1

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