Math 216 (Section 50)
Written Homework #7 – Solutions
Fall 2007
1
(
4.1–4
)
Let us first rewrite
x
(3)
-
2
t
x
+
1
t
2
x
+
5
t
3
x
=
ln
t
t
3
.
Let us define
x
1
:=
x,
x
2
:=
x
=
x
1
,
x
3
:=
x
=
x
2
.
Therefore the given differential equations are rewritten as follows:
x
1
=
x
2
x
2
=
x
3
x
3
-
2
t
x
3
+
1
t
2
x
2
+
5
t
3
x
1
=
ln
t
t
3
=
⇒
x
1
=
x
2
x
2
=
x
3
x
3
=
2
t
x
3
-
1
t
2
x
2
-
5
t
3
x
1
+
ln
t
t
3
.
2
(
4.1–8
)
Let us define
x
1
:=
x,
x
2
:=
x
=
x
1
,
y
1
:=
y,
y
2
:=
y
=
y
1
.
Note that then
x
=
x
2
and
y
=
y
2
. Therefore the given system of differential equations are
rewritten as follows:
x
+ 3
x
+ 4
x
-
2
y
= 0
y
+ 2
y
-
3
x
+
y
= cos
t
=
⇒
x
1
=
x
2
y
1
=
y
2
x
2
+ 3
x
2
+ 4
x
1
-
2
y
1
= 0
x
2
+ 2
y
2
-
3
x
1
+
y
1
= cos
t
=
⇒
x
1
=
x
2
y
1
=
y
2
x
2
=
-
3
x
2
-
4
x
1
+ 2
y
1
y
2
=
-
2
y
2
+ 3
x
1
-
y
1
+ cos
t
3
(
4.1–9
)
Let us define
x
1
:=
x,
x
2
:=
x
=
x
1
,
y
1
:=
y,
y
2
:=
y
=
y
1
.
z
1
:=
z,
z
2
:=
z
=
z
1
.
Note that then
x
=
x
2
,
y
=
y
2
, and
z
=
z
2
.
Therefore the given system of differential
equations are rewritten as follows:
x
= 3
x
-
y
+ 2
z
y
=
x
+
y
-
4
z
z
= 5
x
-
y
-
z
=
⇒
x
1
=
x
2
y
1
=
y
2
z
1
=
z
2
x
2
= 3
x
1
-
y
1
+ 2
z
1
y
2
=
x
1
+
y
1
-
4
z
1
z
2
= 5
x
1
-
y
1
-
z
1
1
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4
(
4.1–16
)
Given
x
= 8
y
and
y
=
-
2
x
,
x
= 8
y
=
-
16
x
=
⇒
x
+ 16
x
= 0
.

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- Fall '07
- Ringenberg
- Cos, e2t
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