Here is the situation:
For which $ \omega \in \mathbb{R} $ does the boundary value problem $$ y^{''}+ \omega^2y =0, \mbox{ } y(0) =0 \mbox{ } \mbox{and} \mbox{ } y(1)=1 $$ have no solution, a unique solution or infinite solutions?
My work:
So first of all we consider the characteristic polynomial, right? So we have : $\lambda^2 + \omega^2$. So where is this zero? We get $\lambda = \pm i \cdot \omega$. Now $ y_1(x) = cos(\omega \cdot x) $and $ y_2(x) = i \cdot sin (\omega \cdot x)$. We conclude $ y(x) = A \cdot cos(\omega \cdot x) + B \cdot i \cdot sin (\omega \cdot x)$. Is this right? Or do you see a mistake? How can I find the number of solutions?
Apply the boundary values to $y(x)=A\cos(\omega x)+B\sin(\omega x)$ to get
$$\begin{align} y(0)=0&=A\cos(0)+B\sin(0)=A\\ y(1)=1&=A\cos(\omega)+B\sin(\omega)\\ \end{align}$$
Since $A$ equals zero the second equation get
$$1=A\sin(\omega)\Leftrightarrow A=\frac1{\sin(\omega)}$$
From hereon you can see the three cases
$(1)$ There is no solution for $\omega=k\pi$ with $k\in\mathbb{Z}$ since the RHS is not defined in this case.
$(2)$ There is exactly one unique solution for either $\omega=\frac{\pi}2+2k\pi$ or $\omega=-\frac{\pi}2+2k\pi$ with $k\in\mathbb{Z}$ which are given by $y_1=A=1$ and $y_2=A=-1$ respectively.
$(3)$ There are infinite solutions for $\omega\notin\left\{k\pi,\frac{\pi}2+2k\pi,-\frac{\pi}2+2k\pi\right\}$ for $k\in\mathbb{Z}$ namely $y(x)=A\sin(\omega x)$.