The characteristic equation

\begin{equation}

r^2 - (2\alpha-1)r + \alpha(\alpha-1) = 0

\end{equation}

factors as $(r - \alpha)(r - (\alpha - 1))$, so the general solution to the ODE is given by

\begin{equation}

y = A e^{\alpha t} + B e^{(\alpha-1)t}

\end{equation}

where $A, B \in \mathbb{R}$.

We consider the following cases:

- $\alpha < 0$: Both exponentials will be decaying, so each solution tends to zero as $t \to \infty$.
- $\alpha = 0$ or $\alpha = 1$: Each $y = c$ for constant $c$ is a solution, so there exist solutions that neither tend to zero nor become unbounded as $t \to \infty$.
- $0 < \alpha < 1$: One exponential is growing and the other decaying, so there exist nonzero solutions that tend to zero as well as solutions that tend to infinity.
- $\alpha > 1$: Both exponentials will be growing. The larger of the two, $Ae^{\alpha t}$, dominates as $t \to \infty$, so $y$ is unbounded unless $A = 0$. If $A$ vanishes identically, then all nonzero solutions $Be^{(\alpha-1)t}$ again become unbounded.

We conclude that the answer is: (i) $\alpha < 0$, and (ii) $\alpha > 1$.