MAT 3237
Differential Equations
Section 4.2
The Inverse Transforms
and the Transform of
Derivatives
http://myhome.spu.edu/lauw
Preview
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Inverse Transform
Transform of Derivatives
Solve IVP
A Tale of Two Worlds
L
f (t )
F (s )
D.E.( I .V .P.)
Algebraic Equation
Solution
t  world
L
1
Solution
s  world
Theorem 4.1
f (t )
1
t
n
e
at
f (t )
F (s)
1
sin kt
s
n!
n 1
cos kt
s
n  1,2,3,...
1
sa
F (s)
k
s2  k 2
s
s2  k 2
Inverse Transform
If L
 f (t )  F ( s), then f (t )  L F ( s)
1
Inverse Transform
If L
 f (t )  F ( s), then f (t )  L F ( s)
1
L 1 is also linear
L
1
aF ( s)  bG ( s)  aL F ( s)  bL G (s)
1
1
Example 0
L
1
2
 3
s 
L1 aF (s)  bG(s)  a L1 F (s)  b L1 G(s)
f (t )
1
t
n
e at
F (s)
f (t )
1
sin kt
s
n!
cos kt
s n 1
n  1,2,3,...
1
sa
F (s)
k
s2  k 2
s
s2  k 2
Example 1
L
1
1
 3
s 
L1 aF (s)  bG(s)  a L1 F (s)  b L1 G(s)
f (t )
1
t
n
e at
F (s)
f (t )
1
sin kt
s
n!
cos kt
s n 1
n  1,2,3,...
1
sa
F (s)
k
s2  k 2
s
s2  k 2
Example 2
L
1
 2 


s  2
f (t )
1
t
n
e at
F (s)
f (t )
1
sin kt
s
n!
cos kt
s n 1
n  1,2,3,...
1
sa
F (s)
k
s2  k 2
s
s2  k 2
Example 3
L
1
 2 
 2

s  9
f (t )
1
t
n
e at
F (s)
f (t )
1
sin kt
s
n!
cos kt
s n 1
n  1,2,3,...
1
sa
F (s)
k
s2  k 2
s
s2  k 2
Transform of Derivatives
Use integration by parts, we can get the
following formula:
 f (t )  sL  f (t )  f (0)
L  f (t )  s 2 L  f (t )  sf (0)  f (0)
L
OR
L  y(t )  sY ( s )  y (0)
L  y(t )  s 2Y ( s )  sy (0)  y (0)
Theoretical Background
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Laplace transforms do not exist for all
functions
If 𝑓(𝑡) satisfy certain “nice conditions”, its
Laplace transform exists (Theorem 4.2)
Under the same “nice conditions”,
lim
b 
f b
e
sb
0
L
 f (t )  sL  f (t )  f (0)
 f (t )
L

  e  st f (t )dt
0
b
 lim  e  st f (t )dt
b 
dv 
0


 lim 
b 



u

 du 



v
Example 5
4t



Solve y  3 y  2 y  e ; y(0)  1, y(0)  5
L  y(t )  sY ( s )  y (0)
L  y(t )  s 2Y ( s )  sy (0)  y (0)
Example 5
4t



Solve y  3 y  2 y  e ; y(0)  1, y(0)  5
To Save time,....
s 2  6s  9
16 1
25 1
1 1



( s  1)( s  2)( s  4)
5 s  1 6 s  2 30 s  4
Special Rules of Partial
Fractions
Partial Fractions of the following form are
very common
P( s )
, a, b  0, a  b, deg( P)  4
2
2
( s  a)( s  b)
Normally, we use
P( s )
As  B Cs  D
 2
 2
2
2
( s  a)( s  b) s  a
s b
Special Rules of Partial
Fractions
In order to reduce the complexity, we can
use the following rules:
h  ks
B
D
 2
 2
2
2
( s  a )( s  b) s  a s  b
2
hs  ks
As
Cs
 2
 2
2
2
( s  a )( s  b) s  a s  b
3
Why?
Special Rules of Partial
Fractions
h  ks 2
As  B Cs  D

 2
2
2
2
( s  a)( s  b) s  a
s b

 As  B   s 2  b    Cs  D   s 2  a 
( s 2  a)( s 2  b)



h  ks 2   As  B  s 2  b   Cs  D  s 2  a

  A  C  s 3  ( B  D) s 2   Ab  Ca  s   Bb  Da 
 AC  0
Compare coefficients, we get 
 Ab  Ca  0
Sub. C   A into Ab  Ca  0, we get A  b  a   0
Example 6
s2 1
( s 2  1)( s 2  4)
Next
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4.3.1