Solution: We will use the Laplace transformation to solve this problem. Let us recall that the Laplace transformation of $f$ is given by \[ F(s) \coloneqq \mathcal{L} \{ f (t)\} = \int _0^{\infty} f(t) e^{-st}\mathrm{d} t. \] Also, we have \begin{align*} \mathcal{L} \{ f'(t) \} & = sF(s) - f(0) \\ \mathcal{L} \{ f''(t) \} & = s^2F(s) - sf(0) - f'(0). \end{align*} The given differential equation is \[ y'' + 4y = f(t) = \begin{cases} t, &\text{ if } 0 \leq t \leq 2,\\ 2, &\text{ if } 2 < t < \infty . \end{cases} \]

Let us find the Laplace transformation of $f(t)$. \begin{align*} \mathcal{L} \{ f(t) \} & = \int _0^{\infty} f(t) e^{-s t}\mathrm{d} t\\ & = \textcolor{blue}{\int _0^{2} t e^{-s t}\mathrm{d} t} +\textcolor{magenta}{ \int _2^{\infty} 2 e^{-st}\mathrm{d} t}\\[1ex] & = \textcolor{blue}{\left[ -\frac{t}{s}e^{-ts} \right] _0^2 - \left[ \frac{1}{s^2}e^{-ts} \right]_0^2 } + \textcolor{magenta}{\left[- \frac{2e^{-ts}}{s} \right]_2^{\infty} } \\[1ex] & = \textcolor{blue}{-\frac{2}{s}e^{-s} - \left( \frac{e^{-2s}}{s^2} - \frac{1}{s^2} \right) } - \textcolor{magenta}{\left( 0 - \frac{2e^{e^{-2s}}}{s} \right) } \\[1ex] & = \textcolor{blue}{\frac{1}{s^2} - \frac{(2s+1)}{s^2}e^{-2s}} + \textcolor{magenta}{\frac{2}{s} e^{-2s}} \\[1ex] & = \frac{1}{s^2} - \frac{1}{s^2}e^{-2s}. \end{align*}

We now take the Laplace transformation of the given differential equation and write $\mathcal{L} \{ y(t) \} = Y(s) $. \begin{align*} & s^2 Y(s) - sy(0) - y'(0) + 4Y(s) = \frac{1}{s^2} - \frac{1}{s^2}e^{-2s} \\ \implies & s^2 Y(s) + 4Y(s) = \frac{1}{s^2} - \frac{1}{s^2}e^{-2s} \\ \implies & Y(s) = \frac{1}{s^2(s^2+4)} - \frac{e^{-2s}}{s^2 (s^2 + 4)} \\ \implies & y(t) = \mathcal{L} ^{-1} \left\{ \frac{1}{s^2(s^2+4)} - \frac{e^{-2s}}{s^2 (s^2 + 4)} \right\} . \end{align*}

So, we need to find the inverse Laplace transformation. Consider the first term. \begin{align*} \mathcal{L} ^{-1} \left\{ \frac{1}{s^2}\cdot \frac{1}{(s^2+4)} \right\}. \end{align*} We have \[ \mathcal{L} ^{-1} \left\{ \frac{1}{s^2 + 2^2} \right\} = \frac{1}{2}\sin 2t. \] Now we recall that \[ \int_0^t f(\tau)\mathrm{d} \tau = \mathcal{L} ^{-1} \left\{ \frac{1}{s}F(s) \right\} . \] Applying the above formula twice, we get \begin{align*} \mathcal{L} ^{-1} \left\{ \frac{1}{s}\cdot \frac{1}{s^2 + 4} \right\} & = \int_0^t \frac{1}{2}\sin 2\tau \mathrm{d} t\\ & = \left[-\frac{1}{4}\cos 2\tau\right]_0^t \\ & = \frac{1}{4}\left( 1 - \cos 2t \right) . \end{align*} Now again applying the same, we get \begin{align*} \mathcal{L} ^{-1} \left\{ \frac{1}{s}\cdot \frac{1}{s\left( s^2 + 4 \right) } \right\} & = \frac{1}{4} \int_0^t (1 - \cos 2t) \mathrm{d} t\\ & = \frac{1}{4}\left[ t - \frac{\sin 2\tau}{2} \right]_0^t \\ & = \frac{t}{4} - \frac{\sin 2t}{t^3}. \end{align*} Thus, \[ \mathcal{L} ^{-1} \left\{ \frac{1}{s^2(s^2+4)} \right\} = \frac{t}{4} - \frac{\sin 2t}{t^3}. \]

We now need to obtain the inverse Laplace transformation of $\frac{e^{-2s}}{s^2 (s^2 + 4)}$. Recall that \[ \mathcal{L} ^{-1} \left\{ \frac{e^{-as}}F(s) \right\} = u(t - a)f(t-a), \] where \[ u(t -a ) = \begin{cases} 0, &\text{ if } t < a ;\\ 1, &\text{ if } t > a. \end{cases} \] Therefore, \begin{align*} \mathcal{L} ^{-1} \left\{ \frac{e^{-2s}}{s^2(s^2 + 4)} \right\} & = u(t-2) \left( \frac{t-2}{4} - \frac{\sin 2(t-2)}{(t-2)^3} \right) . \end{align*} Therefore, \begin{align*} y(t) & = \mathcal{L} ^{-1} \left\{ \frac{1}{s^2(s^2+4)} - \frac{e^{-2s}}{s^2 (s^2 + 4)} \right\} \\ & = \frac{t}{4} - \frac{\sin 2t}{t^3} - u(t-2) \left( \frac{t-2}{4} - \frac{\sin 2(t-2)}{(t-2)^3} \right). \end{align*} We need to find $y\left( \frac{\pi}{2} \right) $. Note that $\frac{\pi}{2} < 2$, so $u(t-2) = 0$. Therefore, \begin{align*} \alpha = y\left( \frac{\pi }{2} \right) & = \frac{\pi }{8}. \end{align*} Hence, \[ \frac{4}{\pi }\alpha = \frac{4}{\pi } \times \frac{\pi}{8} = \frac{1}{2} = 0.5. \]