Differential equations using Laplace transforms

Quote:

Originally Posted by cristi92

$y''+9y=r(t)=\left\{\begin{matrix} 8sin(t),t\in (0,\pi) \\ 0,t \in (\pi,+\infty) \end{matrix}\right. ,y(0)=0,y'(0)=4$

How can I express r(t) using Dirac delta function or Heaviside step function?
I know that Heaviside is the antiderivative of Dirac.

The Heaviside function, H(x), has the value 0 for x< 0, 1 for x> 0. Here, you have to move that "cut point" to x= 1 and reverse ">" and "<". We can do that using
H(1- x). When x< 1, 1- x> 1 so H(1- x)= 1. When x> 1, 1- x< 0 so H(1- x)= 0. You want 8sin(t) for t< 1, 0 for t> 1 so you want 8sin(t)H(1- t).

Quote:

$H(x)=\left\{\begin{matrix} 0, x<0 \\ 1,x\geqslant 0 \end{matrix}\right.$

$\delta(x)=\left\{\begin{matrix} +\infty, x=0 \\ 0,x\neq 0 \end{matrix}\right.$

EDIT:

$r(t)=8sin(t)*(1-u(t-\pi))$
$L(r(t))(s)=L(8sin(t))(s)-L(8sin(t)*u(t-\pi))$
$L(8sin(t)*u(t-\pi))(s)=8*e^{-\pi s}*L(sin(t+\pi))=-8e^{-\pi s}\frac{1}{s^{2}+1} =>$
$L(r(t))(s)=8\frac{1}{s^{2}+1}+8e^{-\pi s}\frac{1}{s^{2}+1}$
then i got
$Y(s)=\frac{8e^{-\pi s}+4s^{2}+12}{(s^{2}+1)(s^{2}+9)}$

I believe that I have to use the Frequency shift property to get rid of $e^{-\pi s}$ , but how?
$L(e^{at}f(t))(s)=F(s-a)$