Calcule as transformadas de Laplace abaixo.
\mathcal{L}\left{2 e^{-3 t} t{4}+e{2 t-7} t^{2} \sin ^{2}(t)\right}

Passo 1

O enunciado pede a transformada de Laplace da soma
2e^{-3t}t^{4} \;+\; e^{2t-7}\,t^{2}\sin^{2}(t).
O conteúdo que vamos usar inclui:
  1. A propriedade de deslocamento na frequência
\mathcal{L}\{e^{at}f(t)\}(s)=F(s-a),
  1. A transformada de potências t^{n},
\mathcal{L}\{t^{n}\}(s)=\frac{n!}{s^{n+1}},
  1. A identidade \sin^{2}(t)=\tfrac{1-\cos(2t)}{2},
  2. A fórmula geral
\mathcal{L}\{t^{n}f(t)\} = (+1)^{n}\frac{d^{n}}{ds^{n}}F(s),
onde F(s)=\mathcal{L}\{f(t)\}(s).

Passo 2

Calcular \mathcal{L}\{2e^{-3t}t^{4}\}. Primeiro, sem o fator 2,
\mathcal{L}\{e^{-3t}t^{4}\}(s) =\mathcal{L}\{t^{4}\}(s+3) =\frac{4!}{(s+3)^{5}} =\frac{24}{(s+3)^{5}}.
Multiplicando por 2, obtemos
\mathcal{L}\{2e^{-3t}t^{4}\}(s) =\frac{48}{(s+3)^{5}}.

Passo 3

Agora o termo e^{2t-7}t^{2}\sin^{2}(t). Note que
e^{2t-7}=e^{-7}\,e^{2t},
logo
\mathcal{L}\{e^{2t-7}t^{2}\sin^{2}(t)\} =e^{-7}\,\mathcal{L}\{e^{2t}t^{2}\sin^{2}(t)\} =e^{-7}\,G(s-2),
onde definimos
G(s)=\mathcal{L}\{t^{2}\sin^{2}(t)\}(s).

Passo 4

Calcular G(s)=\mathcal{L}\{t^{2}\sin^{2}(t)\}(s) usando \sin^{2}(t)=\tfrac{1-\cos(2t)}{2}:
G(s) =\frac{1}{2}\,\mathcal{L}\{t^{2}\}(s) -\frac{1}{2}\,\mathcal{L}\{t^{2}\cos(2t)\}(s).
Temos
\mathcal{L}\{t^{2}\}(s)=\frac{2!}{s^{3}}=\frac{2}{s^{3}},
e
\mathcal{L}\{t^{2}\cos(2t)\}(s) =\frac{d^{2}}{ds^{2}}\!\Bigl[\frac{s}{s^{2}+4}\Bigr].
Calculando a derivada segunda, chega-se a
\frac{d^{2}}{ds^{2}}\!\Bigl[\frac{s}{s^{2}+4}\Bigr] =\frac{2\,s\,(s^{2}-12)}{(s^{2}+4)^{3}}.
Logo
G(s) =\frac{1}{2}\cdot\frac{2}{s^{3}} -\frac{1}{2}\cdot\frac{2\,s\,(s^{2}-12)}{(s^{2}+4)^{3}} =\frac{1}{s^{3}} -\frac{s\,(s^{2}-12)}{(s^{2}+4)^{3}}.

Passo 5

Aplicando o deslocamento s\mapsto s-2 em G(s) e multiplicando por e^{-7}, obtemos
\mathcal{L}\{e^{2t-7}t^{2}\sin^{2}(t)\}(s) =e^{-7}\,\Bigl[ \frac{1}{(s-2)^{3}} -\frac{(s-2)\bigl[(s-2)^{2}-12\bigr]}{\bigl((s-2)^{2}+4\bigr)^{3}} \Bigr].

Resposta

\boxed{ \mathcal{L}\{2e^{-3t}t^{4}+e^{2t-7}t^{2}\sin^{2}(t)\}(s) =\frac{48}{(s+3)^{5}} \;+\;e^{-7}\,\Biggl[ \frac{1}{(s-2)^{3}} -\frac{(s-2)\bigl[(s-2)^{2}-12\bigr]}{\bigl((s-2)^{2}+4\bigr)^{3}} \Biggr] }.

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