Laplace Transform to solve the initial value problem

Question

Using the Laplace Transform to solve the initial value problem:

$$y''+ y = \delta(t-2\pi), y(0) =0 ,y'(0)=1$$

Applying Laplace transformation we have $s(s+1)Y(s)=1+e^{-2\pi s}$. Thus $$Y(s) = \frac{1+e^{-2\pi s}}{s(s+1)} = \frac{1}{s(s+1)} + \frac{e^{-2\pi s}}{s(s+1)} $$

Let $F(s) = \frac{1}{s(s+1)} $. Then inverse laplace transform of $(s)$ is $1-e^{-t}$ and that of $\frac{e^{-2\pi s}}{s(s+1)} $ is $$u(t-2\pi) (1-e^{-(t-2\pi)})$$

Thus the solution is $$y(t) = (1-e^t) +u(t-2\pi) (1-e^{-(t-2\pi)}) $$

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Is the soution correct?

Answer 1

$$y''+ y = \delta(t-2\pi),\quad y(0) =0 ,\quad y'(0)=1$$

Tanking Laplace transform both side,

$$\mathcal{L}\{y''\}+ \mathcal{L}\{y\} = \mathcal{L}\{\delta(t-2\pi)\} \qquad . . . . . (1)$$ Now $$\mathcal{L}\{y\}=\bar{y}(p)$$ $$\mathcal{L}\{y'\}=p~\bar{y}(p)-y(0)=p~\bar{y}(p)$$ $$\mathcal{L}\{y''\}=p^2~\bar{y}(p)-p~y(0)-y'(0)=p^2~\bar{y}(p)-1$$ $$\mathcal{L}\{\delta(t-2\pi)\}=e^{-2\pi p}\qquad \text{as}\quad 2\pi \gt 0$$

From $(1)$, $$p^2~\bar{y}(p)-1+\bar{y}(p)=e^{-2\pi p}\implies(p^2+1)\bar{y}(p)=e^{-2\pi p}+1$$ $$\bar{y}(p)=\frac{e^{-2\pi p}+1}{p^2+1}=\frac{e^{-2\pi p}}{p^2+1}+\frac{1}{p^2+1}$$ Therefore $$ y(t)=\mathcal{L}^{-1}\bar{y}(p)=\mathcal{L}^{-1}\{\frac{e^{-2\pi p}}{p^2+1}\}+\mathcal{L}^{-1}\{\frac{1}{p^2+1}\}= \begin{cases} 2\sin t \quad \text{if} \ t\gt 2\pi\\ \sin t \quad \text{if} \ t \lt 2\pi\end{cases}$$

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• (Second shifting property ) If $\mathcal{L}^{-1}\{F(p)\}=f(t)$, then $$\mathcal{L}^{-1}\{e^{-ap}F(p)\}=g(t)$$where $g(t)= \begin{cases} f(t-a) \quad \text{if} \ t\gt a\\ 0 \quad \text{if} \ t \lt a\end{cases}$

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