Solution.

Part 1

\(\displaystyle {\mathcal {L}\{f(t)\} = \mathcal {L}\{\cosh at\} = \mathcal {L}\Bigg \{\frac {e^{at} + e^{-at}}{2}\Bigg \}}\)

So that \begin {align*} \mathcal {L}\{\cosh at\} & = \frac {1}{2}\mathcal {L}\{e^{at}\} + \frac {1}{2}\mathcal {L}\{e^{-at}\}\\\\ & = \frac {1}{2}\cdot \frac {1}{s - a} + \frac {1}{2}\cdot \frac {1}{s + a}\\\\ & = \frac {1}{2}\Big [\frac {s + a + s - a}{s^2 - a^2}\Big ]\\ & = \frac {s}{s^2 - a^2} \end {align*}

\[\therefore \hspace {1cm} \mathcal {L}\{\cosh at\} = \frac {s}{s^2 - a^2}\hspace {0.4cm}, \hspace {0.4cm} s>a\]

The companion result for \(\sinh \) follows the same way:

\begin {align*} \mathcal {L}\{f(t)\} & = \mathcal {L}\{\sinh at\}\\ & = \mathcal {L}\Bigg \{ \frac {e^{at} - e^{-at}}{2}\Bigg \}\\\\ & = \frac {1}{2}\int ^{\infty }_0 e^{-st}\Big [e^{-(s-a)t} - e^{-(s+a)t}\Big ]dt\\\\ & = \frac {1}{2}\Bigg [\frac {1}{(s - a)} - \frac {1}{(s+a)}\Bigg ]\\ \end {align*}

Part 2

Let \(\displaystyle {\frac {1}{(s -a)(s-b)} = \frac {A}{s -a} + \frac {B}{s -b}\implies B = \frac {-1}{a - b}\hspace {0.3cm},\hspace {0.3cm}A =\frac {1}{a - b}}\)

Hence \(\displaystyle {F(s) = \frac {1}{a-b}\Bigg ( \frac {1}{s - a} - \frac {1}{s-b}\Bigg ) }\)

\begin {align*} f(t) & = \mathcal {L}^{-1}\{F(s)\} = \frac {1}{a-b}\mathcal {L}^{-1}\Bigg \{\frac {1}{s - a}- \frac {1}{s - b}\Bigg \}\\\\ & = \frac {1}{a -b}\Big [\mathcal {L}^{-1}\bigg \{\frac {1}{s - a}\bigg \} - \mathcal {L}^{-1}\bigg \{\frac {1}{s-b}\bigg \}\Big ]\\\\ & = \frac {1}{a - b}\hspace {0.1cm} \Big [e^{at} - e^{bt}\Big ]\\\\ \end {align*}