This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
(
as
+
b
)(
s
2
+ 1) + (
cs
+
d
)(
s
2

1) =
s
2
for all
s
. By setting
s
= 1 and
s
=

1, respectively in the last equation, we
obtain the pair of equations
2(
a
+
b
) = 1
,
2(

a
+
b
) = 1
,
and therefore
a
= 0 and
b
= 1
/
2
.
If we set
s
= 0, then
b

d
= 0 so
d
= 1
/
2
.
Finally, equating the coeﬃcients of the cubic terms on each side, we ﬁnd that
a
+
c
= 0, so
c
= 0
.
Thus
Y
(
s
) =
1
/
2
s
2

1
+
1
/
2
s
2
+ 1
,
and from line 4 and 6 of Table 5.1, the solution of the initial value problem is
y
=
sinh
t
+ sin
t
2
.
2
Let
u
c
(
t
) be the unit step function
u
c
(
t
) =
‰
0
,
t < c
1
, t
≥
c.
Then we have the following useful relations
L{
u
c
(
t
)
f
(
t

c
)
}
=
e

cs
F
(
s
)
,
L{
e
ct
f
(
t
)
}
=
F
(
s

c
)
.
In particular,
L{
u
c
(
t
)
}
=
e

cs
s
.
Let
δ
(
t
) be the Dirac function, then
Z
∞
∞
δ
(
t

c
)
f
(
t
)
dt
=
f
(
c
)
.
If
c >
0, then from above it follows
L{
δ
(
t

c
)
f
(
t
)
}
=
e

cs
f
(
c
)
.
In particular,
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 01/28/2011 for the course MATH 150 taught by Professor T.qian during the Spring '09 term at HKUST.
 Spring '09
 T.Qian
 Equations

Click to edit the document details