Answer
$$
y^{\prime \prime}+4 y^{\prime}+5 y=0, \quad y(0)=1, \quad y^{\prime}(0)=0
$$
The general solution of the given initial value problem is
$$ y(t) =e^{-2t}( \cos t+ 2 \sin t)$$
$y(t)$ has decaying oscillation.
Work Step by Step
$$
y^{\prime \prime}+4 y^{\prime}+5 y=0, \quad y(0)=1, \quad y^{\prime}(0)=0 \quad (i)
$$
We assume that $ y = e^{rt}$, and it then follows that $r$ must be a root of the characteristic equation
$$
r^{2}+4r+5=0,$$
so its roots are
$$
r_{1,\:2}=\frac{-4\pm \sqrt{4^2-4\cdot \:1\cdot \:5}}{2\cdot \:1}
$$
Thus the possible values of $r$ are
$$r_{1}=-2+i ,\quad r_{2}=-2-i .$$
Therefore two solutions of Eq. (1) are
$$
y_{1}(t) = e^{-2+i } t=e^{-2t}(\cos t + i \sin t)
$$
and
$$
y_{2}(t) = e^{-2-i } t=e^{-2t}(\cos t - i \sin t)
$$
Thus the general solution of the differential equation is
$$
y(t) =e^{-2t}( c_{1}\cos t+ c_{2} \sin t) \quad (ii)
$$
where $ c_{1} $ and $c_{2}$ are arbitrary constants.
To apply the first initial condition, we set $t = 0$ in Eq. (ii); this gives
$ y(0) = c_{1} =1. $
For the second initial condition we must differentiate Eq. (ii) as follows
$$
y^{\prime}(t) =e^{-2t}(( c_{2} -2c_{1} )\cos t+ ( -2c_{2} -c_{1} ) \sin t)
$$
and then set $t = 0 $. In this way we find that
$$
y^{\prime}(0) =( c_{2} -2c_{1} )=0
$$
from which $ c_{2} = 2 $.
Using these values of $c_{1}=1$ and $c_{2}=2 $ in Eq. (ii), we obtain
$$
y(t) =e^{-2t}( \cos t+ 2 \sin t)
$$
as the solution of the initial value problem (i).
Since $ \sin t $ and $ \cos t $ are bounded and $ e^{-2t} $ goes to 0 as $ t → ∞ $, $y$ goes to 0 as $ t → ∞$.
From the graph we can tell that the function $y(t)$ has decaying oscillation.
