*This preview shows
page 1. Sign up
to
view the full content.*

Exercises 4.3
So,
α
=
−
1,
β
= 2, and a general solution is given by
y
(
t
)=
c
1
e
−
t
cos 2
t
+
c
2
e
−
t
sin 2
t.
15.
First, we fnd the roots o± the auxiliary equation.
r
2
+10
r
+41=0
⇒
r
=
−
10
±
p
10
2
−
4(1)(41)
2
=
−
5
±
4
i.
These are complex numbers with
α
=
−
5and
β
= 4. Hence, a general solution to the given
di²erential equation is
y
(
t
)=
c
1
e
−
5
t
cos 4
t
+
c
2
e
−
5
t
sin 4
t.
17.
The auxiliary equation in this problem,
r
2
−
r
+ 7 = 0, has the roots
r
=
1
±
p
1
2
−
4(1)(7)
2
=
1
±
√
−
27
2
=
1
2
±
3
√
3
2
i.
There±ore, a general solution is
y
(
t
)=
c
1
e
t/
2
cos
3
√
3
2
t
+
c
2
e
t/
2
sin
3
√
3
2
t
.
19.
The auxiliary equation,
r
3
+
r
2
+3
r
−
5 = 0, is a cubic equation. Since any cubic equation has
a real root, frst we examine the divisors o± the ±ree coeﬃcient, 5, to fnd integer real roots (i±

This is the end of the preview. Sign up
to
access the rest of the document.