# Using Partial Fractions to Find Inverse Laplace Transforms

concept

When using the Laplace Transform to solve differential equations we won't get problems as nice as \(\frac{5}{s} + \frac{2}{s-3}\). We'll get the delightful \(\frac{7s-15}{s^2-3s}\), so it's important that we can take a rational expression and reduce it to terms that exist in our Laplace tables.
Partial fraction expansion allows us to do just this. It's a process that requires quite a bit of algebra but will give us terms that can be directly inverted using our table. Which is neat.

Partial fractions is clunky to talk about in words but it's straightforward to show how the process works. So we'll use a few example to go over the method.

example

Expand \(\frac{2s + 10}{s(s + 10)}\) using partial fractions

Our first step is to factor the denominator (which I've already done, you're welcome). Then we write a fraction for each factor with that factor as the denominator: \(\frac{2s + 10}{s(s + 10)} = \frac{A}{s} + \frac{B}{s + 10}\) Where \(A\) and \(B\) are both constants. Later on I'll explain how you know what kind of numerator to use (it's pretty straightforward). Now we need to find \(A\) and \(B\) by multiplying both sides by \(s(s + 10)\) and equating equal powers of \(s\) on both sides. \(2s + 10 = As + 10A + Bs\) So we can see that \(A = 1\) (since \(10A = 10\)). \(A + B = 2 \implies B = 1\) So \(\frac{2s + 10}{s(s + 10)} = \frac{1}{s} + \frac{1}{s+10}\)example

Expand \(\frac{3s^2 - 12s + 11}{(s-3)(s-2)(s-1)}\) using partial fractions.

Again we write out a fraction for each denominator factor and stick an unknown numerator on top. \(\frac{3s^2 - 12s + 11}{(s-3)(s-2)(s-1)} = \frac{A}{s-3} + \frac{B}{s-2} + \frac{C}{s-1}\) Now we need to find the values of \(A\), \(B\) and \(C\) (which are all constants). We do that by multiplying both sides by \((s-3)(s-2)(s-1)\) and then equating terms with equal powers of \(s\). \(\begin{align} 3s^2 - 12s + 11 & = A(s-2)(s-1) + B(s-3)(s-1) + C(s-3)(s-2) \\ & = As^2 - 3As + 2A + Bs^2 - 4Bs + 3B + Cs^2 - 5Cs + 6C \\ & = s^2[A + B + C] + s[-2A -4B -5C] + [2A + 3B + 6C] \\ \end{align}\) So we get the simultaneous equations: \(A + B + C = 3\) \(-2A -4B -5C = -12\) \(2A + 3B + 6C = 11\) I'll assume you can solve that and we get \(A = B = C = 1\) So \(\frac{3s^2 - 12s + 11}{(s-3)(s-2)(s-1)} = \frac{1}{s-3} + \frac{1}{s-2} + \frac{1}{s-1}\) And we can see that the inverse Laplace transform is going to be: \(f(t) = e^{3t} + e^2t + e^t\) Now in both the above examples the denominator has factored into linear functions of \(s\), that's because I'm super nice to you and wanted to ease you into this.
The terms we use in partial fraction decomposition depend on the factors in the denominator.

fact

The following is a table of terms to be used for corresponding denominators in partial fraction expansions

Factor in denominator | Term in partial fraction decomposition |
---|---|

\(as + b\) | \(\frac{A}{as + b}\) |

\((as+b)^k\) | \(\frac{A_1}{as+b} + \frac{A_2}{(as+b)^2} + \cdots + \frac{A_k}{(as+b)^k}\) |

\(as^2 + bs + c\) | \(\frac{As + B}{as^2 + bs + c}\) |

\((as^2 + bs + c)^k\) | \(\frac{A_1s + B_1}{as^2 + bs + c} + \frac{A_2s + B_2}{(as^2 + bs + c)^2} + \cdots + \frac{A_ks + B_k}{(as^2 + bs + c)^k}\) |

So let's take our new terms our for a spin

example

Use partial fractions to expand \(\frac{2-5s}{(s-6)(s^2+11)}\)

From our table we can see that: \(\frac{2-5s}{(s-6)(s^2+11)} = \frac{A}{s-6} + \frac{Bs + C}{s^2 + 11}\) Now we do what we've always done and multiply both sides by the denominator and equate equal powers of \(s\). \(2 - 5s = As^2 + 11A + Bs^2 -6Bs + Cs -6C\) \( = s^2[A + B] + s[-6B + C] + [11A - 6C]\) From this we can see that \(A + B = 0 \implies A = -B\) \(-6B + C = -5\) \(11A - 6C = 2\) This gives: \(A = -\frac{28}{47}\), \(B = \frac{28}{47}\), \(C = -\frac{67}{47}\) So we can rewrite our problem as: \( \frac{1}{47}\left( -\frac{28}{s-6} + \frac{28s - 67}{s^2 + 11} \right) \) \( = \frac{1}{47}\left( -\frac{28}{s-6} + \frac{28s}{s^2 + 11} - \frac{67}{s^2 + 11} \right) \) And the inverse transform can be easily seen to be: \(f(t) = \frac{1}{47}\left( -28e^{6t} + 28\cos(\sqrt{11}t)-\frac{67}{\sqrt{11}}\sin(\sqrt{11}t) \right) \) Yep, it's a right mess, when you solve real differential equations with this method it almost certainly will be, with square roots and awkward fractions all over the place.
practice problems