Use o Teorema 4.3 para obter a transformada inversa indicada.
mathscr{L}^{-1}\left\{\left(\frac{2}{s}-\frac{1}{s^{3}}\right)^{2}\right\}

Passo 1

O enunciado pede para encontrar
\mathscr{L}^{-1}\Bigl(\bigl(\tfrac{2}{s}-\tfrac{1}{s^{3}}\bigr)^{2}\Bigr)
ou seja, precisamos usar o Teorema 4.3 (convolução) para inverter o quadrado de uma transformada de Laplace.

Passo 2

Primeiro identificamos a função cuja transformada gera \tfrac{2}{s}-\tfrac{1}{s^{3}}. Sabemos que
\mathscr{L}\{2\}=\tfrac{2}{s} e \mathscr{L}\bigl\{\tfrac{t^{2}}{2}\bigr\}=\tfrac{1}{s^{3}}. Logo, definimos
f(t)=2-\tfrac{t^{2}}{2},
de modo que
\mathscr{L}\{f(t)\}=\tfrac{2}{s}-\tfrac{1}{s^{3}}.

Passo 3

Pelo Teorema da Convolução,
\mathscr{L}^{-1}\bigl(F(s)^{2}\bigr) =\bigl(f*f\bigr)(t) =\int_{0}^{t}f(\tau)\,f(t-\tau)\,d\tau.
Aqui F(s)=\mathscr{L}\{f(t)\}, então
\mathscr{L}^{-1}\Bigl(\bigl(\tfrac{2}{s}-\tfrac{1}{s^{3}}\bigr)^{2}\Bigr) =\int_{0}^{t}\Bigl(2-\tfrac{\tau^{2}}{2}\Bigr)\Bigl(2-\tfrac{(t-\tau)^{2}}{2}\Bigr)\,d\tau.

Passo 4

Expandimos o integrando:
\Bigl(2-\tfrac{\tau^{2}}{2}\Bigr)\Bigl(2-\tfrac{(t-\tau)^{2}}{2}\Bigr) =4 -\bigl(\tau^{2}+(t-\tau)^{2}\bigr) +\tfrac{\tau^{2}(t-\tau)^{2}}{4}.
Logo precisamos calcular
\int_{0}^{t}4\,d\tau \;-\;\int_{0}^{t}\tau^{2}d\tau \;-\;\int_{0}^{t}(t-\tau)^{2}d\tau \;+\;\tfrac{1}{4}\int_{0}^{t}\tau^{2}(t-\tau)^{2}d\tau.

Passo 5

Calculando cada integral:
\int_{0}^{t}4\,d\tau=4t, \quad \int_{0}^{t}\tau^{2}d\tau=\tfrac{t^{3}}{3}, \quad \int_{0}^{t}(t-\tau)^{2}d\tau=\tfrac{t^{3}}{3},
e usando a fórmula Beta ou expansão,
\int_{0}^{t}\tau^{2}(t-\tau)^{2}d\tau =t^{5}\!\int_{0}^{1}u^{2}(1-u)^{2}du =t^{5}\,\tfrac{1}{30} =\tfrac{t^{5}}{30}.
Substituindo:
4t \;-\;\tfrac{t^{3}}{3}\;-\;\tfrac{t^{3}}{3}\;+\;\tfrac{1}{4}\cdot\tfrac{t^{5}}{30} =4t-\tfrac{2\,t^{3}}{3}+\tfrac{t^{5}}{120}.

Resposta

\mathscr{L}^{-1}\Bigl(\bigl(\tfrac{2}{s}-\tfrac{1}{s^{3}}\bigr)^{2}\Bigr) =4t-\tfrac{2\,t^{3}}{3}+\tfrac{t^{5}}{120}.

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