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Mathematics

MATH 251

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Spring

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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:
3
s + 1 = s + −s + 1 .
s + 1 s + 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:
s − a b 1
1.(s − a) + b 2, 2.(s − a) + b 2, 3.(s − a)n,
because we know the corresponding Inverse Laplace transforms
▯ ▯ ▯ ▯
1.L−1 s − a = e cos(bt), 2.L −1 b = e sin(bt), (1)
(s − a) + b2 (s − a) + b 2
▯ ▯ ▯ ▯ ▯ ▯ 2
−1 1 at −1 1 at −1 1 t at
3.L s − a = e , L (s − a)2 = te , L (s − a)3 = 2 e , (2)
▯ 1 ▯ t3 ▯ 1 ▯ t4 ▯ 1 ▯ tn
L −1 = e , L −1 = e ,...L −1 = eat
(s − a)4 6 (s − a)5 24 (s − a)n+1 n!
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 = 2 ,
as + 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.
2
a) as + bs + c = 0 has two distinct real roots. Example
?? 1 1
2 = A + B .
s − 3s − 4 s − 4 s + 1 2
where s −3s−4 = (s−4)(s+1). As in the example above, here the rule is that the
two (diﬀerent) standard fractions should be
1 1
, ,
s − S 1 s − S 2
where S and S are the roots of as + bs + c = 0.
1 2
Further examples
?? 1 1 ?? 1 1
2 = A + B , 2 = A + B .
s + s s s + 1 s − 1 s + 1 s − 1
b) as + bs + c = 0 has two repeated real roots. Example
?? 1 1
= A + B .
s − 2s + 1 s − 1 (s − 1)2
2 2
where s − 2s + 1 = (s − 1) . As in the example above, here the rule is that the two
(diﬀerent) standard fractions should be
1 1
, ,
s − S (s − S)2
where S is the repeated root of as + bs + c = 0.
Further examples
?? 1 1 ?? 1 1
2 = A + B 2 , 2 = A + B 2.
s s s s + 6s + 9 s + 3 (s + 3)
Note that for the cases a) and b) you will need to use the ﬁrst formula in (2) for the
Inverse Laplace Transform.
b) as + bs + c = 0 has two complex roots. Example
?? s − 1 2
= A + B .
s − 2s + 5 (s − 1) + 4 (s − 1) + 4
where s − 2s + 5 = (s − 1) + 4. As in the example above, here the rule is that the
two (diﬀerent) standard fractions should be
s − k b
, ,
(s − k) + m 2 (s − k) + m 2
where the denominator (s − k) + m is obtained my completing the squares:
2 ▯ 2 2▯
as + bs + c = a (s − k) + m .
Further examples
√
?? s + 1/2 1 ?? s 5
2 = A 2 + B 2 , 2 = A 2 + B 2 .
4s + 4s + 5 (s + 1/2) + 1 (s + 1/2) + 1 s + 5 s + 5 s + 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. 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.
The ﬁrst preliminary step we have to take here is to decompose the denominator
into a product standard polynomials. These standard polynomial are exactly of two
types
n ▯ 2 2▯
a) (s − a) b) and (s − k) + m .
Examples
(s−1)(s +4s+3) = (s−1)(s+1)(s+3), (s−1)(s +4s+5) = (s−1)((s+2) +1), 2
(s − 1)(s + 4s − 5) = (s − 1)(s − 1)(s + 5) = (s − 1) (s + 5).
Note that in the last example above we have to combine two terms (s − 1) into one
2
(s − 1) , so that all the involved standard polynomials h

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