Answer
$$
y^{\prime \prime}+ y^{\prime}+1.25 y=0
$$
The general solution of that equation is given by
$$ y(t) =e^{-\frac{1}{2}t}( c_{1} \cos t +c_{2} \sin t ) $$
where $ c_{1} $and $c_{2}$ are arbitrary constants.
Work Step by Step
$$
y^{\prime \prime}+ y^{\prime}+1.25 y=0 \quad\quad\quad \quad (1)
$$
We assume that $ y = e^{rt}$, and it then follows that $r$ must be a root of the characteristic equation
$$
100r^2+100r+125=0
$$
so its roots are
$$
\:r_{1,\:2}=\frac{-100\pm \sqrt{100^2-4\cdot \:100\cdot \:125}}{2\cdot \:100}\quad
$$
Thus the possible values of $r$ are
$$r_{1}=-\frac{1}{2}+i ,\quad r_{2}=-\frac{1}{2}-i.$$
Therefore two solutions of Eq. (1) are
$$
y_{1}(t) = e^{(-\frac{1}{2}+i)t} =e^{-\frac{1}{2}t}(\cos t + i \sin t)
$$
and
$$
y_{2}(t) = e^{(-\frac{1}{2}-i)t} =e^{-\frac{1}{2}t}(\cos t - i \sin t)
$$
Thus the general solution of the differential equation is
$$
y(t) =e^{-\frac{1}{2}t}( c_{1} \cos t +c_{2} \sin t )
$$
where $ c_{1} $and $c_{2}$ are arbitrary constants.
