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Properties of Laplace Transform
.

Example 2.2, f(t) = sin bt. Evaluate Lf(t).
.
. . . . .
Solution. Apply the Euler formula and Property 1 again

ibt
L[e ] = L[cos bt] + iL[sin bt],

1 s
= + iL[sin bt],
2
s − ib s + b 2
( )
1 1 s 1 s + ib − s b
L[sin bt] = − = = .
2
2
i s − ib s + b 2 i (s − ib)(s + ib) s + b 2
Thus,
b
L[sin bt] = .
2
s + b 2
. . . .
Let us apply formulas obtained in previous examples to find the following
.
Laplace transforms.

2t 2
Example 2.3, f(t) = (1 + e ) . Evaluate Lf(t).
.
. . . . .
2t 2
2t
4t
Solution. To do this we first note that (1 + e ) = 1 + 2e + e . So,
1 2 1
4t
2t
L[f(t)] = L[1 + 2e + e ] = + + .
s s − 2 s − 4
. . . .
.
2
Example 2.4, f(t) = (cos t + sin t) . Evaluate Lf(t).
.
. . . . .
2
Solution. Since (cos t + sin t) = 1 + 2 sin t cos t = 1 + sin 2t, so

1 2
L[f(t)] = L[1 + sin 2t] = + .
2
s s + 4
. . . .
.

Example 2.5, Use the linearity property of the Laplace transform
2
to determine a) L[3t +4t−2]; b) L[cosh (at)]; c) L[sinh (at)], where
a is a real constant.
. . . . .
.
Solution. a)

2
2
2
L[3t + 4t − 2] = L[3t ] + L[4t] + L[−2] = 3L[t ] + 4L[t] − 2L[1] =
2 1 1 6 4 2
. . . . = 3 · s 3 + 4 · s 2 − 2 · s = s 3 + s 2 − .
s

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