Proof. Integration is linear: \(\int _0^T e^{-st}(af + bg)\,dt = a\int _0^Te^{-st}f\,dt + b\int _0^Te^{-st}g\,dt\), and letting \(T\to \infty \) gives the result. □

Solution.

1.

Since \(\cosh at = \frac 12(e^{at} + e^{-at})\), linearity gives, for \(s>|a|\), \[\mathcal {L}\{\cosh at\} = \frac {1}{2}\cdot \frac {1}{s - a} + \frac {1}{2}\cdot \frac {1}{s + a} = \frac {1}{2}\cdot \frac {(s + a) + (s - a)}{s^2 - a^2} = \frac {s}{s^2 - a^2} .\] In the same way, since \(\sinh at = \frac 12(e^{at} - e^{-at})\), \[\mathcal {L}\{\sinh at\} = \frac {1}{2}\left (\frac {1}{s - a} - \frac {1}{s + a}\right ) = \frac {a}{s^2 - a^2} .\]

2.

By partial fractions, \(\frac {1}{(s -a)(s-b)} = \frac {A}{s -a} + \frac {B}{s -b}\) with \(A = \frac {1}{a - b}\) and \(B = -\frac {1}{a - b}\), so \[f(t) = \frac {1}{a-b}\left (\mathcal {L}^{-1}\left \{\frac {1}{s - a}\right \} - \mathcal {L}^{-1}\left \{\frac {1}{s-b}\right \}\right ) = \frac {e^{at} - e^{bt}}{a - b} .\]

Solution. By linearity and the transforms found so far, \(\mathcal {L}\{3 - 2e^{4t} + 5\sin 2t\} = \frac 3s - \frac {2}{s - 4} + \frac {10}{s^2 + 4}\). For the inverse, split the fraction: \(\frac {2s + 6}{s^2 + 9} = 2\cdot \frac {s}{s^2 + 9} + 2\cdot \frac {3}{s^2 + 9}\), so the inverse transform is \(2\cos 3t + 2\sin 3t\).