# MATH 251 Lecture Notes - Polynomial Long Division, Partial Fraction Decomposition, Junkers J 1

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**preview**shows page 1. to view the full**5 pages of the document.**Partial Fraction Decomposition for Inverse Laplace Trans-

form

Usually partial fractions method starts with polynomial long division in order to

represent a fraction as a sum of a polynomial and an another fraction, where the

degree of the polynomial in the numerator of the new fraction is less than the degree

of the polynomial in its denominator:

s3+ 1

s2+ 1 =s+−s+ 1

s2+ 1 .

We, however, never have to do this polynomial long division, when Partial Fraction

Decomposition is applied to problems from Chapter 6.

Another important fact in Chapter 6 is that we use only the following three types

of fractions:

1.s−a

(s−a)2+b2,2.b

(s−a)2+b2,3.1

(s−a)n,

because we know the corresponding Inverse Laplace transforms

1.L−1s−a

(s−a)2+b2=eat cos(bt),2.L−1b

(s−a)2+b2=eat sin(bt),(1)

3.L−11

s−a=eat,L−11

(s−a)2=teat,L−11

(s−a)3=t2

2eat,(2)

L−11

(s−a)4=t3

6eat,L−11

(s−a)5=t4

24eat,...L−11

(s−a)n+1 =tn

n!eat

We will call fractions 1,2,3 as standard fractions. The Partial Fraction Decomposition

for Inverse Laplace Transform is as follows.

Step 1 Suitable decomposition. The objective of this step is to give the correct

format of the partial fraction decomposition for a given fraction.

Rules of suitable decomposition:

1. Numerator does not matter.

2. Number of standard fractions equals the degree of the denominator.

3. Number of undetermined constants equals the degree of the denominator.

4. All standard fractions involved should be diﬀerent.

Simplest Scenario. When you solve a homogeneous equation ay′′ +by′+cy = 0

you always have to solve

Y=??

as2+bs +c,

where I put ?? in the numerator (because by Rule 1 the numerator does not matter in

Step 1), and in the denominator you will have the characteristic polynomial. Since the

characteristic polynomial is quadratic you will need two diﬀerent standard fractions

(Rule 2 and 4) and two undetermined constants (Rule 3).

There are three cases here.

a) as2+bs +c= 0 has two distinct real roots. Example

??

s2−3s−4=A1

s−4+B1

s+ 1.

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where s2−3s−4 = (s−4)(s+ 1). As in the example above, here the rule is that the

two (diﬀerent) standard fractions should be

1

s−S1

,1

s−S2

,

where S1and S2are the roots of as2+bs +c= 0.

Further examples

??

s2+s=A1

s+B1

s+ 1,??

s2−1=A1

s+ 1 +B1

s−1.

b) as2+bs +c= 0 has two repeated real roots. Example

??

s2−2s+ 1 =A1

s−1+B1

(s−1)2.

where s2−2s+ 1 = (s−1)2. As in the example above, here the rule is that the two

(diﬀerent) standard fractions should be

1

s−S,1

(s−S)2,

where Sis the repeated root of as2+bs +c= 0.

Further examples

??

s2=A1

s+B1

s2,??

s2+ 6s+ 9 =A1

s+ 3 +B1

(s+ 3)2.

Note that for the cases a) and b) you will need to use the ﬁrst formula in (2) for the

Inverse Laplace Transform.

b) as2+bs +c= 0 has two complex roots. Example

??

s2−2s+ 5 =As−1

(s−1)2+ 4 +B2

(s−1)2+ 4.

where s2−2s+ 5 = (s−1)2+ 4. As in the example above, here the rule is that the

two (diﬀerent) standard fractions should be

s−k

(s−k)2+m2,b

(s−k)2+m2,

where the denominator (s−k)2+m2is obtained my completing the squares:

as2+bs +c=a(s−k)2+m2.

Further examples

??

4s2+ 4s+ 5 =As+ 1/2

(s+ 1/2)2+ 1 +B1

(s+ 1/2)2+ 1,??

s2+ 5 =As

s2+ 5 +B√5

s2+ 5.

More Complicated Scenario. When you solve a nonhomogeneous equation ay′′ +

by′+cy =g(t) you will have to deal with a fraction, which denominator is of degree

3 or higher. The same four rules of suitable decomposition still apply here:

1. Numerator does not matter.

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