Solve the initial value problems in Problems and graph each solution function x(t): x"+4x=delta(t)+delta(t-pi) x(0)=x'(0)=0

Solve the initial value problems in Problems and graph each solution function x(t):
$x"+4x=\delta \left(t\right)+\delta \left(t-\pi \right)$
$x\left(0\right)={x}^{\prime }\left(0\right)=0$
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Step 1
Take a Laplace transform of the equation:
$x"+4x=\delta \left(t\right)+\delta \left(t-\pi \right)$
$L\left(x\left(t\right)\right)=X\left(s\right)$

$⇒{s}^{2}X\left(s\right)-sx\left(0\right)-{x}^{\prime }\left(0\right)+4X\left(s\right)=1+{e}^{-\pi s}$
$⇒\left({s}^{2}+4\right)X\left(s\right)=1+{e}^{-\pi s}$
$⇒X\left(s\right)=\frac{\left(1+{e}^{-\pi s}\right)}{{s}^{2}+4}$
$⇒X\left(s\right)=\frac{1}{{s}^{2}+4}+\frac{{e}^{-\pi s}}{{s}^{2}+4}$
Step 2
Apply the inverse Laplace transform:
${L}^{-1}\left\{X\left(s\right)\right\}={L}^{-1}\left\{\frac{1}{{s}^{2}+4}+\frac{{e}^{-\pi s}}{{s}^{2}+4}\right\}$
$={L}^{-1}\left\{\frac{1}{{s}^{2}+4}\right\}+{L}^{-1}\left\{\frac{{e}^{-\pi s}}{{s}^{2}+4}\right\}$
$=\frac{1}{2}\left\{{L}^{-1}\left\{\frac{2}{{s}^{2}+{2}^{2}}\right\}+{L}^{-1}\left\{{e}^{-\pi s}\frac{2}{{s}^{2}+{2}^{2}}\right\}\right\}$
$=\frac{1}{2}{L}^{-1}\left\{\frac{2}{{s}^{2}+{2}^{2}}\right\}+\frac{1}{2}{L}^{-1}\left\{{e}^{-\pi s}\frac{2}{{s}^{2}+{2}^{2}}\right\}$
Using: ${L}^{-1}\left\{{e}^{-cs}F\left(s\right)=u\left(t-c\right)f\left(t-c\right)\right\}$
$=\frac{1}{2}\mathrm{sin}\left(2t\right)+\frac{1}{2}u\left(t-\pi \right)\mathrm{sin}\left(2\left(t-\pi \right)\right)$
$=\frac{1}{2}\mathrm{sin}\left(2t\right)+\frac{1}{2}u\left(t-\pi \right)\mathrm{sin}\left(2t\right)$