Answer
$$
y^{\prime \prime}+2 y^{\prime}+1.25 y=0
$$
The general solution of that equation is given by
$$
y(t) =e^{-t}( c_{1} \cos\frac{t}{2} +c_{2} \sin\frac{t}{2} )
$$
where $ c_{1} $and $c_{2}$ are arbitrary constants.
Work Step by Step
$$
y^{\prime \prime}+2 y^{\prime}+1.25 y=0 \quad\quad\quad \quad (1)
$$
We assume that $ y = e^{rt}$, and it then follows that $r$ must be a root of the characteristic equation
$$
r^{2}+2 r+1.25=100 r^{2}+200 r+125=0,
$$
so its roots are
$$
\:r_{1,\:2}=\frac{-200\pm \sqrt{200^2-4\cdot \:100\cdot \:125}}{2\cdot \:100}
$$
Thus the possible values of $r$ are
$$r_{1}=-1+i\frac{1}{2} ,\quad r_{2}=-1-i\frac{1}{2} .$$
Therefore two solutions of Eq. (1) are
$$
y_{1}(t) = e^{(-1+i\frac{1}{2})t} =e^{-t}(\cos \frac{t}{2} + i \sin \frac{t}{2})
$$
and
$$
y_{1}(t) = e^{(-1-i\frac{1}{2})t} =e^{-t}(\cos \frac{t}{2} - i \sin \frac{t}{2})
$$
Thus the general solution of the differential equation is
$$
y(t) =e^{-t}( c_{1} \cos\frac{t}{2} +c_{2} \sin\frac{t}{2} )
$$
where $ c_{1} $and $c_{2}$ are arbitrary constants.
