Find the Laplace transform
$$Y(s) = \mathcal{L(y)}$$
$$y'' + 4y = \begin{cases} 1, & 0\le t < \pi\\ 0, & \pi\le t< \infty \end{cases} $$ $$y(0) = 1$$ $$y'(0) = 0$$
I start by working with the left part of the equation: $$y'' + 4y = \mathcal{L}(y)(s^2+1)+sy(0)+y'(0)=\mathcal{L}(y)(s^2+1)+s$$ $$\mathcal{L}(y)(s^2+1)+s = \mathcal{L}(\begin{cases} 1, & 0\le t < \pi\\ 0, & \pi\le t< \infty \end{cases}) $$
But I have no idea where to go from here.