Question

Use operador Transformado de Laplace para determinar as soluções das seguintes equações diferenciais que verifiquem as condições iniciais dadas.

b) y" + 4y' + 3y = 0 , { y(0) = 0
{y'(0)= 1 , t >= 0;

Answer 1

É dado o PVI:

$\begin{cases}y''(t)+4y'(t)+3y(t)=0\\y(0)=0\\y'(0)=1,~t\ge0\end{cases}$

Aplicando a Transformada de Laplace na EDO:

$y''(t)+4y'(t)+3y(t)=0\\\\ \mathcal{L}\{y''(t)+4y'(t)+3y(t)\}=\mathcal{L}\{0\}=0\\\\ \mathcal{L}\{y''(t)\}+\mathcal{L}\{4y'(t)\}+\mathcal{L}\{3y(t)\}=0\\\\ \mathcal{L}\{y''(t)\}+4\mathcal{L}\{y'(t)\}+3\mathcal{L}\{y(t)\}=0$

Mas, devemos saber que, se $\mathcal{L}\{y(t)\}=Y(s)$:

$\mathcal{L}\{y''(t)\}=s^2Y(s)-s\underbrace{y(0)}_{0}-\underbrace{y'(0)}_{1}\\\\ \Longrightarrow \mathcal{L}\{y''(t)\}=s^2Y(s)-1\\\\\\\ \mathcal{L}\{y'(t)\}=sY(s)-s\underbrace{y(0)}_{0}\\\\ \Longrightarrow \mathcal{L}\{y'(t)\}=sY(s)$

Substituindo:

$s^2Y(s)-1+4sY(s)+3Y(s)=0\\\\ s^2Y(s)+4sY(s)+3Y(s)=1\\\\ (s^2+4s+3)Y(s)=1\\\\ Y(s)=\dfrac{1}{s^2+4s+3}$

Agora, basta aplicarmos a inversa da Transformada de Laplace:

$\mathcal{L}^{-1}\{Y(s)\}=\mathcal{L}^{-1}\left\{\dfrac{1}{s^2+4s+3}\right\}\\\\ y(t)=\mathcal{L}^{-1}\left\{\dfrac{1}{s^2+4s+4-1}\right\}\\\\ y(t)=\mathcal{L}^{-1}\left\{\dfrac{1}{(s+2)^2-1}\right\}\\\\ y(t)=e^{-2t}\mathcal{L}^{-1}\left\{\dfrac{1}{s^2-1}\right\}\\\\ \boxed{y(t)=e^{-2t}\sinh(t)}$