WrittenHW07-Solutions - Math 216(Section 50 1(4.14 Let us...

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

4 ( 4.1–16 )
Given x = 8 y and y = - 2 x ,
x = 8 y = - 16 x = ⇒ x + 16 x = 0 .