MATH 231, Worksheet
Date: April 27, 2012
Find the inverse Laplace transform for each function.
1
s2 (s2 + 4)
Answer: The partial fraction expansion for F (s) is
1. F (s) =
1
A B
Cs + D
=
+ 2+ 2
.
s2 (s2 + 4)
s
s
s +4
Multiplying through by the common denominator yields the equation
1 = As(s2 + 4) + B(s2 + 4) + (Cs + D)s2
= (A + C)s3 + (B + D)s2 + 4As + 4B.
Since this must hold for all real s, the coefficients of the various powers of s must be equal:
s3 :
s2 :
s1 :
s0 :
0
0
0
1
=
=
=
=
A+C
B+D
4A
4B
C
D
A
B
⇒
=
=
=
=
0
−1/4
0
1/4
Thus,
L−1
1
s2 (s2 + 4)
=
1 −1
L
4
1
s2
1
− L−1
8
2
s2 + 4
=
1
1
t − cos(2t).
4
8
s
+ 6s + 11
Answer: First, note that s2 + 6s + 11 is an irreducible quadratic, having nonreal roots. Thus,
we complete the square:
2. F (s) =
s2
−1
L
s
s2 + 6s + 11
−1
= L
−1
= L
−1
s 3 −
s2 + 2 s7→s+3 s2 + 2 s7→s+3
s 3
− √ L−1
s2 + 2 s7→s+3
2
= L
= L
−1
s
(s2 + 6s + 9) + 2
(s + 3) − 3
(s + 3)2 + 2
( √
2 s2 + 2 s7→s+3
)
√
√
3
= e−3t cos( 2t) − √ e−3t sin( 2t).
2
s
+ 6s + 11
Answer: Building on our answer to Number 2, we have
3. F (s) = e−s
s2
L−1 e−s
s
2
s + 6s + 11
√
√
3
= u1 (t) e−3(t−1) cos( 2(t − 1)) − √ e−3(t−1) sin( 2(t − 1)) .
2
1
1
+ 2
3
(s − 1)
s + 2s − 8
Answer: We first perform partial fraction decomponsition on the second term. [Note that
such a decomposition of the first term cannot make it simpler.]
4. F (s) =
1
1
A
B
=
=
+
.
+ 2s − 8
(s − 2)(s + 4)
s−2 s+4
s2
We find that A = 1/6 and B = −1/6. Thus,
−1
−1
L
{F (s)} = L
5. F (s) = e−2s
1
1
1
1
1
+ ·
− ·
3
(s − 1)
6 s−2 6 s+4
1
2 + L−1
3
s s7→s−1
6
=
1 −1
L
2
=
1 2 t 1 2t 1 −4t
t e + e − e .
2
6
6
1
s−2
1
− L−1
6
1
s+4
1
1
+ e−s 2
3
(s − 1)
s + 2s − 8
Answer: [Note: I changed the problem from es to e−s in the 2nd term.] First, if we look
carefully back at the work of the previous problem, we note that
−1
L
1
(s − 1)3
1
= t2 et
2
and
−1
L
1
2
s + 2s − 8
=
1 2t
(e − e−4t ).
6
Using Entry 14 from Table 6.2.1, we get
L−1 {F (s)} =
1
1
u2 (t)(t − 2)2 et−2 + u1 (t) e2(t−1) − e−4(t−1) .
2
6
e−πs/2
s2 + 9
Answer: First, we deal with (s2 + 9)−1 ; that is, F (s) without its exponential factor. We
have
1
1
3
1
L−1 2
= L−1 2
= sin(3t).
s +9
3
s +9
3
6. F (s) =
The exponential factor in F (s) calls for using Entry 14, again, giving
L−1 {F (s)} =
7. F (s) =
1
uπ (t) sin(3(t − π)).
3
(s + 1)e−3s
(s2 + 2s + 2)2
Answer: [Note that I changed e3s to e−3s in this problem.] This is not one I’d expect
students this semester to do, as it uses an entry, Entry 19, we have not discussed, nor have
we seen examples of its use. If you wish to give it a try, the correct answer is
1
L−1 {F (s)} = u3 (t)(t − 3)e−(t−3) sin(t − 3).
2