Problem 02
By using the linearity property, show that

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

Problem 03
Find the Laplace transform of   $f(t) = \sin bt$.
 

Problem 03
$\displaystyle \mathcal{L} \left\{f(t)\right\} = \int_0^\infty e^{-st} f(t) \, dt$

$\displaystyle \mathcal{L} (\sin bt) = \int_0^\infty e^{-st} \sin bt \, dt$
 

For   $\displaystyle \int_0^\infty e^{-st} \sin bt \, dt$.

Using integration by parts:   $\displaystyle \int u\,dv = uv - \int v\, du$.   Let

Problem 02
Find the Laplace transform of   $f(t) = e^{at}$.
 

Solution 02
$\displaystyle \mathcal{L} \left\{f(t)\right\} = \int_0^\infty e^{-st} f(t) \, dt$

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

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

$\displaystyle \mathcal{L} (e^{at}) = \int_0^\infty e^{-(s - a)t} \, dt$

$\displaystyle \mathcal{L} (e^{at}) = -\dfrac{1}{s - a} \int_0^\infty e^{-(s - a)t} \, [ \, -(s - a) \, dt \, ]$