6.1. s2 + 2s + 10 = (s + 1)2 + 32 , and
s2
s+1
5
2s − 3
2(s + 1) − 5
3
=2·
− ·
,
=
2
2
2
2
+ 2s + 10
(s + 1) + 3
(s + 1) + 3
3 (s + 1)2 + 32
It follows that the inverse Laplace transform of F is 2e−t cos 3t − 35 e−t sin 3t.
6.2. Laplace transform of the equation is
s2 Y −s−1−4sY +4+4Y = 0,
Y =
1
s−2−1
1
s−3
=
,
=
−
s2 − 4s + 4
(s − 2)2
s − 2 (s − 2)2
hence y(t) = e2t − te2t .
6.3. Let us find at first the inverse Laplace transform of H(s) =
s2
s−2
s2 −4s+3 :
s−2
A
B
s−2
=
=
+
,
− 4s + 3
(s − 3)(s − 1)
s−3 s−1
where
A(s − 1) + B(s − 3) = s − 2.
If we substitute s = 3, we get 2A = 1, hence A = 1/2. If we substitute
s =
1,
s−2
1
1
1
we get −2B = −1, hence B = 1/2, and we have s2 −4s+3 = 2 s−3 + s−1 ,
hence the inverse Laplace transform h(t) of H(s) is 12 (e3t + et ). Consequently,
the inverse Laplace transform of F (s) is
1
u2 (t) · h(t − 2) = u2 (t) · (e3t−6 + et−2 ).
2
6.4. Laplace transform of the equation is
s2 Y + 4Y =
since sin t = sin(t − 2π).
Hence
Y =
(s2
1
e−2πs
−
,
s2 + 1 s2 + 1
1
e−2πs
− 2
.
2
+ 1)(s + 4) (s + 1)(s2 + 4)
Let us find the inverse Laplace transform of H(s) =
for a decomposition
1
(s2 +1)(s2 +4) .
We are looking
1
As + B
Cs + D
= 2
+ 2
,
(s2 + 1)(s2 + 4)
s +1
s +4
so that (As + B)(s2 + 4) + (Cs + D)(s2 + 1) = 1, which is equivalent to
As3 + 4As + Bs2 + 4B + Cs3 + Cs + Ds2 + D = 1,
(A + C)s3 + (B + D)s2 + (4A + C)s + 4B + D = 1,
1
Equalities A + C = 0 and 4A + C = 0 imply that A = C = 0. From B + D = 0
we get D = −B; and them from 4B + D = 1 we get 4B − B = 1, i.e., B = 1/3
and D = −1/3. Therefore,
1
1
1
1
=
−
,
(s2 + 1)(s2 + 4)
3 s2 + 1 s2 + 4
and the inverse Laplace transform of H(s) is
h(t) =
1
1
sin t − sin 2t.
3
6
Then
y(t) = h(t)−u2π (t)h(t−2π) =
1
1
sin t− sin 2t−u2π (t)
3
6
Since sin is 2π-periodic, we have
1
3 sin t −
y(t) =
1
6
1
1
sin(t − 2π) − sin 2(t − 2π) .
3
6
sin 2t for 0 ≤ t ≤ 2pi,
0 for t > 2pi.
6.5. Laplace transform of the equation is
s2 Y − s + 2sY − 2 + 2Y = e−πt ,
hence
Y =
We have
s2
s + 2 + e−πt
.
s2 + 2s + 2
(s + 1) + 1
s+2
=
,
+ 2s + 2
(s + 1)2 + 1
hence its inverse Laplace transform is e−t (cos t + sin t). Inverse Laplace trans1
form of s2 +2s+2
is e−t sin t. Consequently,
y(t) = e−t (cos t + sin t) + uπ (t)e−(t−π) sin(t − π).
2