Answer
$$\alpha=2$$
Work Step by Step
$$y''-y'-2y=0,\quad{y}(0)=\alpha,\quad{y'}(0)=2$$Let $y=e^{\lambda{t}}$ so that $(\ln{y})'=\lambda$.
$${\lambda}^2-{\lambda}-2=0$$
$$(\lambda+1)(\lambda-2)=0$$
$$\lambda_{1,2}=-1,2$$
The general solution is ${y}=c_{1}e^{-t}+c_{2}e^{2t}$. Substituting in the constraints, we obtain
$c_{1}+c_{2}=\alpha$ and $-c_{1}e+2c_{2}=2\Rightarrow{c}_{1}=\alpha-c_{2}\Rightarrow3{c}_{2}-\alpha=2\Rightarrow{c}_{2}=\frac{\alpha+2}{3}\Rightarrow{c}_{1}=\frac{2(\alpha-1)}{3}$
$$\therefore{y}=\frac{2(\alpha-1)e^{-t}+(\alpha+2)e^{2t}}{3}$$
As $t\rightarrow+\infty$, $e^{-t}\rightarrow0$ and $e^{2t}\rightarrow+\infty\Rightarrow(\alpha+2)e^{2t}\rightarrow+\infty$.
Thus, for the solution to approach zero, $\alpha+2=0$.
$$\therefore\alpha=2$$
