# Use Laplace transforms to solve the initial value problems.y"+4y=2\delta(t-\pi), y(0)=1 , y'(0)=0

Use Laplace transforms to solve the initial value problems.
$$y"+4y=2\delta(t-\pi), y(0)=1 , y'(0)=0$$

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We know that $$L\left\{\delta(t-t_0)\right\}=e^{-s t_0}$$
and $$L^{-1}\left\{e^{-as} F(s)\right\}=f(t-a)u(t-a)$$
Now , given that, $$y"+4y=2\delta(t-\pi); y(0)=1 . y'(0)=0$$
Taking Laplace transformation on both sides we get,
$$L\left\{y"+4y\right\}=2L\left\{\delta(t-\pi)\right\}$$
$$\Rightarrow L\left\{y"\right\}+4L\left\{y\right\}=2L\left\{\delta(t-\pi)\right\}$$
$$\Rightarrow sY(s)-sy(0)-y'(0)+4y(s)=2e^{-\pi s}$$
$$\Rightarrow s^2Y(s)-s-0+4y(s)=2e^{-\pi s} \left[\because y(s)=L\left\{y\right\}\right]$$
$$\displaystyle\Rightarrow{y}{\left({s}\right)}{\left[{s}^{{2}}+{4}\right]}={2}{e}^{{-\pi{s}}}+{s}$$
$$\displaystyle\Rightarrow{y}{\left({s}\right)}={2}{\frac{{{e}^{{-\pi{s}}}}}{{{s}^{{2}}+{4}}}}+{\frac{{{s}}}{{{s}^{{2}}+{4}}}}$$
Now taking inverse Laplace transformation,
$$y(d)=L^{-1}\left\{\frac{2e^{-\pi s}}{s^2+4}\right\}+L^{-1}\left\{\frac{s}{s^2+4}\right\}$$
$$\therefore y(d)=L^{-1}\left\{e^{-\pi s} \frac{2}{s^2+2^2}\right\}+L^{-1}\left\{\frac{s}{s^2+2^2}\right\}\ \ \ (2)$$
Now, $$L^{-1}\left\{e^{-\pi s} \frac{2}{s^2+2^2}\right\}$$
Let, $$\displaystyle{F}{\left({s}\right)}={\frac{{{2}}}{{{s}^{{2}}+{2}^{{2}}}}},\ \text{ Then }\ {f{{\left({t}\right)}}}={\sin{{2}}}{t}$$
$$\displaystyle\therefore{f{{\left({t}-\pi\right)}}}={\sin{{2}}}{\left({t}-\pi\right)}$$
$$L^{-1}\left\{e^{-\pi s} \frac{2}{s^2+2^2}\right\}=f(t-2)u(t-a)$$
$$\displaystyle={\sin{{\left({2}{t}-{2}\pi\right)}}}{u}{\left({t}-\pi\right)}$$
$$\displaystyle={\sin{{2}}}{\left({t}-\pi\right)}{u}{\left({t}-\pi\right)}$$
and $$L^{-1}\left\{\frac{s}{s^2+2^2}\right\}=\cos 2t$$
From (2) we get
$$\displaystyle{y}{\left({t}\right)}={\sin{{2}}}{\left({t}-\pi\right)}{u}{\left({t}-\pi\right)}+{\cos{{2}}}{t}$$
$$\displaystyle\Rightarrow{y}{\left({t}\right)}={\sin{{2}}}{t}{u}{\left({t}-\pi\right)}+{\cos{{2}}}{t}$$