The answer to a transformed equation is a fraction in s that is not in the table, and partial fractions is how it gets there. Each distinct linear factor gets a term, a factor repeated m times gets m terms, and an irreducible quadratic gets a numerator B·s + C. The coefficients are exact fractions found by substituting roots and comparing powers of s, and each term then inverts through the table.
Core ideas
Three ideas: the shape of the answer is fixed before any arithmetic, repeated factors need every power, and a quadratic needs a two-part numerator.
Factor the denominator first. Every distinct linear factor s − a contributes A/(s − a). Every irreducible quadratic contributes (B·s + C) over that quadratic. Only then do the coefficients get computed, and the work is finding numbers for a form already written down.
If (s + 1)² divides the denominator, the form has both B/(s + 1) and C/(s + 1)². Leaving out the first power removes a coefficient the numerator needs, and the equations for the others then have no solution.
1/(s − a) is the transform of e^(at), 1/(s − a)² of t·e^(at), and a quadratic written as (s + 1)² + 4 has the cosine and sine entries shifted by e^(−t). Finding the coefficients is the arithmetic; inverting is reading the table from right to left.
The material
The form for each kind of factor, the two ways of finding coefficients, and the inverse of each term.
The previous lesson ended with fractions such as (s + 3)/((s + 1)(s + 2)) and handled them with one trick: multiply by a factor and set s to its root. That trick covers one case. This lesson takes the cases in order of difficulty, starting from the rule that decides which terms to write.
Let F be a fraction in s whose numerator has lower degree than its denominator. Each factor (s − a)^m of the denominator contributes the m terms A₁/(s − a) + A₂/(s − a)² + ⋯ + Aₘ/(s − a)^m, and each irreducible quadratic factor q contributes (B·s + C)/q.
Here A₁, …, Aₘ, B and C are the numbers to find. The lab refuses a fraction whose numerator is not lower in degree, because the table has no entry for a polynomial in s.
For (s + 3)/((s + 1)(s + 2)) the form is A/(s + 1) + B/(s + 2). Multiply both sides by the whole denominator, so that s + 3 = A·(s + 2) + B·(s + 1). This is an identity in s, which holds at every value. At s = −1 the term with B vanishes and 2 = A. At s = −2 the term with A vanishes and 1 = −B. The lab prints 2/(s + 1) − 1/(s + 2), the same coefficients the previous lesson found.
For 1/(s·(s + 1)²) the form has three terms, and the worked example below finds them. The reason the middle one cannot be dropped is arithmetic. Suppose the form were only A/s + C/(s + 1)². Multiplying out gives 1 = A·(s + 1)² + C·s. The s² coefficient on the right is A, which must be 0 to match the left side; yet at s = 0 the identity gives A = 1. The two demands contradict each other, so no such A and C exist. The term B/(s + 1) supplies the missing s².
Substituting roots finds only some coefficients: s = 0 gives A and s = −1 gives C, but no value of s isolates B. The remaining coefficient comes from comparing the powers of s on both sides of the identity, here the s² terms, which gives 0 = A + B.
For 1/(s·(s² + 2s + 5)) the quadratic has no real roots, since 2² − 4·5 = −16 is negative. The form is A/s + (B·s + C)/(s² + 2s + 5). At s = 0 the identity 1 = A·(s² + 2s + 5) + (B·s + C)·s gives 1 = 5A, so A = 1/5. The s² terms give 0 = A + B, so B = −1/5, and the s terms give 0 = 2A + C, so C = −2/5.
Completing the square, s² + 2s + 5 = (s + 1)² + 4, shows how this inverts. Write the numerator s + 2 as (s + 1) + 1. The part with s + 1 over (s + 1)² + 4 is the cosine entry shifted by e^(−t). The leftover 1 over (s + 1)² + 4 is half of the sine entry, since that entry has 2 on top. With the factor −1/5 in front, the inverse is 1/5 − (1/5)·e^(−t)·cos(2t) − (1/10)·e^(−t)·sin(2t), which is what the lab prints, with the fraction written as (1/5)/s − (s/5 + 2/5)/((s + 1)² + 4).
The lab does not trust any of this arithmetic. After it finds the coefficients and writes the inverse, it transforms the inverse back and compares the result with the original fraction exactly. If they differ it reports an error and prints no answer; for the three presets they agree. A decomposition is unique, so a set of coefficients that reproduces the fraction is the set.
Two limits of the lab, named here because they decide what you can type. A quadratic factor whose frequency is irrational, such as s² + 2, is refused with the factor named, since its inverse would hold cos(√2·t). And an irreducible factor of degree three or more is refused. Neither makes the method wrong; they bound what the lab prints.
Interactive
Transforms, partial fractions and solutions, exactly
Method
Factor, write the form, find the numbers, invert each term, transform back. The form comes before any arithmetic.
The numerator must have lower degree than the denominator. If it does not, divide first; the polynomial part is not a table entry.
One term for each power of each linear factor, from the first up to its multiplicity, and a numerator B·s + C over each irreducible quadratic.
Multiply through by the whole denominator. The result is an identity between polynomials in s.
Substitute each root to isolate one coefficient. Compare powers of s for whatever is left.
Replace each term by its table entry, completing the square for a quadratic. Substitute one value of s into both sides as a spot check.
The second preset. The form has three terms because s appears once and s + 1 twice.
Spot-check at s = 1: the left side is 1/(1·4) = 1/4, and the right side is 1 − 1/2 − 1/4 = 1/4. The lab prints the same three terms and the same inverse.
The t·e^(−t) is what the repeated factor buys. A single factor s + 1 would give only e^(−t); the square gives that and the same exponential multiplied by t.
You should be able to write the form from the factored denominator, find each coefficient as an exact fraction by substituting roots and comparing powers, complete the square for a quadratic, and invert every term.
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