Question

Obtenha a transformada de Laplace da função g(t)= s e n ( 2 t ) t

Answer 1

A transformada de laplace da função g(t) = tsen(2t) é:

                 $\large\displaystyle\begin{gathered} \mathcal{L}\left\{t\sin(2t)\right\} = \frac{4s}{(s^2+4)^2}\end{gathered}$

Desejamos calcular a transformada de laplace da função g(t) = tsen(2t).

Para calcular a transformada da função dada, temos a seguinte propriedade:

             $\large\displaystyle\begin{gathered} \mathcal{L}\left\{t^ng(t)\right\} = (-1)^n\frac{d^n}{ds^n}\mathcal{L}\left\{g(t)\right\} \end{gathered}$

Aplicando na sua transformada, temos:

$\large\displaystyle\begin{gathered} \mathcal{L}\left\{t^ng(t)\right\} = (-1)^n\frac{d^n}{ds^n}\mathcal{L}\left\{g(t)\right\} \end{gathered}$

$\large\displaystyle\begin{gathered} \mathcal{L}\left\{t\sin(2t)\right\} = -\frac{d}{ds}\mathcal{L}\left\{\sin(2t)\right\} \end{gathered}$

Lembrando que a transformada do sen(kt) já é tabelada. A mesma é dada por:

                   $\large\displaystyle\begin{gathered} \mathcal{L}\left\{\sin(kt)\right\}=\frac{k}{s^2+k^2} \end{gathered}$

Ficando então:

$\large\displaystyle\begin{gathered} \mathcal{L}\left\{t\sin(2t)\right\} = -\frac{d}{ds}\mathcal{L}\left\{\sin(2t)\right\} \end{gathered}$

$\large\displaystyle\begin{gathered} \mathcal{L}\left\{t\sin(2t)\right\} = -\frac{d}{ds}\left( \frac{2}{s^2+2^2} \right) \end{gathered}$

$\large\displaystyle\begin{gathered} \mathcal{L}\left\{t\sin(2t)\right\} = -\frac{d}{ds}\left( \frac{2}{s^2+4} \right) \end{gathered}$

Agora, devemos derivar aquele "trem" hahah, para isso devemos aplicar a derivada do quociente. Dada por:

                   $\large\displaystyle\begin{gathered} \left( \frac{f}{g}\right)'= \frac{f'\cdot g-f\cdot g'}{g^2} \end{gathered}$

• Logo:

$\large\displaystyle\begin{gathered} \mathcal{L}\left\{t\sin(2t)\right\} = -\frac{d}{ds}\left( \frac{2}{s^2+4} \right) \end{gathered}$

$\large\displaystyle\begin{gathered} \mathcal{L}\left\{t\sin(2t)\right\} = -\left( \frac{(2)'\cdot (s^2+4)-2\cdot (s^2+4)'}{(s^2+4)^2} \right) \end{gathered}$

$\large\displaystyle\begin{gathered} \mathcal{L}\left\{t\sin(2t)\right\} = -\left( \frac{0\cdot (s^2+4)-2\cdot 2s}{(s^2+4)^2} \right) \end{gathered}$

$\large\displaystyle\begin{gathered} \mathcal{L}\left\{t\sin(2t)\right\} = -\left( \frac{-4s}{(s^2+4)^2} \right) \end{gathered}$

$\large\displaystyle\text{ \begin{gathered} \therefore\boxed{\boxed{\green{\mathcal{L}\left\{t\sin(2t)\right\} = \frac{4s}{(s^2+4)^2}}}}\ \checkmark\end{gathered} }$

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Answer 2

Resposta:

Explicação passo a passo: