
3
e
−
3
+
20
3
e
3
5.
4
3
e
3

20
3
e
−
3
pacheco (jnp926) – Homework 6 – staron – (52840)
2
Explanation:
At
x
= 1, therefore,
y
0
(1) =
4
3
e
3
+
20
3
e
−
3
.
004
10.0points
If
y
0
is the particular solution of the differ
ential equation
dy
dx
+ 3
y
+ 3
e
4
x
= 0
such that
y
(0) = 5, find the value of
y
0
(1).
1.
y
0
(1) =
3
7
e
4
+
32
7
e
−
3
2.
y
0
(1) =

3
7
e
4
+
38
7
e
−
3
correct
3.
y
0
(1) =

3
7
e
4

32
7
e
−
3
4.
y
0
(1) =

3
7
e
−
4

38
7
e
3
5.
y
0
(1) =
3
7
e
−
4
+
38
7
e
3
Explanation: