Question

Use a transformada de Laplace para resolver o P.V.I. proposto:
y" - y' - 6y = 0;
y(0) = 1
y'(0) = -1

Answer 1

• O resultado desse problema de valor inicial é igual a:

             $\Large\displaystyle\text{ \begin{gathered}y=\frac{4e^{-2t}+e^{3t}}{5}\end{gathered} }$

Para resolver um PVI através da transformada de Laplace, devemos aplicar a transformada de Laplace em ambos os lados da equação diferencial. Ficando então:

$\Large\displaystyle\begin{gathered} \mathcal{L}\left\{y''\right\} - \mathcal{L}\left\{ y'\right\} - \mathcal{L}\left\{6y\right\} = \mathcal{L}\left\{0 \right\}\end{gathered}$

Pela lineariedade, temos que

$\Large\displaystyle\begin{gathered} \mathcal{L}\left\{y''\right\} - \mathcal{L}\left\{ y'\right\} - 6\mathcal{L}\left\{y\right\} = \mathcal{L}\left\{0 \right\}\end{gathered}$

Agora, vale ressaltar as seguintes propriedades da transformada de Laplace:

$\Large\displaystyle\text{ \begin{gathered} \mathcal{L}\left\{\frac{d^2}{dt^2} (y)\right\}= s^2\cdot \mathcal{L}\left\{ y \right\}-s\cdot y(0)-y'(0)\ \ \red{ (I)}\end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered} \mathcal{L}\left\{\frac{d}{dt} (y)\right\}= s\cdot\mathcal{L}\left\{ y \right\}-y(0)\ \ \ \red{ (II)}\end{gathered} }$

Sabendo disso, logo:

$\Large\displaystyle\begin{gathered} \left[s^2\mathcal{L}\{y\}-sy(0)-y'(0)\right]- \left[ s\mathcal{L}\{y\}-y(0)\right]- 6\mathcal{L}\left\{y\right\} =0\end{gathered}$

$\Large\displaystyle\begin{gathered} \left[s^2\mathcal{L}\{y\}-s+1\right]- \left[ s\mathcal{L}\{y\}-1\right]- 6\mathcal{L}\left\{y\right\} =0\end{gathered}$

$\Large\displaystyle\begin{gathered} \mathcal{L}\{y\}\cdot \left[s^2 -s-6 \right] =s-2\end{gathered}$

$\Large\displaystyle\begin{gathered} \mathcal{L}\{y\}=\frac{s-2}{s^2 -s-6 }\end{gathered}$

$\Large\displaystyle\text{ \begin{gathered}\boxed{y=\mathcal{L}^{-1}\left\{\frac{s-2}{s^2 -s-6 }\right\}}\ \ \red{(III)}\end{gathered} }$

Feito isso, vamos agora abrir em frações parciais, ficando então:

$\Large\displaystyle\text{ \begin{gathered}y=\mathcal{L}^{-1}\left\{\frac{s-2}{\left(s+2\right)\left(s-3\right)}\right\}\end{gathered} }$

$\Large\displaystyle\begin{gathered}\frac{s-2}{\left(s+2\right)\left(s-3\right)}=\frac{A}{(s+2)}+\frac{B}{(s-3)} \end{gathered}$

$\Large\displaystyle\begin{gathered}\frac{s-2}{\left(s+2\right)\left(s-3\right)}=\frac{A(s-3)+B(s+2)}{(s+2)(s-3)} \end{gathered}$

$\Large\displaystyle\begin{gathered}A(s-3)+B(s+2)= s-2\end{gathered}$

• Adotando s como -2, temos que:

$\Large\displaystyle\begin{gathered}-5A= -4\end{gathered}$

$\Large\displaystyle\text{ \begin{gathered}\therefore \blue{\boxed{A= \frac{4}{5}} }\end{gathered} }$

• Adotando s como 3, temos que:

$\Large\displaystyle\begin{gathered}5B= 1\end{gathered}$

$\Large\displaystyle\text{ \begin{gathered}\therefore \blue{\boxed{B= \frac{1}{5}}}\end{gathered} }$

Vamos agora substituir na transformada inversa, ficando então:

$\Large\displaystyle\begin{gathered}\frac{s-2}{\left(s+2\right)\left(s-3\right)}=\frac{4}{5(s+2)}+\frac{1}{5(s-3)} \end{gathered}$

$\Large\displaystyle\text{ \begin{gathered}y=\mathcal{L}^{-1}\left\{\frac{4}{5(s+2)}+\frac{1}{5(s-3)} \right\}\end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered}y=\frac{4}{5}\cdot \mathcal{L}^{-1}\left\{\frac{1}{s+2} \right\}+\frac{1}{5}\cdot \mathcal{L}^{-1}\left\{\frac{1}{s-3} \right\}\end{gathered} }$

$\Large\displaystyle\begin{gathered}y=\frac{4}{5}\cdot e^{-2t}+\frac{1}{5}\cdot e^{3t}\end{gathered}$

$\Large\displaystyle\text{ \begin{gathered}\green{\underline{\boxed{\boxed{y=\frac{4e^{-2t}+e^{3t}}{5}}}}}\ \ (\checkmark )\end{gathered} }$

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