The solution of the initial value problem y'' + 4y=2delta(t - pi/4), y(0)=0, y'(0)=0 and draw the graph of the solution.

Let the differential equation $y^{″}+4y^{\prime }=2\delta (t-4\pi ).$
Applying the Laplace transform to the differential equation,
$L{y}^{″}+4y=L2\delta (t-\pi /4)$
$Ly+4Ly=2L\delta (t-\pi /4)$
By using $L{f}^n(t)=s^{n}F(s)-s^{n-1}f(0)-\cdots -f^{n-1}(0)$ and
$L\delta (t-t_0)=e^{-s{t}_0}$
$\Rightarrow [s^{2}Y(s)-sy(0)-y^{\prime }(0)]+4Y(s)=2e^{-(\pi /4)s}$
Using the initial conditios $y(0)=0and{y}^{\prime }(0)=1,$
$s^{2}Y(s)+4Y(s)=2e^{-(\pi /4)s}$
$\Rightarrow Y(s)[s^2+4]=2e^{-(\pi /4)s}$
$\Rightarrow Y(s)=2\frac{e^{-(\pi /4)s}}{s^2+4}$
Now, applying the inverse Laplace transformation on both sides,
$\Rightarrow L^{-1}Y(s)=L^{-1}\left\{2\frac{e^{-(\pi /4)s}}{s^2+4}\right\}$
By using $L^{-1}{e}^{cs}G(s)=u_c(t)g(t-c)$ and properties of inverse Laplace,
$\Rightarrow y(t)=u_{\pi /4}(t)\sin (2(t-\pi /4))$
Therefore, the solution of the initial value problem
$y^{″}+4y=2\delta (t-\pi /4),y(0)=0,y^{\prime }(0)=0$ is
$y(t)=u_{\pi /4}(t)\sin (2(t-\pi /4))$
By using $u_{\pi /4}=\{\begin{array}{ll}1& t\ge \pi /4\\ 0& t<\pi /4\end{array}$
For $t<\pi /4,theny(t)=0$
For $t\ge \pi /4,theny(t)=\sin (2(t-\pi /4))$
Thuse, the graph of the solution $y(t)=u_{\pi /4}(t)\sin (2(t-\pi /4)fort<\pi /4andt\ge \pi /4$ is as below: