Problem 01 | Laplace Transform of Derivatives

$f(t) = t^3$       ..........       $f(0) = 0$

$f'(t) = 3t^2$       ..........       $f'(0) = 0$

$f''(t) = 6t$       ..........       $f''(0) = 0$

$f'''(t) = 6$
 

$\mathcal{L} \left\{ f'''(t) \right\} = s^3 \mathcal{L} \left\{ f(t) \right\} - s^2 f(0) - s \, f'(0) - f''(0)$

$\mathcal{L} (6) = s^3 \mathcal{L} (t^3) - s^2 f(0) - s \, f'(0) - f''(0)$

$6 \, \mathcal{L} (1) = s^3 \mathcal{L} (t^3) - s^2(0) - s(0) - (0)$

$6 \left( \dfrac{1}{s} \right) = s^3 \mathcal{L} (t^3)$

$s^3 \mathcal{L} (t^3) = \dfrac{6}{s}$

$\mathcal{L} (t^3) = \dfrac{6}{s^4}$           answer