# To determine the solution of the initial value problem {y}{''}+{4}{y}=delta{left({t}-piright)}-delta{left({t}-{2}piright)},{y}{left({0}right)}={0},{y}'{left({0}right)}={0} and draw the graph of the solution.

To determine the solution of the initial value problem
$y{}^{″}+4y=\delta \left(t-\pi \right)-\delta \left(t-2\pi \right),y\left(0\right)=0,{y}^{\prime }\left(0\right)=0$
and draw the graph of the solution.
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Decision:
Let the differential equation $y{}^{″}+4y=\delta \left(t-\pi \right)-\delta \left(t-2\pi \right).$
Applying the Laplace transform to the differential equation
$L\left\{y{}^{″}+4y\right\}=L\left\{\delta \left(t-\pi \right)-\delta \left(t-2\pi \right)\right\}$
$L\left\{y{}^{″}\right\}+4L\left\{y\right\}=L\left\{\delta \left(t-\pi \right)\right\}-L\left\{\delta \left(t-2\pi \right)\right\}$
By using $L\left\{{f}^{n}\left(t\right)\right\}={s}^{n}F\left(s\right)-{s}^{n-1}f\left(0\right)-\dots -{f}^{n-1}\left(0\right)$ and
$L\left\{\delta \left(t-{t}_{0}\right)\right\}={e}^{s{t}_{0}},$
$⇒\left[{s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)\right]+4Y\left(s\right)={e}^{-\pi s}-{e}^{-2\pi s}$
Using the initial conditions
${s}^{2}Y\left(s\right)+4Y\left(s\right)={e}^{-\pi s}-{e}^{-2\pi s}$
$⇒Y\left(s\right)\left[{s}^{2}+4\right]={e}^{-\pi s}-{e}^{-2\pi s}$
$⇒Y\left(s\right)=\frac{{e}^{-\pi s}-{e}^{-2\pi s}}{{s}^{2}+4}$
Now, applying the inverse Laplace transformation on both sides,
$⇒{L}^{-1}\left\{Y\left(s\right)\right\}={L}^{-1}\left\{\frac{{e}^{-\pi s}}{{s}^{2}+4}-\frac{{e}^{-2\pi s}}{{s}^{2}+4}\right\}$
$⇒{L}^{-1}\left\{Y\left(s\right)\right\}=\frac{1}{2}{L}^{-1}\left\{\frac{2{e}^{-\pi s}}{{s}^{2}+{2}^{2}}\right\}-\frac{1}{2}{L}^{-1}\left\{\frac{2{e}^{-2\pi s}}{{s}^{2}+{2}^{2}}\right\}$
By using ${L}^{-1}\left\{{e}^{cs}G\left(s\right)\right\}={u}_{c}\left(t\right)g\left(t-c\right)$ and properties of inverse Laplace,