Resolva os PVI’s a seguir utilizando transformada de Laplace:
y^{\prime \prime}+2 y^{\prime}+5 y=e^{-t} \operatorname{sen} t, y(0)=0, y^{\prime}(0)=1

Passo 1

O enunciado dá o PVI
y'' + 2y' + 5y = e^{-t}\sin t,\quad y(0)=0,\;y'(0)=1
e pede a solução via transformada de Laplace. O conteúdo envolve: propriedades
\mathcal{L}\{y'\}, \mathcal{L}\{y''\}, a transformada de
e^{at}\sin(bt) e frações parciais para inverter Y(s).

Passo 2

Aplicamos \mathcal{L} em ambos os lados:
\mathcal{L}\{y''\} + 2\,\mathcal{L}\{y'\} + 5\,\mathcal{L}\{y\} = \mathcal{L}\{e^{-t}\sin t\}.
Usando
\mathcal{L}\{y''\}=s^2Y(s)-s\,y(0)-y'(0),
\mathcal{L}\{y'\}=sY(s)-y(0),
\mathcal{L}\{y\}=Y(s) e
\mathcal{L}\{e^{-t}\sin t\}=\frac{1}{(s+1)^2+1}, obtemos
\bigl(s^2Y(s)-1\bigr) + 2\bigl(sY(s)\bigr) + 5Y(s) = \frac{1}{(s+1)^2+1}.

Passo 3

Agrupando termos em Y(s):
(s^2+2s+5)Y(s) - 1 = \frac{1}{(s+1)^2+1},
logo
Y(s) = \frac{1}{s^2+2s+5} + \frac{1}{(s^2+2s+5)\bigl((s+1)^2+1\bigr)}.
Como s^2+2s+5=(s+1)^2+4, escreve-se
Y(s) = \frac{1}{(s+1)^2+4} + \frac{1}{\bigl((s+1)^2+4\bigr)\bigl((s+1)^2+1\bigr)}.

Passo 4

Fazemos a decomposição
\frac{1}{\bigl((s+1)^2+4\bigr)\bigl((s+1)^2+1\bigr)} = \frac{A}{(s+1)^2+1} + \frac{B}{(s+1)^2+4}.
Exigindo
A((s+1)^2+4)+B((s+1)^2+1)=1
A+B=0,\quad 4A+B=1 \;\Longrightarrow\;A=\tfrac13,\;B=-\tfrac13.
Assim
Y(s) = \frac{1}{(s+1)^2+4} + \frac{\tfrac13}{(s+1)^2+1} - \frac{\tfrac13}{(s+1)^2+4} = \frac{2/3}{(s+1)^2+4} + \frac{1/3}{(s+1)^2+1}.

Passo 5

Invertemos termo a termo usando
\mathcal{L}^{-1}\{\tfrac1{(s+1)^2+a^2}\} = e^{-t}\tfrac{1}{a}\sin(at):
\mathcal{L}^{-1}\!\Bigl\{\tfrac{1}{(s+1)^2+4}\Bigr\} = \tfrac12e^{-t}\sin(2t), \quad \mathcal{L}^{-1}\!\Bigl\{\tfrac{1}{(s+1)^2+1}\Bigr\} = e^{-t}\sin t.
Logo
y(t) = \frac{2}{3}\!\Bigl(\tfrac12e^{-t}\sin(2t)\Bigr) + \frac{1}{3}\!\bigl(e^{-t}\sin t\bigr) = \frac{1}{3}e^{-t}\bigl(\sin(2t)+\sin t\bigr).

Resposta

y(t) = \frac{1}{3}e^{-t}\bigl(\sin(2t) + \sin t\bigr).

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