Let $()$ be the solution of the initial value problem
$$y''+4y=\begin{cases} t, & 0\leq t\leq 2\\ 2, & 2<t<\infty \end{cases}.$$ Also, it is given that $$y(0)=0, y'(0)=0.$$
Given that $\alpha=y\big(\frac{\pi}{2}\big)$, show that $4\pi\alpha=0.5$ (round off to two decimal places).
I started by taking the Laplace transform of the given equation to note that (Let $Y(s)$ be the Laplace transform of $y(t)$): $$s^2Y(s)-sy(0)-y'(0)+4Y(s)=\int_0^2 te^{-st} dt+\int_2^{\infty}2e^{-st}dt$$ which means that $$(s^2+4)Y(s)=-\frac{t}{s}e^{-st}\big|_0^2-\frac1{s^2}e^{-st}\big|_0^2-\frac{2}se^{-st}\big|_2^{\infty}$$ $$=-\frac2se^{-2s}-\frac1{s^2}e^{-2s}+\frac1{s^2}+\frac2se^{-2s}$$ $$=\frac1{s^2}-\frac{1}{s^2}e^{-2s}.$$ This implies that $$Y(s)=\frac{1}{(s^2+4)}\bigg(\frac1{s^2}-\frac1{s^2}e^{-2s}\bigg).$$
How to proceed forward.
$$ Y(s) = K(s)-e^{-2s}K(s) $$ where $$K(s)= \frac{1}{s^2(s^2+4)} =\frac{1/4}{s^2}-\frac{1/4}{s^2+4} $$ Using inverse Laplace table $$ y(t) = k(t) - k(t-2)u(t-2) $$
where $ k(t) = \frac14t - \frac{1}{8} \sin(2t) $.
It follows $y(\pi/2)=\frac14 (\pi/2)$ which is different to what is claimed?