In each of Problems 9 through 24, using the linearity of L − 1 , partial fraction expansions, and Table 5.3.1 to find the inverse Laplace transform of the given function: s 3 − 2 s 2 − 6 s − 6 ( s 2 + 2 s + 2 ) s 2 TABLE 5. 3. 1 Elementary Laplace transforms. f ( t ) = L − 1 { F ( s ) } F ( s ) = L { f ( t ) } e a t cos b t s − a ( s − a ) 2 + b 2 , s > a t n , n = positive integer n ! s n + 1 , s > 0
In each of Problems 9 through 24, using the linearity of L − 1 , partial fraction expansions, and Table 5.3.1 to find the inverse Laplace transform of the given function: s 3 − 2 s 2 − 6 s − 6 ( s 2 + 2 s + 2 ) s 2 TABLE 5. 3. 1 Elementary Laplace transforms. f ( t ) = L − 1 { F ( s ) } F ( s ) = L { f ( t ) } e a t cos b t s − a ( s − a ) 2 + b 2 , s > a t n , n = positive integer n ! s n + 1 , s > 0
In each of Problems 9 through 24, using the linearity of
L
−
1
, partial fraction expansions, and Table 5.3.1 to find the inverse Laplace transform of the given function:
s
3
−
2
s
2
−
6
s
−
6
(
s
2
+
2
s
+
2
)
s
2
TABLE 5. 3. 1
Elementary Laplace transforms.
f
(
t
)
=
L
−
1
{
F
(
s
)
}
F
(
s
)
=
L
{
f
(
t
)
}
e
a
t
cos
b
t
s
−
a
(
s
−
a
)
2
+
b
2
,
s
>
a
t
n
,
n
=
positive
integer
n
!
s
n
+
1
,
s
>
0
A Problem Solving Approach To Mathematics For Elementary School Teachers (13th Edition)
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