Exercises 4.4
To choose
s
, we find the roots of the characteristic equation, which is
r
2
+ 9 = 0.
Since
r
=
±
3
i
are its simple roots, we take
s
= 1. Thus
y
p
(
t
) =
t
(
A
3
t
3
+
A
2
t
2
+
A
1
t
+
A
0
)
cos 3
t
+
t
(
B
3
t
3
+
B
2
t
2
+
B
1
t
+
B
0
)
sin 3
t.
29.
The characteristic equation
r
2
−
6
r
+ 9 = (
r
−
3)
2
= 0 has a double root
r
= 3. Therefore, a
particular solution is of the form
y
p
(
t
) =
t
2
(
A
6
t
6
+
A
5
t
5
+
A
4
t
4
+
A
3
t
3
+
A
2
t
2
+
A
1
t
+
A
0
)
e
3
t
.
31.
From the form of the righthand side, we conclude that a particular solution should be of the
form
y
p
(
t
) =
t
s
(
A
3
t
3
+
A
2
t
2
+
A
1
t
+
A
0
)
cos
t
+
(
B
3
t
3
+
B
2
t
2
+
B
1
t
+
B
0
)
sin
t e
−
t
.
Since
r
=
−
1
±
i
are simple roots of the characteristic equation,
r
2
+ 2
r
+ 2 = 0, we should
take
s
= 1. Therefore,
y
p
(
t
) =
t
(
A
3
t
3
+
A
2
t
2
+
A
1
t
+
A
0
)
cos
t
+
(
B
3
t
3
+
B
2
t
2
+
B
1
t
+
B
0
)
sin
t e
−
t
.
33.
The righthand side of the equation suggests that
y
p
(
t
) =
t
s
(
A
cos
t
+
B
sin
t
). By inspection,
we see that
r
=
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Cos, Zagreb, Highways in Croatia, G protein coupled receptors, yp

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