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Interactive Differential Equations: A Step-by-Step Approach to Methods & Modeling

Geoffrey W. Cox, Ph.D., ?

Subsection 10.2.3 Partial Fraction Decomposition

In the Laplace Transform Method, we often encounter rational functions with polynomial denominators that can be factored into simpler linear or quadratic terms. In such cases, partial fraction decomposition becomes a powerful technique. By breaking the original rational function into simpler fractions, each corresponding to a known form in the Laplace Transform Table, we can efficiently compute the inverse Laplace transform.
The key idea is to express a complex rational function as a sum of simpler fractions, each with a simpler denominator. For example, if the denominator can be factored as (s−a)(s−b), we can rewrite the function as:
f(s)(s−a)(s−b)=As−a+Bs−b.
Once expressed in this way, we can directly apply the inverse Laplace transform to each term. Let’s explore a series of examples to demonstrate this approach.

Example 22. Find the Inverse Laplce Transform.

Solution 1. Y(s)=2s+5(s+1)(s+4). Find L−1{Y(s)}
First, decompose Y(s) into partial fractions:
2s+5(s+1)(s+4)=As+1+Bs+4.
Multiply both sides by (s+1)(s+4),
2s+5=A(s+4)+B(s+1).
and solve for A and B by selecting convenient values for s:
s=−1:s=−4:
2(−1)+5= A(−1+4)+B(−1+1)3= A(3)+B(0)⟹A=12(−4)+5= A(−4+4)+B(−4+1)−3= A(0)+B(−3)⟹B=1
Therefore, the partial fraction decomposition is:
2s+5(s+1)(s+4)=1s+1+1s+4.
Applying the inverse Laplace transform to each term:
L−1{Y(s)}=L−1{1s+1}+L−1{1s+4}=e−t+e−4t.
Solution 2. P(s)=ss2−s+6. Find L−1{P(s)}
First, factor the quadratic denominator:
s2−s−6=(s+2)(s−3).
Now decompose P(s) into partial fractions:
s(s+2)(s−3)=As+2+Bs−3.
Multiply both sides by (s+2)(s−3),
s=A(s−3)+B(s+2).
and solve for A and B by selecting convenient values for s:
s=−2:s=3:
−2= A(−2−3)+B(−2+2)−2= A(−5)+B(0)⟹A=2/53= A(3−3)+B(3+2)3= A(0)+B(5)⟹B=3/5
The partial fraction decomposition is:
s(s+2)(s−3)=2/5s+2+3/5s−3.
Therefore:
L−1{P(s)}=25e−2t+35e3t.
Solution 3. Q(s)=3s+4s2−2s+1. Find L−1{Q(s)}
Factor the quadratic denominator:
s2−2s+1=(s−1)2.
Decompose Q(s) into partial fractions:
3s+4(s−1)2=As−1+B(s−1)2.
Multiply both sides by (s−1)2,
s=A(s−1)+B.
and solve for A and B by selecting convenient values for s:
s=1:s=0:
3(1)+4= A(1−1)+B7= A(0)+B⟹B=73(0)+4= A(0−1)+B4= −A+7⟹A=3
The partial fraction decomposition is:
3s−1+7(s−1)2.
Therefore:
L−1{Q(s)}=3et+7tet.
Partial fraction decomposition also works for more complex rational functions with higher-degree polynomials in the denominator. The key is to factor the denominator as much as possible, then apply the decomposition to each term. Let’s examine a more complex case involving cubic and quartic denominators.

Example 23. Find L−1{R(s)} and L−1{S(s)}.

R(s)=2s2+5s+1s3−s,S(s)=4s3+3s2−2s+1s4−1.
Solution 1. L−1{R(s)}
First, factor the cubic denominator:
s3−s=s(s−1)(s+1).
Now decompose R(s) into partial fractions:
R(s)=2s2+5s+1s(s−1)(s+1)=As+Bs−1+Cs+1
and multiply both sides by s(s−1)(s+1) to get
2s2+5s+1=A(s−1)(s+1)+B(s)(s+1)+C(s)(s−1).
and solve for A,B, and C by selecting convenient values for s:
s=0:s=1:s=−1:
1= A(−1)(1)+B(0)(1)+C(0)(−1)1= −A8= A(0)(2)+B(1)(2)+C(1)(0)8= 2B−2= A(1)(0)+B(−1)(0)+C(−1)(−2)−2= 2C
A= −1B= 4C= −1
Therefore,
L−1{R(s)}= L−1{−1s+4s−1+−1s+1}.r(t)= −1+4et−e−t.
Solution 2. L−1{S(s)}
Factor the quartic denominator:
s4−1=(s2−1)(s2+1)=(s−1)(s+1)(s2+1).
Decompose S(s) into partial fractions:
4s3+3s2−2s+1(s−1)(s+1)(s2+1)=As−1+Bs+1+Cs+Ds2+1.
and multiply both sides by (s−1)(s+1)(s2+1) to get
4s3+3s2−2s+1=A(s+1)(s2+ 1)+B(s−1)(s2+1)+ (Cs+D)(s−1)(s+1).
and solve for A,B,C, and D by selecting convenient values for s:
s=1:32s=−1:32s=0:32s=−2:32
6= A(2)(2)+B(0)(2)+(C(1)+D)(0)(2)6= 4A⟹A=322= A(0)(2)+B(−2)(2)+(C(−1)+D)(−2)(0)2= −4B⟹B=−121= A(1)(1)+B(−1)(1)+(C(0)+D)(−1)(1)1= 32+12−D⟹D=1−15= A(−1)(3)+B(−3)(3)+(C(−2)+D)(−3)(−1)−15= −3A−9B+−6C+3D−15= −92+92+−6C+3(1)⟹C=3
Therefore,
L−1{S(s)}= L−1{32s−1−12s+1+3s+1s2+1}.s(t)= 32et−12e−t+3cos⁡(t)+sin⁡(t).
Partial fraction decomposition simplifies complex rational functions, breaking them into manageable terms that correspond to known inverse Laplace transforms. In the next section, we will explore advanced techniques, including handling repeated roots and higher-order polynomials.

Reading Questions Check-Point Questions

1. Partial fraction decomposition can only be used for quadratic denominators.

    Partial fraction decomposition can only be used for quadratic denominators

  • True.

  • False. Partial fraction decomposition can be used for any polynomial denominator, including linear, quadratic, cubic, and higher degrees.

  • False.

  • False. Partial fraction decomposition can be used for any polynomial denominator, including linear, quadratic, cubic, and higher degrees.

2. The purpose of partial fraction decomposition is to break down complex rational functions into __________ that can be easily matched to known inverse Laplace transforms.

    The purpose of partial fraction decomposition is to break down complex rational functions into __________ that can be easily matched to known inverse Laplace transforms
  • simpler fractions
  • Correct! Partial fraction decomposition simplifies the rational function into fractions that match known Laplace forms.
  • more complex fractions
  • No, the goal is to simplify the function, not to make it more complex.
  • linear equations
  • No, the decomposition simplifies fractions, not into linear equations.

3. What is the first step in computing L−1{s+6s2+7s+6}?

    What is the first step in computing L−1{s+6s2+7s+6}?
  • Find a common denominator.
  • No, you need to factor the denominator first before any further decomposition.
  • Factor the denominator.
  • Correct! The first step is to factor the quadratic denominator s2+5s+6.
  • Cancel out the 6 in the numerator and denominator.
  • No, canceling out the 6 in the the function is not a valid operation.
  • Cancel out the s in the numerator and denominator.
  • No, canceling out the s in the the function is not a valid operation.

4. The partial fraction decomposition of 2s+3(s+2)(s+3) is XXs+2 + XXs+3.

    The partial fraction decomposition of 2s+3(s+2)(s+3) is XXs+2 + XXs+3
  • 1, 1
  • Correct! The partial fraction decomposition is 1s+2+1s+3.
  • 2, 3
  • Incorrect. The correct coefficients for the partial fraction decomposition are both 1.
  • 0, 1
  • No, the coefficients should be 1 for both terms in the decomposition.

5. When using partial fraction decomposition for a cubic denominator, the function is always decomposed into three terms.

  • True.

  • False. The number of terms in the decomposition depends on the factorization of the denominator, not just its degree.

  • False.

  • False. The number of terms in the decomposition depends on the factorization of the denominator, not just its degree.

6. Which of the following is the correct partial fraction decomposition of 3s+4(s−1)2?

    Which of the following is the correct partial fraction decomposition of 3s+4(s−1)2?

  • True.

  • False.

7. In partial fraction decomposition, the numerator of each term must have a degree less than the denominator.

  • True.

  • True. In partial fraction decomposition, each term’s numerator must have a lower degree than the corresponding denominator.

  • False.

  • True. In partial fraction decomposition, each term’s numerator must have a lower degree than the corresponding denominator.
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