Exercises 4.3
49.
(a)
To find the roots of the auxiliary equation,
p
(
r
) := 3
r
3
+ 18
r
2
+ 13
r
−
19 = 0, one can
use Newton’s method or intermediate value theorem. We note that
p
(
−
5) =
−
9
<
0
,
p
(
−
4) = 25
>
0
,
p
(
−
2) = 3
>
0
,
p
(
−
1) =
−
17
<
0
,
p
(0) =
−
19
<
0
,
p
(1) = 15
>
0
.
Therefore, the roots of
p
(
r
) belong to the intervals [
−
5
,
−
4], [
−
2
,
−
1], and [0
,
1], and
we can take
r
=
−
5,
r
=
−
2, and
r
= 0 as initial quesses.
Approximation yields
r
1
≈ −
4
.
832,
r
2
≈ −
1
.
869, and
r
3
≈
0
.
701. So, a general solution is given by
y
(
t
) =
c
1
e
r
1
t
+
c
2
e
r
2
t
+
c
3
e
r
3
t
=
c
1
e
−
4
.
832
t
+
c
2
e
−
1
.
869
t
+
c
3
e
0
.
701
t
.
(b)
The auxiliary equation,
r
4
−
5
r
2
+ 5 = 0, is of quadratic type. The substitution
s
=
r
2
yields
s
2
−
5
s
+ 5 = 0
⇒
s
=
5
±
√
5
2
⇒
r
=
±
√
s
=
±
5
±
√
5
2
.
Therefore,
r
1
=
5
−
√
5
2
≈
1
.
176
,
r
2
=
5 +
√
5
2
≈
1
.
902
,
r
3
=
−
r
1
,
and
r
4
=
−
r
2
are the roots of the auxiliary equation, and a general solution to
y
(ıv)
−
5
y
+ 5
y
= 0 is
given by
y
(
t
) =
c
1
e
r
1
t
+
c
2
e
−
r
1
t
+
c
3
e
r
2
t
+
c
4
e
−
r
2
t
.
(c)
We can use numerical tools to find the roots of the auxiliary fifth degree polynomial
equation
r
5
−
3
r
4
−
5
r
3
+ 15
r
2
+ 4
r
−
12 = 0. Alternatively, one can involve the rational