power of t

Problem 03
Find the Laplace transform of   $f(t) = t^2 \cos 3t$.
 

Solution 03
$\mathcal{L} \left\{ t^n f(t) \right\} = (-1)^n \dfrac{d^n}{ds^n} F(s)$
 

$\mathcal{L} (\cos 3t) = \dfrac{s}{s^2 + 3^2}$

$\mathcal{L} (\cos 3t) = \dfrac{s}{s^2 + 9}$
 

$\mathcal{L} (t^2 \cos 3t) = (-1)^2 \dfrac{d^2}{ds^2} \left[ \dfrac{s}{s^2 + 9} \right]$

$\mathcal{L} (t^2 \cos 3t) = \dfrac{d}{ds} \left[ \dfrac{(s^2 + 9)(1) - s(2s)}{(s^2 + 9)^2} \right]$

Problem 02
Find the Laplace transform of   $f(t) = t \sin 3t$.
 

Solution 02
$\mathcal{L} \left\{ t^n f(t) \right\} = (-1)^n \dfrac{d^n}{ds^n} F(s)$
 

$\mathcal{L} (\sin 3t) = \dfrac{3}{s^2 + 3^2}$

$\mathcal{L} (\sin 3t) = \dfrac{3}{s^2 + 9}$
 

$\mathcal{L} (t \sin 3t) = (-1)^1 \dfrac{d}{ds} \left[ \dfrac{3}{s^2 + 9} \right]$

$\mathcal{L} (t \sin 3t) = -\left[ \dfrac{-3(2s)}{(s^2 + 9)^2} \right]$

$\mathcal{L} (t \sin 3t) = -\dfrac{-6s}{(s^2 + 9)^2}$

Problem 01
Find the Laplace transform of   $f(t) = t \cos 2t$.
 

Solution 01
$\mathcal{L} \left\{ t^n f(t) \right\} = (-1)^n \dfrac{d^n}{ds^n} F(s)$
 

$\mathcal{L} (\cos 2t) = \dfrac{s}{s^2 + 2^2}$

$\mathcal{L} (\cos 2t) = \dfrac{s}{s^2 + 4}$
 

$\mathcal{L} (t \cos 2t) = (-1)^1 \dfrac{d}{ds} \left[ \dfrac{s}{s^2 + 4} \right]$

$\mathcal{L} (t \cos 2t) = -\left[ \dfrac{(s^2 + 4)(1) - s(2s)}{(s^2 + 4)^2} \right]$

$\mathcal{L} (t \cos 2t) = -\dfrac{s^2 + 4 - 2s^2}{(s^2 + 4)^2}$