Problem 02 | Linearity Property of Laplace Transform

$\displaystyle \mathcal{L}\left\{ f(t) \right\} = \int_0^\infty e^{st} f(t) \, dt$

$\displaystyle \mathcal{L}(\cosh at) = \int_0^\infty e^{st} \cosh at \, dt$
 

But
$\cosh at = \dfrac{e^{at} + e^{-at}}{2}$
 

Thus,
$\displaystyle \mathcal{L}(\cosh at) = \int_0^\infty e^{st} \left( \dfrac{e^{at} + e^{-at}}{2} \right) \, dt$

$\displaystyle \mathcal{L}(\cosh at) = \dfrac{1}{2}\int_0^\infty e^{st}(e^{at}) \, dt + \dfrac{1}{2}\int_0^\infty e^{st}(e^{-at}) \, dt$

$\mathcal{L}(\cosh at) = \frac{1}{2}\mathcal{L}(e^{at}) + \frac{1}{2}\mathcal{L}(e^{-at})$
 

From the table of Laplace transform,
$\mathcal{L}(e^{at}) = \dfrac{1}{s - a}$   and   $\mathcal{L}(e^{-at}) = \dfrac{1}{s + a}$
 

Hence,
$\mathcal{L}(\cosh at) = \dfrac{1}{2}\left( \dfrac{1}{s - a} \right) + \dfrac{1}{2}\left( \dfrac{1}{s + a} \right)$

$\mathcal{L}(\cosh at) = \dfrac{1}{2}\left( \dfrac{1}{s - a} + \dfrac{1}{s + a} \right)$

$\mathcal{L}(\cosh at) = \dfrac{1}{2}\left[ \dfrac{(s + a) + (s - a)}{(s - a)(s + a)} \right]$

$\mathcal{L}(\cosh at) = \dfrac{1}{2}\left( \dfrac{s + a + s - a}{s^2 - a^2} \right)$

$\mathcal{L}(\cosh at) = \dfrac{1}{2}\left( \dfrac{2s}{s^2 - a^2} \right)$

$\mathcal{L}(\cosh at) = \dfrac{s}{s^2 - a^2}$       okay