Exercises 4.2
(d)
By the definition 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. Factoring 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
±
(
−
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
+
y
−
6
y
+ 4
y
= 0 is given by
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 differential equation is
r
3
−
6
r
2
−
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Factoring, Equations, Quadratic equation, Elementary algebra, Quintic equation, auxiliary equation

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