Use a transformada de Laplace para resolver o problema de valor inicial dado:
y^{\prime \prime \prime}+2 y^{\prime \prime}-y^{\prime}-2 y=\sin (3 t), y(0)=0, y^{\prime}(0)=0, y^{\prime \prime}(0)=1

Passo 1

O enunciado pede para resolver por transformada de Laplace o problema de valor inicial
y''' + 2y'' - y' - 2y = \sin(3t), \quad y(0)=0,\;y'(0)=0,\;y''(0)=1.
Vamos usar as propriedades
  1. \mathcal{L}\{y'\}=sY(s)-y(0),
  2. \mathcal{L}\{y''\}=s^2Y(s)-s\,y(0)-y'(0),
  3. \mathcal{L}\{y'''\}=s^3Y(s)-s^2y(0)-s\,y'(0)-y''(0),
    e a transformada \mathcal{L}\{\sin(3t)\}=\tfrac{3}{s^2+9}.

Passo 2

Tomando transformada em ambos os lados, com Y(s)=\mathcal{L}\{y(t)\}:
\begin{aligned} \mathcal{L}\{y'''\}+2\mathcal{L}\{y''\}-\mathcal{L}\{y'\}-2\mathcal{L}\{y\} &= \frac{3}{s^2+9},\\\ \bigl(s^3Y(s)-1\bigr)+2\bigl(s^2Y(s)\bigr)-\bigl(sY(s)\bigr)-2Y(s) &= \frac{3}{s^2+9}. \end{aligned}
Reunindo termos em Y(s):
Y(s)\bigl(s^3+2s^2 - s - 2\bigr) \;-\;1 \;=\;\frac{3}{s^2+9}.

Passo 3

Isolando Y(s) e fatorando o polinômio do denominador:
s^3+2s^2-s-2 = (s+2)(s-1)(s+1),
Y(s) = \frac{1 + \tfrac{3}{s^2+9}}{(s+2)(s-1)(s+1)} = \frac{s^2+12}{(s^2+9)\,(s+2)(s-1)(s+1)}.

Passo 4

Fazemos a decomposição em frações parciais:
\frac{s^2+12}{(s^2+9)(s-1)(s+1)(s+2)} = \frac{A}{s-1} + \frac{B}{s+1} + \frac{C}{s+2} + \frac{D\,s+E}{s^2+9}.
Calculando constantes por confronto ou substituição de valores:
A=\frac{13}{60},\quad B=-\frac{13}{20},\quad C=\frac{16}{39},\quad D=\frac{3}{130},\quad E=-\frac{3}{65}.

Passo 5

Agora aplicamos a transformada inversa termo a termo:
\mathcal{L}^{-1}\Bigl\{\tfrac{A}{s-1}\Bigr\}=A\,e^{t},\quad \mathcal{L}^{-1}\Bigl\{\tfrac{B}{s+1}\Bigr\}=B\,e^{-t},\quad \mathcal{L}^{-1}\Bigl\{\tfrac{C}{s+2}\Bigr\}=C\,e^{-2t},
\mathcal{L}^{-1}\Bigl\{\tfrac{D\,s}{s^2+9}\Bigr\}=D\cos(3t),\quad \mathcal{L}^{-1}\Bigl\{\tfrac{E}{s^2+9}\Bigr\}=\frac{E}{3}\,\sin(3t).
Logo
y(t) = \frac{13}{60}\,e^{t} - \frac{13}{20}\,e^{-t} + \frac{16}{39}\,e^{-2t} + \frac{3}{130}\cos(3t) - \frac{1}{65}\sin(3t).

Resposta

y(t) = \frac{13}{60}\,e^{t} - \frac{13}{20}\,e^{-t} + \frac{16}{39}\,e^{-2t} + \frac{3}{130}\cos(3t) - \frac{1}{65}\sin(3t).

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