Answer
$$
y^{\prime \prime}+6 y^{\prime}+13 y=0
$$
The general solution of that equation is given by
$$ y(t) =e^{-3t}( c_{1} \cos 2t+c_{2} \sin 2t) $$
where $ c_{1} $ and $c_{2}$ are arbitrary constants.
Work Step by Step
$$
y^{\prime \prime}+6 y^{\prime}+13 y=0 \quad (I)
$$
We assume that $ y = e^{rt}$, and it then follows that $r$ must be a root of the characteristic equation
$$
r^{2}+6r+13=0,$$
so its roots are
$$
\:r_{1,\:2}=\frac{-6\pm \sqrt{6^2-4\cdot \:1\cdot \:13}}{2\cdot \:1}
$$
Thus the possible values of $r$ are
$$r_{1}=-3+2i ,\quad r_{2}=-3-2i,.$$
Therefore two solutions of Eq. (I) are
$$
y_{1}(t) = e^{(-3+2i)t}= e^{-3t}(\cos 2t + i \sin 2t)
$$
and
$$
y_{2}(t) = e^{(-3-2i)t}= e^{-3t}(\cos 2t - i \sin 2t)
$$
Thus the general solution of the given differential equation is
$$
y(t) =e^{-3t}( c_{1} \cos 2t+c_{2} \sin 2t)
$$
where $ c_{1} $ and $c_{2}$ are arbitrary constants.
