laplace criterion was formulated by simon de laplace and
Answers
Step-by-step explanation:
Laplace Integral. The integral R ∞
0
g(t)e
−stdt is called the Laplace
integral of the function g(t). It is defined by limN→∞ R N
0
g(t)e
−stdt and
depends on variable s. The ideas will be illustrated for g(t) = 1, g(t) = t
and g(t) = t
2
, producing the integral formulas in Table 1.
R ∞
0
(1)e
−stdt = −(1/s)e
−st
t=∞
t=0 Laplace integral of g(t) = 1.
= 1/s Assumed s > 0.
R ∞
0
(t)e
−stdt =
R ∞
0 − d
ds (e
−st)dt Laplace integral of g(t) = t.
= −
d
ds
R ∞
0
(1)e
−stdt Use R
d
dsF(t, s)dt =
d
ds
R
F(t, s)dt.
= − d
ds (1/s) Use L(1) = 1/s.
= 1/s2 Differentiate.
R ∞
0
(t
2
)e
−stdt =
R ∞
0 −
d
ds (te−st)dt Laplace integral of g(t) = t
2
.
= −
d
ds
R ∞
0
(t)e
−stdt
= −
d
ds (1/s2
) Use L(t) = 1/s2
.
= 2/s3
Table 1. The Laplace integral R ∞
0
g(t)e
−stdt for g(t) = 1, t and t
2
.
R ∞
0
(1)e
−st dt =
1
s
R ∞
0
(t)e
−st dt =
1
s
2
R ∞
0
(t
2
)e
−st dt =
2
s
3
In summary, L(t
n
) = n!
s
1+n
An Illustration. The ideas of the Laplace method will be illus-
trated for the solution y(t) = −t of the problem y
0 = −1, y(0) = 0. The
method, entirely different from variation of parameters or undetermined
coefficients, uses basic calculus and college algebra; see Table 2.
Table 2. Laplace method details for the illustration y
0 = −1, y(0) = 0.
y
0
(t)e
−st = −e
−st Multiply y
0 = −1 by e
−st
.
R ∞
0
y
0
(t)e
−stdt =
R ∞
0 −e
−stdt Integrate t = 0 to t = ∞.
R ∞
0
y
0
(t)e
−stdt = −1/s Use Table 1.
s
R ∞
0
y(t)e
−stdt − y(0) = −1/s Integrate by parts on the left.
R ∞
0
y(t)e
−stdt = −1/s2 Use y(0) = 0 and divide.
R ∞
0
y(t)e
−stdt =
R ∞
0
(−t)e
−stdt Use Table 1.
y(t) = −t Apply Lerch’s cancellation law.
carry on