Problem 02 | Linearity Property of Laplace Transform

Problem 02
By using the linearity property, show that

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

 

Solution 02
$f(t) = \cosh at$

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

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

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

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

Linearity Property | Laplace Transform

Linearity Property
If   $a$   and   $b$   are constants while   $f(t)$   and   $g(t)$   are functions of   $t$   whose Laplace transform exists, then
 

$\mathcal{L} \left\{ a \, f(t) + b \, g(t) \right\} = a \, \mathcal{L} \left\{ f(t) \right\} + b \, \mathcal{L} \left\{ g(t) \right\}$

 

Proof of Linearity Property
$\displaystyle \mathcal{L} \left\{ a \, f(t) + b \, g(t) \right\} = \int_0^\infty e^{-st}\left[ a \, f(t) + b \, g(t) \right] \, dt$