Question

Determine a transformada de Laplace da função g(t) = t^2 cos t, sabendo que ℒ [ cos t] =s/s^2+1

Answer 1

• A transformada de laplace da função g(t) = t² cos t é:

                   $\large\displaystyle\begin{gathered} \mathcal{L}\left\{ t^2 \cos t\right\} = \frac{2s(s^2-3)}{(s^2+1)^3} \end{gathered}$

Desejamos calcular a transformada de laplace da função g(t) = t² cos t.

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

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

Aplicando na sua transformada, temos:

$\large\displaystyle\begin{gathered} \mathcal{L}\left\{ t^2 \cos t\right\} = (-1)^2 \frac{d^2}{ds^2}\mathcal{L}\left\{\cos t\right\} \end{gathered}$

$\large\displaystyle\begin{gathered} \mathcal{L}\left\{ t^2 \cos t\right\} = \frac{d^2}{ds^2}\left( \frac{s}{s^2+1} \right)\end{gathered}$

Aplicamos então 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:

$\large\displaystyle\begin{gathered} \mathcal{L}\left\{ t^2 \cos t\right\} = \frac{d^2}{ds^2}\left( \frac{s}{s^2+1} \right)\end{gathered}$

$\large\displaystyle\begin{gathered} \mathcal{L}\left\{ t^2 \cos t\right\} = \frac{d}{ds}\left[ \frac{d}{ds}\left(\frac{s}{s^2+1} \right)\right]\end{gathered}$

$\large\displaystyle\begin{gathered} \mathcal{L}\left\{ t^2 \cos t\right\} = \frac{d}{ds}\left[ \left(\frac{(s)'\cdot (s^2+1)-s\cdot (s^2+1)'}{(s^2+1)^2} \right)\right]\end{gathered}$

$\large\displaystyle\begin{gathered} \mathcal{L}\left\{ t^2 \cos t\right\} = \frac{d}{ds}\left[ \left(\frac{1\cdot (s^2+1)-s\cdot 2s}{(s^2+1)^2} \right)\right]\end{gathered}$

$\large\displaystyle\begin{gathered} \mathcal{L}\left\{ t^2 \cos t\right\} = \frac{d}{ds}\left[ \left(\frac{s^2+1-2s^2}{(s^2+1)^2} \right)\right]\end{gathered}$

$\large\displaystyle\begin{gathered} \mathcal{L}\left\{ t^2 \cos t\right\} = \frac{d}{ds}\left[ \left(\frac{-s^2+1}{(s^2+1)^2} \right)\right]\end{gathered}$

Resolvendo a outra derivada.

$\large\displaystyle\begin{gathered} \mathcal{L}\left\{ t^2 \cos t\right\} = \left(\frac{(-s^2+1)'\cdot (s^2+1)^2-(-s^2+1)\cdot (s^2+1)^2'}{[(s^2+1)^2]^2} \right)\end{gathered}$

$\large\displaystyle\begin{gathered} \mathcal{L}\left\{ t^2 \cos t\right\} = \left(\frac{-2s(s^2+1)^2-4s(-s^2+1)\left(s^2+1\right)}{(s^2+1)^4} \right)\end{gathered}$

$\large\displaystyle\text{ \begin{gathered} \mathcal{L}\left\{ t^2 \cos t\right\} = \left(\frac{-2s\cancel{(s^2+1)^2}-4s(-s^2+1)\cancel{\left(s^2+1\right)}}{(s^2+1)^{\not{4}}} \right)\end{gathered} }$

$\large\displaystyle\begin{gathered} \mathcal{L}\left\{ t^2 \cos t\right\} = \left(\frac{2s(s^2+1) - 4s(1-s^2)}{(s^2+1)^3} \right)\end{gathered}$

$\large\displaystyle\begin{gathered} \mathcal{L}\left\{ t^2 \cos t\right\} = \left(\frac{-2s^3-2s-4s+4s^3}{(s^2+1)^3} \right)\end{gathered}$

$\large\displaystyle\begin{gathered} \mathcal{L}\left\{ t^2 \cos t\right\} = \left(\frac{2s^3-6s}{(s^2+1)^3} \right)\end{gathered}$

$\large\displaystyle\text{ \begin{gathered} \therefore \boxed{\boxed{\green{\mathcal{L}\left\{ t^2 \cos t\right\} = \frac{2s(s^2-3)}{(s^2+1)^3} }}}\ \checkmark\end{gathered} }$

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