differential equation solved by Laplace transform

We solve the differential equation

$$y"(t)+y(t)=f(t)$$

with initial condition

$$y(0)=0,\:y'(0)=0$$

where

$$f(t)=u(t-2\pi)-u(t-\pi)$$
$$\mathcal{L}(y(t)(s)=Y(s)$$
$$\mathcal{L}(y'(t)(s)=sY(s)$$
$$\mathcal{L}(y"\;(t)(s)=s^2Y(s)$$

We have

$$\mathcal{L}(y"(t)+y(t))(s)=s^2Y(s)+Y(s)=F(s)=\frac{1}{s}(e^{-2\pi s}-e^{-\pi s})$$

We have

$$Y(s)=\frac{1}{s^2+1}F(s)$$

We have from convolution theorem

$$y(t)=\sin*f(t)=\int_{0}^{t}f(\tau)\sin(t-\tau)d\tau$$
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