Question

Use a transformada de laplace.
y''−10y'+9y=5t, y(0)= −1, y'(0)=2

Answer 1

• O resultado desse problema de valor inicial, calculado através da transformada de Laplace, é igual a:

     $\Large\displaystyle\text{ \begin{gathered} y=\frac{50H(t)}{81}+\frac{5t}{9}-2e^t+\frac{31e^{9t}}{81}\end{gathered} }$

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

$\Large\displaystyle\begin{gathered}\mathcal{L}\{y''\}-\mathcal{L}\{10y'\}+\mathcal{L}\{9y\}=\mathcal{L}\{5t\}\end{gathered}$

E pela lineariedade, temos que:

$\Large\displaystyle\begin{gathered} \mathcal{L}\{y''\}-10\mathcal{L}\{y'\}+9\mathcal{L}\{y\}=5\mathcal{L}\{t\}\end{gathered}$

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

$\Large\displaystyle\begin{gathered} \mathcal{L}\{y''\}=s^2\cdot \mathcal{L}\{y\}-s\cdot y(0)-y'(0)\ \ \ \green{\sf(I)}.\end{gathered}$

$\Large\displaystyle\begin{gathered} \mathcal{L}\{y'\}=s\cdot \mathcal{L}\{y\}-y(0)\ \ \ \green{\sf(II)}.\end{gathered}$

Vale resaltar também que a transformada de t já é tabelada, e é igual a 1/s². Sabendo disso, logo:$\Large\displaystyle\begin{gathered} \left[s^2 \mathcal{L}\{y\}-s y(0)-y'(0)\right]-10\left[s\mathcal{L}\{y\}-y(0)\right]+9\mathcal{L}\{y\}=\frac{5}{s^2}\end{gathered}$ $\Large\displaystyle\begin{gathered} \left[s^2 \mathcal{L}\{y\}+s-2\right]-10\left[s\mathcal{L}\{y\}+1\right]+9\mathcal{L}\{y\}=\frac{5}{s^2}\end{gathered}$

Colocando $\large\displaystyle\begin{gathered} \mathcal{L}\{y\}\end{gathered}$ em evidência, temos que:

$\Large\displaystyle\begin{gathered} \mathcal{L}\{y\}\cdot \left[s^2-10s+9\right]=\frac{5}{s^2}-s+12 \end{gathered}$

$\Large\displaystyle\begin{gathered} \mathcal{L}\{y\}\cdot \left[s^2-10s+9\right]=\frac{5-s^3+12s^2}{s^2}\end{gathered}$

$\Large\displaystyle\begin{gathered} \mathcal{L}\{y\}=\frac{5-s^3+12s^2}{s^2\left(s^2-10s+9\right)}\end{gathered}$

Vamos agora isolar o y, ficando então:

$\Large\displaystyle\text{ \begin{gathered} y=\mathcal{L}^{-1}\left\{\frac{5-s^3+12s^2}{s^2\left(s^2-10s+9\right)}\right\}\end{gathered} }$

Para resolver isso é bem tranquilo, vamos abrir em frações parciais.

 $\Large\displaystyle\begin{gathered} \frac{5-s^3+12s^2}{s^2\left(s^2-10s+9\right)}=\frac{5-s^3+12s^2}{s^2\left(s-1\right)\left(s-9\right)}\end{gathered}$

 $\Large\displaystyle\begin{gathered} \frac{5-s^3+12s^2}{s^2\left(s^2-10s+9\right)}=\frac{A}{s} + \frac{B}{s^2}+\frac{C}{s-1}+\frac{D}{s-9}\end{gathered}$

Irei deixar a resolução para encontrar o A , B , C e D em anexo, por causa do limite de caracteres. Mas resolvendo aquele sistema, temos que:

$\Large\displaystyle\text{ \begin{gathered} y=\mathcal{L}^{-1}\left\{\frac{50}{81s}+\frac{5}{9s^2}-\frac{2}{s-1}+\frac{31}{81(s-9)}\right\}\end{gathered} }$

E pela lineariedade:

$\large\displaystyle\text{ \begin{gathered} \frac{50}{81}\mathcal{L}^{-1}\left\{\frac{1}{s}\right\}+\frac{5}{9}\mathcal{L}^{-1}\left\{\frac{1}{s^2}\right\}-2\mathcal{L}^{-1}\left\{\frac{1}{s-1}\right\}+\frac{31}{81}\mathcal{L}^{-1}\left\{\frac{1}{s-9}\right\}\end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered} \therefore \green{\underline{\boxed{ y=\frac{50H(t)}{81}+\frac{5t}{9}-2e^t+\frac{31e^{9t}}{81}}}}\ \ (\checkmark).\end{gathered} }$

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