Begin2.DVI

Example 12-12. Laplace transform

Find the Laplace transform of eαt

Solution

By definition L

{

eαt ; t→s

}

=

∫∞

0

eαte−st dt =

∫∞

0

e−(s−α)tdt Scale the integral

to obtain

L

{

eαt

}

=

1

s−α

∫∞

0

e−udu, u = (s−α)t

to obtain

L

{

eαt

}

=^1

s−α

[

−e−u

]∞

0 =

1

s−α

Tlim→∞

[

−e−T−(−e^0 )

]

which produces the result

L

{

eαt

}

=

1

s−α provided that s > α

Using various integration techniques one can verify the following short table of

Laplace transforms.

Short Table of Laplace Transforms

Function f(t) Laplace Transform L{ f(t)}=F(s)

f(t) = L−^1 {F(s)} F(s)

(^11) s
t s^12
t^2 s2! 3
tn−^1 (ns−n1)! n= 1, 2 , 3 ,...
eα t s−^1 α
t eα t (s−^1 α) 2
tn−^1 eα t ((sn−−α1)!)n n= 1, 2 , 3 ,...
tk−^1 eα t (sΓ(−αk))k, k > 0
sin(α t) s (^2) +αα 2
cos(α t) s (^2) +sα 2
sinh(α t) s (^2) −αα 2
cosh(α t) s (^2) −sα 2