Answer
$$\beta=-1$$
Work Step by Step
$$4y''-y=0,\quad{y}(0)=2,\quad{y'}(0)=\beta$$Let $y=e^{\lambda{t}}$ so that $(\ln{y})'=\lambda$.
$$4{\lambda}^2-1=0$$
$$\lambda_{1,2}=\pm\frac{1}{2}$$
The general solution is ${y}=c_{1}e^{\frac{t}{2}}+c_{2}e^{-\frac{t}{2}}$. Substituting in the constraints, we obtain
$c_{1}+c_{2}=2$ and $\frac{c_{1}-c_{2}}{2}=\beta\Rightarrow{c}_{1}=2\beta+c_{2}\Rightarrow{c}_{2}=1-\beta\Rightarrow{c}_{1}=\beta+1$
$$\therefore{y}=(\beta+1)e^{\frac{t}{2}}+(1-\beta)e^{-\frac{t}{2}}$$
As $t\rightarrow+\infty$, $e^{-\frac{t}{2}}\rightarrow0$ and $e^{\frac{t}{2}}\rightarrow+\infty\Rightarrow(\beta+1)e^{\frac{t}{2}}\rightarrow+\infty$.
Thus, for the solution to approach zero, $\beta+1=0$.
$$\therefore\beta=-1$$
