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