588 Higher Engineering Mathematics
Problem 1. Determine (a) L{2t
4 e
3t
}
(b) L{4e 3t cos 5t }.
(a) From (i) of Table 62.1,
L{2t
4 e
3t
} = 2L{t
4 e
3t
} = 2
4!
(s − 3) 4+1
=
2(4)(3)(2)
(s − 3) 5 =
48
(s − 3)
5
(b) From (iii) of Table 62.1,
L{4e
3t cos 5t } = 4L{e
3t cos 5t }
= 4
s − 3
(s − 3) 2 + 5 2
=
4(s − 3)
s 2 − 6s + 9 + 25
=
4(s − 3)
s 2 − 6s +34
Problem 2. Determine (a) L{e −2t sin 3t }
(b) L{3e θ cosh 4θ}.
(a) From (ii) of Table 62.1,
L{e
−2t sin 3t }=
3
(s − (−2)) 2 + 3 2 =
3
(s +2) 2 + 9
=
3
s 2 + 4s + 4 + 9
=
3
s 2 + 4s + 13
(b) From (v) of Table 62.1,
L{3e
θ cosh 4θ} = 3L{e
θ cosh 4θ}=
3(s − 1)
(s − 1) 2 − 4 2
=
3(s − 1)
s 2 −2s +1−16
=
3(s − 1)
s 2 − 2s −15
Problem 3. Determine the Laplace transforms of
(a) 5e −3t sinh 2t (b) 2e 3t (4 cos 2t − 5 sin2t ).
(a) From (iv) of Table 62.1,
L{5e
−3t sinh 2t } = 5L{e
−3t sinh 2t }
= 5
2
(s − (−3)) 2 − 2 2
=
10
(s + 3) 2 − 2 2 =
10
s 2 + 6s+9 − 4
=
10
s 2 + 6s + 5
(b) L{2e 3t (4 cos 2t − 5 sin 2t )}
= 8L{e
3t cos 2t } − 10L{e
3t sin 2t }
=
8(s − 3)
(s − 3) 2 + 2 2 −
10(2)
(s − 3) 2 + 2 2
from (iii) and (ii) of Table 62.1
=
8(s − 3) − 10(2)
(s − 3) 2 + 2 2 =
8s − 44
s 2 − 6s + 13
Problem 4. Show that
L
3e
−
1
2 x sin
2 x
=
48
(2s + 1)(4s 2 + 4s + 17)
Since cos 2x = 1 −2 sin 2 x, sin 2 x =
1
2
(1 − cos 2x).
Hence,
L
3e
−
1
2 x sin 2 x
= L
3e
−
1
2 x 1
2
(1 − cos 2x)
=
3
2
L
e
−
1
2 x
−
3
2
L
e
−
1
2 x cos 2x
=
3
2
⎛
⎜
⎜
⎝
1
s −
−
1
2
⎞
⎟
⎟
⎠ −
3
2
⎛
⎜
⎜
⎜
⎝
s −
−
1
2
s −
−
1
2
2
+2 2
⎞
⎟
⎟
⎟
⎠
from (iii) of Table 61.1 (page 584) and (iii)
of Table 62.1 above,
=
3
2
s +
1
2
−
3
s +
1
2
2
s +
1
2
2
+ 2 2
=
3
2s + 1
−
6s + 3
4
s 2 + s +
1
4
+ 4