The Inverse Laplace Transform and Transforms of Derivatives - Ÿ4.2 1. Inverse Laplace Transform:
Let FŸs c£fŸt ¤. We say fŸt is the inverse Laplace transform of FŸs and write fŸt c "1 £FŸs ¤.
c "1 £F¤ is also a linear operator, i.e.,
a. c "1 £FŸs GŸs ¤ c "1 £FŸs ¤ c "1 £GŸs ¤
b. c "1 £cFŸs ¤ cc "1 £FŸs ¤
Example Find the inverse Laplace transform fŸt of FŸs where
Ÿa FŸs 13 , s 0
s
Ÿb FŸs 1 , s "2
s2
Ÿc FŸs 1 , s0
9s 2 2
2
Ÿs 2 " 1 , s0
Ÿd FŸs s2 " 2 , s 0
Ÿe FŸs s5
s 3
s 2 6s 9
s
Ÿf FŸs Ÿg FŸs Ÿs " 1 Ÿs " 2 Ÿs 4 Ÿs " 1 Ÿs 2 4 Ÿh FŸs 2s 1
Ÿs 2 1 Ÿs 2 2 a.
FŸs 13 1 2!3 1 c£t 2 ¤ c
2! s
2
s
1 t 2 , fŸt 1 t 2
2
2
b.
FŸs 1 c£e "2t ¤, fŸt e "2t
s2
c.
1
1
FŸs 9s 2 2
9 s2 1 c sin
3 2
1
2
9
2
t
3
9 s2 c
2
2
3
9
2
t
3
1 sin
3 2
2
3
1
2
3
s2 2
3
2
, fŸt 1 sin
3 2
2
t
3
3t "
1 cos
2 3
3t
d.
s
FŸs s2 " 2 2
s 3
3
s c sin
fŸt sin
3t
3t "
"
2
"
3
1
2 3 s2 3
1 c cos
2 3
1 cos
2 3
3t
2
c sin
3t
e.
Ÿs 2 " 1 FŸs s5
2
4
2
s " 2s5 1 1s " 23 1 4!5
4! s
s
s
c£1¤ " c£t 2 ¤ 1 c£t 4 ¤ c 1 " t 2 1 t 4 , fŸt 1 " t 2 1 t 4
24
24
4!
f. First write FŸs as a sum of partial fractions. Use TI-89/F2/expand(expression).
1
16
25
1
s 2 6s 9
"
5Ÿs " 1 6Ÿs " 2 30Ÿs 4 Ÿs " 1 Ÿs " 2 Ÿs 4 " 16 c£e t ¤ 25 c£e 2t ¤ 1 c£e "4t ¤ c " 16 e t 25 e 2t 1 e "4t
5
5
6
30
6
30
25
1
16
t
2t
"4t
fŸt "
e e e
5
6
30
g. First write FŸs as a sum of partial fractions.
1
1
s
" 1 s2 " 4 " 1 2s
1 24
FŸs 5 s 4
5 s 4
5 s 4
5Ÿs " 1 5Ÿs " 1 Ÿs " 1 Ÿs 2 4 1 c£e t ¤ " 1 c£cosŸ2t ¤ 2 c£sinŸ2t ¤ c 1 e t " 1 cosŸ2t 2 sinŸ2t 5
5
5
5
4
4
1
1
2
t
fŸt e " cosŸ2t sinŸ2t 5
5
4
h. First write FŸs as a sum of partial fractions.
FŸs 2 2s 12
2s2 1 " 2s2 1 22s 2 1 " 22s " 2 1
s 2
s 2
s 2
s 1
s 1
s 1
Ÿs 1 Ÿs 2 " 1 c sin 2 t
2c£cos t¤ c£sin t¤ " 2c cos 2 t
2
FŸs 2 t " 1 sin 2 t
2
2 t " 1 sin 2 t
2
c 2 cos t sin t " 2 cos
fŸt 2 cos t sin t " 2 cos
2. Transforms of Derivatives:
U
What is c f ? Observe that
c f
U
.
;0 f
u e "st , du "se "st
U
Ÿt e "st dt
.
lim fŸt e "st | b0 " ; Ÿ"s fŸt e "st dt
bv.
U
0
dv f Ÿt dt, v fŸt lim fŸb e "sb " fŸ0 sc£f¤ sc£f¤ " fŸ0 bv.
What is c f
UU
?
c f
UU
Ÿn snc f
sc f
U
U
U
" f Ÿ0 s 2 c£f¤ " sfŸ0 " f Ÿ0 What is c£f Ÿn ¤?
c f
U
U
" s n"1 f Ÿ0 " s n"2 f Ÿ0 ". . . " sfŸ0 " f Ÿ0 3. Solve Initial Value Problems Using Laplace Transform:
For a given initial value problem for an nth-order linear differential equation:
LŸy fŸx ,
yŸ0 ) 0 , y U Ÿ0 ) 1 , . . . , y Ÿn"1 Ÿ0 ) n"1 ,
two steps to find the solution:
a. Find c£y¤, let FŸs c£y¤.
b. y c "1 £FŸs ¤.
Example Solve y UU 3y U " 4y t e "2t , yŸ0 0, y U Ÿ0 "1 .
2
a. Take the Laplace transform of both sides of the equation:
c£y UU 3y U " 4y¤ c£t e "2t ¤
c£y UU ¤ 3c£y U ¤ " 4c£y¤ c£t¤ c£e "2t ¤
s 2 c£y¤ " syŸ0 " y U Ÿ0 3Ÿsc£y¤ " yŸ0 " 4c£y¤ 12 1
s2
s
Ÿs 2 3s " 4 c£y¤ 1 12 1
s2
s
c£y¤ 1
Ÿs 2 3s " 4 1 1 "1
s2
s2
b. Solve for y :
1
Ÿs 2 3s " 4 3
1
23
1
"
" 3 " 12
16s
6Ÿs 2 80Ÿs 4 15Ÿs " 1 4s
c 23 e "4t " 1 e "2t 1 e t " 1 t " 3
16
6
4
15
80
23
3
1
1
1
t
"4t
"2t
y
e " e e " t"
80
16
6
4
15
1 1 "1
s2
s2