For the types of functions we encounter in this course, the Laplace transform is \(1-1\).

Suppose \(F(s)=\mathcal {L}(f(t))\). For example, \(\frac {6}{s^4}=\mathcal {L}(t^3)\). Then we call \(f(t)\) the inverse Laplace transform of \(F(s)\). In our example, \(t^3\) is the inverse Laplace transform of \(\frac {6}{s^4}\). Notation: \(f=\mathcal {L}^{-1}(F)\).

While there is a formula for computing \(\mathcal {L}^{-1}(F)\) we will not use it as it requires the theory of functions of a complex variable (complex analysis). We can however, make use the table (see appendix) to find inverse Laplace transforms.

Suppose we need \(\mathcal {L}^{-1}\left (\dfrac {s}{s^{2}+4}\right )\). By looking through the table of Laplace transforms we find \(\mathcal {L}(\cos {(kt)})=\dfrac {s}{s^2+k^2}\). So \(\mathcal {L}^{-1}\left (\dfrac {s}{s^{2}+4}\right )=\cos {(2t)}\).

Note: Linearity also holds for the inverse Laplace transform! We leave the justification as exercise left for the reader.

Let’s find \(\mathcal {L}^{-1}\left (\dfrac {2s+3}{s^{2}+4s+13}\right )\).

\[\mathcal {L}^{-1}\left (\dfrac {2s+3}{s^{2}+4s+13}\right )=\mathcal {L}^{-1}\left (\dfrac {2(s+2)-1}{(s+2)^{2}+9}\right )=2\mathcal {L}^{-1}\left (\dfrac {s+2}{(s+2)^{2}+9}\right )-\dfrac {1}{3}\mathcal {L}^{-1}\left (\dfrac {3}{(s+2)^{2}+9}\right )\]

  \(\displaystyle =2e^{-2t}\cos {(3t)}-\dfrac {1}{3}e^{-2t}\sin {(3t)}=e^{-2t}\left (2\cos {(3t)}-\dfrac {1}{3}\sin {(3t)}\right )\)

In the last example, the denominator is irreducible over the reals. Our method changes when the denominator factors nicely:

Let’s find \(\mathcal {L}^{-1}\left (\dfrac {2s-1}{s^{2}-5s+6}\right )\).

Using a PFD (partial fraction decomposition) we get
\(\displaystyle \dfrac {2s-1}{s^{2}-5s+6}= \dfrac {2s-1}{(s-2)(s-3)}=\dfrac {5}{s-3}-\dfrac {3}{s-2}\).

Then \(\mathcal {L}^{-1}\left (\dfrac {2s-1}{s^{2}-5s+6}\right )=5\mathcal {L}^{-1}\left (\dfrac {1}{s-3}\right )-3\mathcal {L}^{-1}\left (\dfrac {1}{s-2}\right )=5e^{3t}-3e^{2t}\).

Let’s find \(\mathcal {L}^{-1}\left (\dfrac {s}{s^{2}-8s+16}\right )\). We can setup a PFD as follows:

\[\dfrac {s}{s^{2}-8s+16}=\dfrac {s}{(s-4)^{2}}=\dfrac {A}{s-4}+\dfrac {4}{(s-4)^{2}}.\]

Clearing fractions and equating common-degree coefficients, we get \(A=1\).

So \(\displaystyle \mathcal {L}^{-1}\left (\dfrac {s}{s^{2}-8s+16}\right )=\mathcal {L}^{-1}\left (\dfrac {1}{s-4}\right )+4\mathcal {L}^{-1}\left (\dfrac {1}{(s-4)^{2}}\right )=e^{4t}+4te^{4t}=e^{4t}(1+4t)\).