Introduction to Laplace transforms 585
=
1
s 2 + a 2 [(0) − 1(0 − a)]
=
a
s 2 + a 2 (provided s > 0)
(b) From equation (1),
L{t
2
} =
∞
0
e
−st t
2 dt
=
t 2 e −st
−s
−
2t e −st
s 2 −
2e −st
s 3
∞
0
by integration by parts twice,
=
(0 − 0 − 0) −
0 − 0 −
2
s 3
=
2
s 3 (provided s > 0)
(c) From equation (1),
L{cosh at } = L
1
2
(e
at
+ e
−at
)
,
from Chapter 5
=
1
2
L{e
at
} +
1
2
L{e
−at
},
equations (2) and (3)
=
1
2
1
s − a
+
1
2
1
s − (−a)
from (iii) of Table 61.1
=
1
2
1
s − a
+
1
s + a
=
1
2
(s + a) + (s − a)
(s − a)(s + a)
=
s
s 2 − a 2 (provided s > a)
Problem 4. Determine the Laplace transforms of:
(a) sin 2 t (b) cosh 2 3x.
(a) Since cos 2t = 1 −2sin 2 t then
sin 2 t =
1
2
(1 − cos2t ). Hence,
L{sin
2 t } = L
1
2
(1 − cos 2t )
=
1
2
L{1} −
1
2
L{cos 2t }
=
1
2
1
s
−
1
2
s
s 2 + 2 2
from (i) and (v) of Table 61.1
=
(s 2 + 4) − s 2
2s(s 2 + 4)
=
4
2s(s 2 + 4)
=
2
s(s
2
+ 4)
(b) Since cosh 2x = 2 cosh 2 x − 1 then
cosh 2 x =
1
2
(1 + cosh 2x) from Chapter 5.
Hence cosh 2 3x =
1
2
(1 + cosh 6x)
Thus L{cosh
2 3x} = L
1
2
(1 + cosh 6x)
=
1
2
L{1} +
1
2
L{cosh 6x}
=
1
2
1
s
+
1
2
s
s 2 − 6 2
=
2s 2 − 36
2s(s 2 − 36)
=
s 2 − 18
s(s 2 − 36)
Problem 5. Find the Laplace transform of
3 sin(ωt + α), where ω and α are constants.
Using the compound angle formula for sin(A + B),
from Chapter 17, sin(ωt + α) may be expanded to
(sin ωt cos α + cos ωt sin α). Hence,
L{3sin (ωt + α)}
= L{3(sin ωt cos α + cos ωt sin α)}
= 3 cosαL{sin ωt } + 3 sinαL{cosωt },
since α is a constant
= 3 cosα
ω
s 2 + ω 2
+ 3 sinα
s
s 2 + ω 2
from (iv) and (v) of Table 61.1
=
3
(s 2 + ω 2 )
(ω cos α + s sin α)
Now try the following exercise
Exercise 219 Further problems on an
introduction to Laplace transforms
Determine the Laplace transforms in Problems
1 to 9.