If L f(t)=1/s(s^2+1), then find L[e^-tf(2t)]
Answers
Answer:
f L{f(t)} = F(s), then the inverse Laplace transform of F(s) is
L
−1
{F(s)} = f(t). (1)
The inverse transform L
−1
is a linear operator:
L
−1
{F(s) + G(s)} = L
−1
{F(s)} + L
−1
{G(s)}, (2)
and
L
−1
{cF(s)} = cL
−1
{F(s)}, (3)
for any constant c.
2. Example: The inverse Laplace transform of
U(s) = 1
s
3
+
6
s
2 + 4
,
is
u(t) = L
−1
{U(s)}
=
1
2
L
−1
2
s
3
+ 3L
−1
2
s
2 + 4
=
s
2
2
+ 3 sin 2t. (4)
3. Example: Suppose you want to find the inverse Laplace transform x(t) of
X(s) = 1
(s + 1)4
+
s − 3
(s − 3)2 + 6
.
Just use the shift property (paragraph 11 from the previous set of notes):
x(t) = L
−1
1
(s + 1)4
+ L
−1
s − 3
(s − 3)2 + 6
=
e
−t
t
3
6
+ e
3t
cos √
6t.
4. Example: Let y(t) be the inverse Laplace transform of
Y (s) = e
−3s
s
s
2 + 4
Step-by-step explanation:
If L f(t)=1/s(s^2+1), then find L[e^-tf(2t)]