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Solving Differential Equations with Laplace Transforms: 1. Take the Laplace transform of both sides of the equation. 2. Using the initial conditions, solve the equation for Y(s). 3. Take the inverse Laplace of both sides of the equation to find y(t).
Inverse Laplace Transforms of Rational Functions Determine which Laplace Transform you will use by examining the denominator. Then using algebraic procedures make the numerator fit the form. You may need to do one or more of the following, before the denominator will match one of the forms from the table.
1. Divide, if the degree of a polynomial in the numerator is greater than or equal to the degree of a polynomial in the denominator.
2. If the denominator can be factored -rewrite as 2 or more rational expressions using method of partial fractions.
3. If the denominator cannot be factored – complete the square. Example #1. Use the Laplace transform to solve the given initial-value problem:
tyy 4 , 0)0(,0)0( yy
Solution:
L yy 4 L t
2
2 1)(4)(')0()(
ssYsysysYs
2
2 1)(4)(
ssYsYs
4
11)(
22 ss
sY
using partial fractions we get 422
s
B
s
AsY
Solving for A and B yields to 4
1A and
4
1B
By partial fractions:
4
1
4
11
4
1)(
22 ss
sY =222 2
2
8
11
4
1
ss
)(ty L )(1 sY L
4
1
4
11
4
122
1
ss
4
1L
8
112
1
s L
22
1
2
2
s
so )2sin(8
1
4
1)( ttty
LAPLACE TRANSFORMS AND
NON-STANDARD FUNCTIONS
Denominator is made to
match one of the forms
from the Laplace
Transforms Table (in this
case it is sin(kt))
2
Non-Standard Functions
1. Unit Step Function or Heaviside Function denoted by either u(t-a) or H(t-a) (Turns on at a and stays on):
)( atH )( atu =
or 0 if 1
or 0 if 0
atat
atat
Sample graph
2. Unit Pulse Function (Turns on at a and off at b):
if 0
if 1
- if 0
)()()(,
tb
bta
at
btuatutP ba
Sample graph
3. Dirac Delta Function (Unit Impulse Function):
at
atat
if
if 0)(
Sample graph
3
Laplace transforms of unit step functions and unit pulse functions.
1. Convert unit pulse function to unit step function before taking the Laplace transform.
2. Apply the Second Translation Theorem (STT):
Example #2. Find the Laplace transform of the following function:
tt
tt
tt
tf
4 ,5
41 ,2
10 ,
)(
2
Solution: First we will rewrite f(t) in terms of Heaviside function:
ttuttututtututf 5)4(2)4()1()1()0()( 2
And group the terms, based on the Heaviside function:
26)4(2)1()0(
25)4(2)1()0()(
22
22
ttutttuttu
tttutttuttutf
Notice that 1)0( tu for all 0t so f(t) becomes:
26)4(2)1()( 22 ttutttuttf
Now we are ready to find the Laplace transform of f(t):
L tf L 26)4(2)1( 22 ttutttut
= L 2t +L 22)1( tttu +L 26)4( ttu =
= L 2t + se L 2112 tt + se 4 L 246 t =
= L 2t + se L 23 tt + se 4 L 226 t =
=L 2t -3 se L t - se L 2t +6 se 4 L t +22 se 4 L 1 =
= s
e
s
e
s
e
s
e
s
ssss 4
2
4
323
226232
=
3
4 1132232
s
ssese ss
L asetfatu L atf , 0a
4
Inverse Laplace transform of the unit step functions. Apply the Second Translation Theorem (STT):
Example #3. Find the inverse transform of 2
21
s
esF
s
Solution:
2
2
22
2 111
se
ss
esF s
s
so
tf L )(1 sF L
2
1 1
s
se 2 L
2
1 1
s)2()2( ttut
which can also be written as:
t
tttf
2 ,2
20 ,)(
You try it: 1. Use the Laplace transform to solve the given initial-value problem:
,8152 teyyy 12)0(,2)0( yy
2. Find the Laplace transform of
1 ,6
10 ,2
tt
tttf
3. Find the inverse transform of 22
)1(2)(
2
2
ss
essF
s
Answers:
1. tt eey 2
1
2
5 5
2. s
e
s
e
s
e
ssF
sss
6222
)(233
3. )2cos()2(2)( 2 tetutf t
L ),()()(1 atfatusFe as
)0( a