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

Geoffrey W. Cox, Ph.D., ?

Subsection 10.2.2 Completing the Square

As you progress through the Laplace Transform Method, you will often encounter rational functions that do not immediately match any known form in the table of Laplace transforms. For example, the function
s−1s2−2s+5
does not align with standard forms. A common approach is to rewrite the denominator in the form (s−a)2+b2, allowing it to match known transforms like L7 or L8. This technique, called completing the square, is essential for converting quadratic expressions into forms that are easier to work with in inverse Laplace computations.
Completing the square is particularly useful when the quadratic expression in the denominator has complex roots. As a quick check, recall from solving quadratic equations that if the discriminant
b2−4ac
is negative, the quadratic function has complex roots. This indicates that completing the square is the correct approach, as the following examples will demonstrate.

Example 21. Find the Inverse Laplce Transform.

Solution 1. H(s)=s−1s2−2s+5. Find L−1{H(s)}
The discriminant of the denominator s2−2s+5 is
b2−4ac=(−2)2−4(1)(5)=4−20=−16,
which is negative. Therefore, completing the square is necessary:
s2−2s+5=(s2−2s)+5= (s2−2s+1⏞(s−1)2−1)+5= (s−1)2+4.
Rewriting H(s) gives:
s−1s2−2s+5=s−1(s−1)2+4,
which matches ∫L8 (a=1,b=2). Therefore,
h(t)=L−1{H(s)}=etcos⁡(2t).
Solution 2. K(s)=11s2−6s+14. Find L−1{K(s)}
The discriminant of the denominator s2−6s+14 is:
b2−4ac=(−6)2−4(1)(14)=36−56=−20,
indicating that completing the square is necessary:
s2−6s+14=(s2−6s)+14=(s2−6s+9⏞(s−3)2−9)+14=(s−3)2+5.
Rewriting K(s) as:
11s2−6s+14=11(s−3)2+5⏟Match with L7=1155(s−3)2+5⏟\(L7\)(a=3,b=5).
Therefore,
k(t)=L−1{K(s)}= 115e3tsin⁡(5t).
Solution 3. P(s)=s+3s2+2s+10. Find L−1{P(s)}
Completing the square for the denominator of P(s) gives:
s+3s2+2s+10=s+3(s+1)2+9.
However, the numerator s+3 does not match s+1. To resolve this, we rewrite 3 as 1+2 and group terms:
s+3s2+2s+10= s+1+2(s+1)2+9= s+1(s+1)2+9+2(s+1)2+9= s+1(s+1)2+9⏟∫L8 (a=−1,b=3)+233(s+1)2+9⏟∫L7 (a=−1,b=3),
Now, apply the inverse Laplace transform:
p(t)= L−1{P(s)}= L−1{s+1(s+1)2+9}+23L−1{3(s+1)2+9}= e−tcos⁡(3t)+23e−tsin⁡(3t)
Completing the square is a crucial technique when working with quadratic expressions in the denominator of Laplace transforms. It allows for the backward transformations by rewriting the s-function in a form that aligns with known inverse Laplace transforms. In the next subsection, we will explore another technique: partial fraction decomposition, which helps break down more complex rational functions into simpler parts that can be easily matched to common inverse transforms.

Reading Questions Check-Point Questions

1. To complete the square: s2−s+1=s2−s+X−X+1.

missing number =

2. After completing the square of the denominator: 11s2+18s+400=11(s+9)2+X.

missing number =

3. Completing the square is used when the discriminant of the quadratic expression in the denominator is positive.

    Completing the square is used when the discriminant of the quadratic expression in the denominator is positive

  • True.

  • False. Completing the square is used when the discriminant is negative, indicating complex roots.

  • False.

  • False. Completing the square is used when the discriminant is negative, indicating complex roots.

4. The discriminant of the denominator of −19s2−3s+1 is X.

discriminant =

5. After completing the square of the denominator, 11s2−6s+14=XXXXX.

    After completing the square of the denominator, 11s2−6s+14==XXXXX
  • 11(s+3)2+9
  • Incorrect. This is not the correct result after completing the square.
  • 11(s−1)2+14
  • Incorrect. This is not the correct result after completing the square.
  • 11(s−3)2+5
  • Correct! This is the correct result after completing the square.
  • 11(s−2)2+10
  • Incorrect. This is not the correct result after completing the square.

6. What is the next step to compute L−1{s−1s2−2s+5}?

    What is the next step to compute L−1{s−1s2−2s+5}?
  • Factor the denominator
  • Incorrect. Factoring the denominator is not applicable in this case.
  • Complete the square in the denominator
  • Correct! Completing the square is necessary to match the form in the Laplace transform table.
  • Cancel out the s in the numerator and denominator.
  • No, canceling out the s in the the function is not a valid operation.
  • Differentiate the entire function
  • Incorrect. Differentiation is not needed for this problem.

7. What is the next step needed to compute L−1{s+3(s−1)2+9}?

  • Look-up the inverse Laplace transform in the table.
  • Incorrect. This function is not directly in the table.
  • Factor the denominator.
  • Incorrect. Factoring the denominator is not necessary at this stage.
  • Rewrite the numerator, then split the fraction like so: s−1+4(s−1)2+9=s−1(s−1)2+9+4(s−1)2+9.
  • Correct! The numerator needs an s−1 to match with ∫L8.
  • Split the fraction directly, like so: s+3(s−1)2+9=s(s−1)2+9+3(s−1)2+9
  • Incorrect. The next step is to decompose the function into simpler forms.
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