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Exercises 4.2
(d)
By the defnition oF cosh
t
,
y
3
(
t
)=cosh
t
=
e
t
+
e
−
t
2
=
1
2
e
t
+
1
2
e
−
t
=
1
2
y
1
(
t
)+
1
2
y
2
(
t
)
,
and given Functions are linearly dependent on (
−∞
,
∞
).
41.
The auxiliary equation For this problem is
r
3
+
r
2
−
6
r
+ 4 = 0. ±actoring yields
r
3
+
r
2
−
6
r
+4 =
(
r
3
−
r
2
)
+
(
2
r
2
−
2
r
)
+(
−
4
r
+4)
=
r
2
(
r
−
1) + 2
r
(
r
−
1)
−
4(
r
−
1) = (
r
−
1)(
r
2
+2
r
−
4)
.
Thus the roots oF the auxiliary equation are
r
=1 and
r
=
−
2
±
p
(
−
2)
2
−
4(1)(
−
4)
2
=
−
1
±
√
5
.
ThereFore, the Functions
e
t
,
e
(
−
1
−
√
5)
t
,and
e
(
−
1+
√
5)
t
are solutions to the given equation, and
they are linearly independent on (
−∞
,
∞
) (see Problem 40). Hence, a general solution to
y
0
+
y
0
−
6
y
0
+4
y
=0isg
ivenby
y
(
t
)=
c
1
e
t
+
c
2
e
(
−
1
−
√
5)
t
+
c
3
e
(
−
1+
√
5)
t
.
42.
The auxiliary equation associated with this di²erential equation is
r
3
−
6

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