Fracci Ones Parcial Es

7/30/2019 Fracci Ones Parcial Es

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Fracciones Parciales


2/26

Partial Fractions General

Case

In the solution of ODEs by the Laplace Transform

method, expressions of the form often occur.

Here P(s) and Q(s) are both polynomials.

)(

)(

sP

sQ

These are easy to invert if they are written in partial

fraction form:

)()()()()(

2

2

1

1

n

n

sa

sa

sa

sPsQ

Here )())(()(21 n

ssssP


3/26

Partial Fractions Complex

Roots If1 and 2 are a complex conjugate pair, then we

can avoid complex numbers by combining these

factors into a quadratic expression. Say 1= -a+ib,

2= -a-ib, then:

Then

22

21 )())(( basss

)()()()(

)(

3

3

22

n

n

s

a

s

a

bas

BsA

sP

sQ


4/26

Partial Fractions Repeated

Roots If1 = 2 then the partial fraction form is:

Etc, etc

)()()()()(

)(

3

3

2

1

2

1

1

n

n

s

a

s

a

s

a

s

a

sP

sQ

If1 = 2 = 3 then the partial fraction form is:

)()()()()(

)(3

1

3

2

1

2

1

1

n

n

s

a

s

a

s

a

s

a

sP

sQ


5/26

Method 1 - Cover-up

Rule

Basic idea:

Multiply by a factor (s - i)

Put s = i

Evaluate ai

)()()()(

)(

2

2

1

1

n

n

s

a

s

a

s

a

sP

sQ


6/26

Cover-up Rule

Find the partial fraction form for:

The partial fraction form is:

)1(

1

ss

)1()1(

1

s

B

s

A

ss


7/26

Cover-up Rule

To calculate A, multipy by s and put s=0:

)1()1(

1:

s

BsA

s

s

10

1

1:0Put AAs


8/26

Cover-up Rule

To calculate B, multipy by s+1 and put s=-1:

BAs

s

s )1(1

:)1(

101

1:1Put

BBs

Thus the partial fraction form is:

)1(

11

)1(

1

ssss


9/26

Cover-up Rule

The above procedure can be carried out by

covering up. ForA use:

Now put s = 0, ignoring the covered up items:

)1()1(1

sB

sA

ss

110

1

AA


10/26

Cover-up Rule

Similarly forB, covering up gives:

Now put s = -1, ignoring the covered up items:

11

1

BB

)1()1(

1

s

B

s

A

ss


11/26

Complex Cover-up

Rule

Find the coefficientsA, B, Cin:

Find A using the standard cover-up idea:)1()1()1)(1(

122

s

CBs

s

A

ss

)1()1()1)(1(

122

s

CBs

s

A

ss


12/26

Complex Cover-up

Rule

Put s+1 = 0, that is s = -1:

2

1

)1)1((

12

AA

The coefficients B and Care more difficult to

calculate. A modified version of the cover-up rule

involves using complex factors


13/26

Complex Cover-up

Rule

Multiply the whole equation by (s+i):

i)(

)1(

i)(

)1(

i)(

)1)(1(

122

s

s

CBss

s

As

ss

i)(i)i)((

i)()1(

i)(i)i)()(1(

1

sss

CBss

s

A

ssss


14/26

Complex Cover-up

Rule

Now put (s+i)=0, that is s = -i:

i)(i)(

)1(i))(1(

1

s

CBss

s

A

ss

i)2(

i)()0(

)1(i)2)(1i(

1

CB

i

A

After tidying up:

CB

i2

)1i(


15/26

Complex Cover-up

Rule Equate real and imaginary parts:

Final partial fraction:

2

1,

2

1 CB

)1()1()1)(1(

12

21

21

21

2

s

s

sss


16/26

Method 2 - Substitution of

values

Basic idea

Put s = bi fori=1,2,,n. Here biare

convenient, easy to work with, numbers

Obtain n equations in the n unknown ai

Solve forai

)()()()(

)(

2

2

1

1

ni

n

iii

i

b

a

b

a

b

a

bP

bQ


17/26

Substitution of

values Find the coefficientsA, B, Cin:

)1()1(

122

s

C

s

B

s

A

ss

)1()1(

122

s

C

s

B

s

A

ss

Put s=0:1

)10(

1

BB

Use cover-up forB :)(2s


18/26

Substitution of values

Use cover-up forC:

)1()1(

122

s

C

s

B

s

A

ss

Put s=-1:

1

)1(

12

CC


19/26

Substitution of

values We cannot use cover-up forA. So far we have:

)1(

11

)1(

122

sss

A

ss

One good way to calculateA is to substitute a

convenient value fors. Say s =1:

)11(

1

1

1

1)11(1

122

A1

2

11

2

1 AA


20/26

Substitution of

values Substitute back:

)1(

111

)1(

122

sssss


21/26

Example - substitution of

values Substitute values to work outA and B in:

)1()1(

1

s

B

s

A

ss

Put s=1: BABA

2

1

2

1

212

1

Put s=2: BABA

3

2

3

1

326

1


22/26

Example - substitution of

values Subtract two equations

11

66

1 AB

B

Substitute back

)1(

11

)1(

1

ssss


23/26

Method 3 - Equate coefficients

Basic idea:

Multiply whole equation by P(s)

Equate coefficients of each power ofs

Solve resulting equations forai

i

i

nn

s

sPsP

sPasPasPasQ

)()(where

),(...)()()( 2211


24/26

Example - Equate

coefficients CalculateA and B in:

)1()1(

1

s

B

s

A

ss

Multiply by s(s+1): BssA )1(1

Equate coefficients:

BA0:sFor

A1:1For 1,1 BA


25/26

Example - Equate

coefficients Calculate

A and B in:

Multiply by

(s+1)(s2+1):)1)(()1(1

2

sCBssA

)1()1()1)(1(

122

s

CBs

s

A

ss

BA0:sFor2

Equate coefficients:

)()()(1 2 CAsCBsBA

0:For CBs

2

1,

2

1,

2

1 CBA1:1For CA


26/26

Preferred method for

partial fractions Calculate as many coefficients as possible using

the simple cover-up rule

Calculate the remaining coefficients by:

Substituting values for s

Using the complex cover-up method

Equating coefficients

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