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Exercises 4.4
15.
For this problem, the corresponding homogeneous equation is
y
0
−
5
y
0
+6
y
= 0, which has
the associated auxiliary equation
r
2
−
5
r
+ 6 = 0. The roots of this equation are
r
=3and
r
= 2. Therefore, neither
y
=
e
x
nor
y
=
xe
x
satis±es the homogeneous equation, and in the
expression
y
p
(
x
)=
x
s
(
Ax
+
B
)
e
x
for a particular solution we can take
s
=0. So
y
p
(
x
)=(
Ax
+
B
)
e
x
⇒
y
0
p
(
x
Ax
+
B
+
A
)
e
x
⇒
y
0
p
(
x
Ax
+
B
+2
A
)
e
x
⇒
(
Ax
+
B
A
)
e
x
−
5(
Ax
+
B
+
A
)
e
x
+6(
Ax
+
B
)
e
x
=
xe
x
⇒
(2
Ax
−
3
A
B
)
e
x
=
xe
x
⇒
2
A
=1
,
−
3
A
B
=0
⇒
A
/
2
,
B
=3
/
4
,
and
y
p
(
x
x/
2+3
/
4)
e
x
.
16.
The corresponding homogeneous equation has the auxiliary equation
r
2
−
1 = 0, whose roots
are
r
=
±
1. Thus, in the expression
θ
p
(
t
A
1
t
+
A
0
)cos
t
+(
B
1
t
+
B
0
)sin
t
none of the
terms is a solution to the homogeneous equation. We ±nd
θ
p
(
t
A
1
t
+
A
0
t
B
1
t
+
B
0
t
⇒
θ
0
p
(
t
A
1
cos
t
−
(
A
1
t
+
A
0
t
+
B
1
sin
t
B
1
t
+
B
0
t
=(
B
1
t
+
A
1
+
B
0
t
−
A
1
t
−
A
0
+
B
1
t
⇒
θ
0
p
(
t
B
1
cos
t
−
(
B
1
t
+
B
0
+
A
1
t
−
A
1
sin
t
−
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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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