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Use Laplace Transform to solve initial value problem:
$$y^{\prime \prime}+y=\delta(t-2 \Pi), y(0)=0, y^{\prime}(0)=1$$

Let $$\mathcal {L}\{y(t)\}=y(s)$$

$$\mathcal{L}\left\{y^{\prime \prime}\right\}+\mathcal{L}\{y\}=\mathcal{L}\{\delta(t-2 \pi)\} $$

$$s^{2} y(s)-s y(0)-y^{\prime}(0)+y(s)=e^{-2 \pi s} $$

but $$y(0)=0, y^{\prime}(0)=1$$

$$s^{2}(y(s))-1+y(s)=e^{-2 \pi s} $$

$$y(s)\left[s^{2}+1\right]=e^{-2 \pi s}+1$$

$$\longrightarrow y(s)=\frac{e^{-2 \pi s}+1}{s^{2}+1} $$

$$y(s)=\frac{e^{-2 \pi s}}{s^{2}+1}+\frac{1}{s^{2}+1} $$

Using Laplace inverse we get:

$$y=\sin (t-2 \pi) \cdot u(t-2 \pi)-\sin (t)$$

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