Answer
$\alpha\leq0$ for the solution to tend to zero and $a>0$ for the solution to be unbounded.
Work Step by Step
1) $$y''-(2\alpha-1)y'+\alpha(\alpha-1)y=0$$Let $y-e^{\lambda{t}}$ so that $(\ln{y})'=\lambda$.
$${\lambda}^2-(2\alpha-1){\lambda}+\alpha(\alpha-1)=0$$
$$\lambda_{1,2}=\frac{2\alpha-1\pm\sqrt{{(2\alpha-1)}^2-4\alpha(\alpha-1)}}{2}=\frac{2\alpha-1\pm1}{2}=\alpha,\alpha-1$$
$$\therefore{y}=c_{1}e^{\alpha{t}}+c_{2}e^{(\alpha-1)t}$$
As $t\rightarrow+\infty$, $e^{(\alpha+k)t}(k\geq0)\rightarrow+\infty$ unless $a\leq0$.
$\therefore\alpha\leq0$ for the solution to tend to zero and $a>0$ for the solution to be unbounded.
