find 1. L ^-1 [ 3/ (s-3) ^2+25]
2. L^-1 [1/ (s-3) ^2 +2/(s+1) ^2 +4]
Answers
Step-by-step explanation:
satisfies
L{f}(s) = F(s),
then we say that f(t) is the inverse Laplace transform of F(s) and employ the notation
f(t) = L
−1{F}(t).
Table of inverse Laplace transform
F(s) f(t) = L
−1{F}(t)
1
s
, s > 0 1
1
s − a
, s > a e
at
(n − 1)!
s
n , s > 0 t
n−1
, n = 1, 2, ...
b
s
2 + b
2
, s > 0 sin bt
s
s
2 + b
2
, s > 0 cos bt
(n − 1)!
(s − a)
n , s > a e
att
n−1
, n = 1, 2, ...
b
(s − a)
2 + b
2
, s > a e
at sin bt
s − a
(s − a)
2 + b
2
, s > a e
at cos bt
Example 1. Determine the inverse Laplace transform of the given function.
(a) F(s) = 2
s
3
.
SOLUTION. L
−1
2
s
3
= L
−1
2!
s
3
= t
2
(b) F(s) = 2
s
2+4 .
SOLUTION. L
−1
2
s
2+4
= L
−1
2
s
2+22
= sin 2t.
(c) F(s) = s+1
s
2+2s+10 .
SOLUTION. L
−1
s+1
s
2+2s+10
= L
−1
n
s+1
(s+1)2+9o
= L
−1
n
s+1
(s+1)2+32
o
= e−t
cos 3t.
Theorem 1. (linearity of the inverse transform) Assume that L
−1{F}, L
−1{F1},
and L
−1{F2} exist and are continuous on [0, ∞) and c is any constant. Then
L
−1
{F1 + F2} = L
−1
{F1} + L
−1
{F2}
L
−1
{cF} = cL
−1
{F}.
Example 2. Determine L
−1
n
3
(2s+5)3 +
2s+16
s
2+4s+13 +
3
s
2+4s+8o
.