Question

Usando a definição da Transformada de Laplace é correto afirmar que L{e5t} é igual a :

Answer 1

$\large\begin{array}{lr}\begin{aligned}\rm \pounds\left\{ e^{5t} \right\} &\displaystyle\rm= \int\limits_{0}^{+\infty} e^{-st} \cdot e^{5t}\,dt \\\\&\displaystyle\rm= \lim_{M \to +\infty}\int\limits_{0}^{M} e^{-st} \cdot e^{5t}\,dt \\\\&\displaystyle\rm= \lim_{M \to +\infty}\int\limits_{0}^{M} e^{-st + 5t} \,dt \\\\&\displaystyle\rm= \lim_{M \to +\infty}\int\limits_{0}^{M} e^{-t(s - 5)} \,dt \\\\&\displaystyle\rm= \lim_{M \to +\infty} \left.-\frac{e^{-t(s - 5)}}{s-5} \right|_{0}^{M} \\\\&\displaystyle\rm= \lim_{M \to +\infty} \left[-\frac{e^{-M(s - 5)}}{s-5} \right] - \left[-\frac{e^{0\cdot(s - 5)}}{s-5} \right] \\\\&\displaystyle\rm= \lim_{M \to +\infty} \left[-\frac{e^{-M(s - 5)}}{s-5} \right] + \left[\frac{1}{s-5} \right] \end{aligned} \end{array}$

 

Note que temos dois possíveis casos:

$\large\begin{array}{lr}\rm \forall~ s-5 > 0 \Rightarrow s > 5,\,logo\!:\\\\\displaystyle\rm \lim_{M\to+\infty} -\frac{e^{-M(s-5)} }{s-5} + \frac{1}{s-5} = \frac{1}{s - 5} \\\\\rm \forall~ s-5 < 0 \Rightarrow s < 5,\,logo\!: \\\\\displaystyle\rm \lim_{M\to+\infty} -\frac{e^{-M(s-5)} }{s-5} + \frac{1}{s-5} = -\infty ~~\{diverge\} \end{array}$

 

Portanto:

$\large\begin{array}{lr}\red{\underline{\boxed{\boxed{\displaystyle\rm \therefore\: \pounds\left\{ e^{5t} \right\} = \int\limits_{0}^{+\infty} e^{-st} \cdot e^{5t}\,dt = \frac{1}{s-5}\,, ~~ \forall~s > 5}}}} \\\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\blacksquare \end{array}$