Calcule as transformadas de Laplace abaixo.
\mathcal{L}\left\{-5 t^{2}+e^{-5 t-4} \cos ^{2}(3 t)\right\}

Passo 1

O enunciado pede a transformada de Laplace da função
f(t)=-5\,t^{2}+e^{-5t-4}\cos^{2}(3t).
O conteúdo necessário envolve: linearidade da transformada, a fórmula \displaystyle \mathcal{L}\{t^{n}\}=\frac{n!}{s^{n+1}}, o fator exponencial e^{-at} que desloca s\mapsto s+a, e a identidade trigonométrica \displaystyle \cos^{2}(3t)=\frac{1+\cos(6t)}{2}.

Passo 2

Calculamos primeiro
\mathcal{L}\{-5\,t^{2}\} =-5\;\mathcal{L}\{t^{2}\} =-5\;\frac{2!}{s^{3}} =-5\;\frac{2}{s^{3}} =-\frac{10}{s^{3}}.

Passo 3

Agora a parte \displaystyle e^{-5t-4}\cos^{2}(3t). Podemos escrever
e^{-5t-4} =e^{-4}\,e^{-5t},
logo
\mathcal{L}\{e^{-5t-4}\cos^{2}(3t)\} =e^{-4}\,\mathcal{L}\{e^{-5t}\cos^{2}(3t)\}.
Pela identidade
\cos^{2}(3t)=\frac{1+\cos(6t)}{2}
temos
\mathcal{L}\{e^{-5t}\cos^{2}(3t)\} =\frac{1}{2}\,\mathcal{L}\{e^{-5t}\} +\frac{1}{2}\,\mathcal{L}\{e^{-5t}\cos(6t)\}.
Aplicando o deslocamento s\to s+5:
\mathcal{L}\{e^{-5t}\} =\frac{1}{s+5}, \quad \mathcal{L}\{e^{-5t}\cos(6t)\} =\frac{s+5}{(s+5)^{2}+6^{2}}.
Portanto
\mathcal{L}\{e^{-5t}\cos^{2}(3t)\} =\frac{1}{2}\,\frac{1}{s+5} +\frac{1}{2}\,\frac{s+5}{(s+5)^{2}+36}.
Multiplicando por e^{-4} fica
\mathcal{L}\{e^{-5t-4}\cos^{2}(3t)\} =\frac{e^{-4}}{2}\,\frac{1}{s+5} +\frac{e^{-4}}{2}\,\frac{s+5}{(s+5)^{2}+36}.

Passo 4

Pela linearidade, somamos as duas partes:
\mathcal{L}\{f(t)\} =-\frac{10}{s^{3}} +\frac{e^{-4}}{2}\,\frac{1}{s+5} +\frac{e^{-4}}{2}\,\frac{s+5}{(s+5)^{2}+36}.

Resposta

\boxed{ \mathcal{L}\bigl\{-5t^{2}+e^{-5t-4}\cos^{2}(3t)\bigr\} =-\frac{10}{s^{3}} +\frac{e^{-4}}{2}\,\frac{1}{s+5} +\frac{e^{-4}}{2}\,\frac{s+5}{(s+5)^{2}+36} }.

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