Question

Determine a transformada inversa de Laplace de F(s) = 4/(s+4).

Answer 1

• A transformada inversa de Laplace da função F(s) é igual a:

       $\Large\displaystyle\text{ \begin{gathered} \mathcal{L}^{-1} \left\{ \frac{4}{s+4} \right\} = 4e^{-4t}\end{gathered} }$

Desejamos calcular a transformada inversa de Laplace da função F(s) = 4/(s+4). Para isso, vale ressaltar a transformada da exponencial, dada por

      $\Large\displaystyle\text{ \begin{gathered} \boxed{\mathcal{L} \left\{ e^{\phi t}\right\} = \frac{1}{s -\phi}} \end{gathered} }$

Podemos ainda provar essa transformada ou pela definição, Sendo

 $\Large\displaystyle\text{ \begin{gathered} \mathcal{L} \left\{ e^{\phi t}\right\} = \int _{0^+}^{\infty} e^{\phi t} \cdot e^{-st} dt \end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered} \mathcal{L} \left\{ e^{\phi t}\right\} = \int _{0^+}^{\infty} e^{\phi t -st} dt \end{gathered} }$

Chamando então u = $\phi t -st$ , du = $( \phi -s )dt$ , temos

$\Large\displaystyle\text{ \begin{gathered} \mathcal{L} \left\{ e^{\phi t}\right\} = \frac{1}{\phi -s} \cdot\int _{0^+}^{\infty} e^{u} du \end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered} \mathcal{L} \left\{ e^{\phi t}\right\} = \frac{1}{\phi -s} \cdot \left( e^{\phi t-st} \right)\bigg|_{0^+}^{\infty} \end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered} \mathcal{L} \left\{ e^{\phi t}\right\} = \frac{1}{\phi -s} \cdot \left(e^{\phi \cdot \infty-s\cdot \infty } - e^{\phi \cdot 0 - s\cdot 0}\right) \end{gathered} }$

Pela propriedade da multiplicação de bases iguais, temos que

$\Large\displaystyle\text{ \begin{gathered} \mathcal{L} \left\{ e^{\phi t}\right\} = \frac{1}{\phi -s} \cdot \left(e^{\phi \cdot \infty }\cdot \frac{1}{\infty }- 1\right) \end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered} \mathcal{L} \left\{ e^{\phi t}\right\} = \frac{1}{\phi -s} \cdot \left(- 1\right) \end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered} \boxed{\mathcal{L} \left\{ e^{\phi t}\right\} = \frac{1}{s -\phi}} \end{gathered} }$

• Com isso, resolvendo sua questão, temos que

$\Large\displaystyle\text{ \begin{gathered} \mathcal{L}^{-1} \left\{ \frac{4}{s+4} \right\} = 4\mathcal{L}^{-1} \left\{ \frac{1}{s-(-4)} \right\} \end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered} \green{\boxed{\mathcal{L}^{-1} \left\{ \frac{4}{s+4} \right\} = 4e^{-4t}}}\ \ (\checkmark ) \end{gathered} }$

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