366 Chapter 6 Laplace Transform
L(f)=F(s)=∫∞
0e−stf(t)dt
L(cf(t))=cL(f(t))
L(f(t)+g(t))=L(f(t))+L(g(t))
L(f′(t))=−f( 0 )+sF(s)
L(f′′(t))=−f′( 0 )−sf( 0 )+s^2 F(s)
L(f(n)(t))=−f(n−^1 )( 0 )−sf(n−^2 )( 0 )−···−sn−^1 f( 0 )+snF(s)
L(ebtf(t))=F(s−b)
L(∫t
0f(t′)dt′)
=^1 sF(s) L( 1
tf(t))
=∫s∞F(s′)ds′
L(tf(t))=−dFdsTable 1 Properties of the Laplace transformor
L[∫t0f(t′)dt′]
=^1 sL(
f(t))
. (5)
Differentiation and integration with respect tosmay produce transforma-
tions of previously inaccessible functions. We need the two formulas
−de−st
ds=te−st,∫∞
se−s′tds′=^1
te−stto derive the results
L(
tf(t))
=−dF(s)
ds, L(
1
tf(t))
=
∫∞
sF(s′)ds′. (6)(Note that, unlessf( 0 )=0, the transform off(t)/twill not exist.) Examples
of the use of these formulas are
L(
tsin(ωt))
=−dsd(
ω
s^2 +ω^2)
=(s (^22) +sωω (^2) ) 2 ,
L
(
sin(t)
t)
=
∫∞
sds′
s′^2 + 1 =π
2 −tan− (^1) (s)=tan− 1
(
1
s)
.
Significant properties of the Laplace transform are summarized in Table 1.
When a problem is solved by use of Laplace transforms, a prime difficulty
is computation of the corresponding function oft. Methods for computing
the “inverse transform”f(t)=L−^1 (F(s))include integration in the complex
plane, convolution, partial fractions (discussed in Section 6.2), and tables of