Solve $$\begin{cases} y(t)=z''\\ z(t)=y'' \end{cases}$$
$y(0) = z(0) = 0$
$y(\pi/2) = z(\pi/2) = 1 $
My attempt:
$$\begin{cases} y(t)=z''\\ z(t)=y'' \end{cases}\Rightarrow y(t)=y^{(4)}$$
The roots of corresponding characteristic equation are $1,-1,i,-i$. That's why
$$y(t)=c_{1}e^{t}+c_{2}e^{-t}+c_{3}\cos t+c_{4}\sin t$$
Using initial conditions, we get $$\begin{cases} 0=c_{1}+c_{2}+c_{3}\\ 1=c_{1}e^{\pi/2}+c_{2}e^{-\pi/2}+c_{4} \end{cases}$$
My question: how to solve this system of 2 equations in 4 variables?
The answer is supposed to be $$y(t)=z(t)=\frac{e^{t}-e^{-t}}{e^{\pi/2}-e^{-\pi/2}}$$
Note that you have four initial conditions, giving four equations. As $z = y''$, in addition to $y(0) = 0$ and $y(\frac \pi 2) = 1$, you have $$ y''(0) = 0 \iff c_1 + c_2 -c_3 = 0 $$ and $$ y''(\textstyle\frac{\pi}{2}) = 0 \iff c_1e^{\frac \pi 2} + c_2e^{-\frac \pi 2} - c_4 = 0 $$ This (together with your two equations above) gives $$ c_1 + c_2 = c_3 = 0, \qquad c_1e^{\frac \pi 2} + c_2e^{-\frac\pi 2} = c_4 = 0 $$ So $$ c_1 = -c_2 = \frac 1{e^{\frac\pi 2} - e^{-\frac{\pi}2}}, \qquad c_3 = c_4 = 0 $$