Engineering Mathematics
Thursday, 13 November 2014
Laplace Transforms
Let
f
(
t
)
be
a
function
of
t
defined
for
all
positive
values
of
t
.
Then
the
Laplace
Transform
of
f
(
t
)
is
,
L
f
(
t
)
=
∫
0
∞
e
−
st
f
(
t
)
dt
s
is
a
parameter
which
may
be
real
or
complex
number
.
Transforms
of
Elementary
Functions
L
1
=
1
s
proof
L
t
n
=
n
!
s
n
+
1
proof
L
e
at
=
1
s
−
a
proof
L
sin
at
=
a
s
2
+
a
2
proof
L
cos
at
=
s
s
2
+
a
2
proof
L
sinh
at
=
a
s
2
−
a
2
proof
L
cosh
at
=
s
s
2
−
a
2
proof
Properties
of
Laplace
Transforms
Linearity
First Shifting
Change of Scale
Solved
Problems
1
.
e
at
t
n
solution
2
.
e
at
sin
bt
solution
3
.
e
at
cos
bt
solution
4
.
e
at
sinh
bt
solution
5
.
e
at
cosh
bt
solution
6
.
sin
2
t
sin
3
t
solution
7
.
cos
2
2
t
solution
8
.
sin
3
2
t
solution
9
.
e
−
3
t
(
2
cos
5
t
−
3
sin
5
t
)
solution
10
.
e
2
t
cos
2
t
solution
11
.
√
te
3
t
solution
12
.
f
(
t
)
=
t
/
τ
,
when
0
<
t
<
τ
1
,
when
t
>
τ
solution
13
.
f
(
t
)
=
1
,
0
<
t
≤
1
t
,
1
<
t
≤
2
0
,
t
>
2
solution
14
.
√
t
−
1
√
t
3
solution
15
.
cos
√
t
√
t
solution
16
.
sin
at
t
solution
17
.
e
2
t
+
4
t
3
−
2
sin
3
t
+
3
cos
3
t
solution
18
.
1
+
2
√
t
+
3
√
t
solution
19
.
3
cosh
5
t
−
4
sinh
5
t
solution
20
.
cos
(
at
+
b
)
solution
21
.
sin
t
−
cos
t
2
solution
22
.
sin
2
t
cos
3
t
solution
23
.
sin
√
t
solution
24
.
sin
5
t
solution
25
.
cos
3
2
t
solution
26
.
e
−
at
sinh
bt
solution
27
.
e
2
t
3
t
5
−
cos
4
t
solution
28
.
e
−
3
t
sin
5
t
sin
3
t
solution
29
.
e
−
t
sin
2
t
solution
30
.
e
2
t
sin
4
t
solution
31
.
cosh
at
sin
at
solution
32
.
sinh
3
t
cos
2
t
solution
33
.
t
2
e
2
t
solution
34
.
1
+
te
−
t
3
solution
35
.
t
√
1
+
sin
t
solution
36
.
f
(
t
)
=
4
,
0
≤
t
≤
1
3
,
t
>
1
solution
37
.
f
(
t
)
=
sin
t
,
0
<
t
<
π
0
,
t
>
π
solution
38
.
f
(
x
)
=
sin
(
x
−
π
/
3
),
x
>
π
/
3
0
,
x
<
π
/
3
solution
39
.
f
(
t
)
=
cos
(
t
−
2
π
/
3
),
t
>
2
π
/
3
0
,
t
<
2
π
/
3
solution
40
.
f
(
t
)
=
t
2
,
0
<
t
<
2
t
−
1
,
2
<
t
<
3
7
,
t
>
3
solution
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