Problem 03 | Laplace Transform of Derivatives

$f'(t) = 5e^{5t}$
 

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

$\mathcal{L} (5e^{5t}) = s \, \mathcal{L} (e^{5t}) - 1$

$1 = s \, \mathcal{L} (e^{5t}) - \mathcal{L} (5e^{5t})$

$s \, \mathcal{L} (e^{5t}) - 5 \, \mathcal{L} (e^{5t}) = 1$

$(s - 5) \, \mathcal{L} (e^{5t}) = 1$

$\mathcal{L} (e^{5t}) = \dfrac{1}{s - 5}$           answer