Consider the initial value problem

y3y10y=0,y(0)=α,y(0)=4\displaystyle {y}{''}-{3}{y}'-{10}{y}={0},\quad{y}{\left({0}\right)}=\alpha,\quad{y}'{\left({0}\right)}={4}

Find the value of α\displaystyle \alpha so that the solution to the initial value problem approaches zero as t\displaystyle {t}\to\infty

α\displaystyle \alpha =