Chapter 4
21. (a)
With
y
(
t
)=
e
rt
,
y
0
(
t
)=
re
rt
, the equation becomes
are
rt
+
be
rt
=(
ar
+
b
)
e
rt
=0
.
Since the function
e
rt
is never zero on (
−∞
,
∞
), to satisfy the above equation we must
have
ar
+
b
=0
.
(b)
Solving the characteristic equation,
ar
+
b
= 0, obtained in part (a), we get
r
=
−
b/a
.So
y
(
t
)=
e
rt
=
e
−
bt/a
, and a general solution is given by
y
=
ce
−
bt/a
,where
c
is an arbitrary
constant.
23.
We form the characteristic equation, 5
r
+ 4 = 0, and Fnd its root
r
=
−
4
/
5. Therefore,
y
(
t
)=
ce
−
4
t/
5
is a general solution to the given equation.
25.
The characteristic equation, 6
r
−
13 = 0, has the root
r
=13
/
6. Therefore, a general solution
is given by
w
(
t
)=
ce
13
t/
6
.
27.
Assuming that
y
1
(
t
)=
e
−
t
cos 2
t
and
y
2
(
t
)=
e
−
t
sin 2
t
are linearly dependent on (0
,
1), we
conclude that, for some constant
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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