Answer
$$
9 y^{\prime \prime}+6y^{\prime }+82 y=0, \quad y(0)=-1, \quad y^{\prime}(0)=2
$$
The general solution of the given initial value problem is
$$ y(t) =e^{-\frac{1}{3}t}( - \cos3t +\frac{5}{9} \sin 3t) $$
$y → 0$ as $ t → ∞$
$ y(t) $ has decaying oscillation.
Work Step by Step
$$
9 y^{\prime \prime}+6y^{\prime }+82 y=0, \quad y(0)=-1, \quad y^{\prime}(0)=2 \quad (1)
$$
We assume that $ y = e^{rt}$, and it then follows that $r$ must be a root of the characteristic equation
$$
9r^{2}+6r+82=0,$$
so its roots are
$$
\:r_{1,\:2}=\frac{-6\pm \sqrt{6^2-4\cdot \:9\cdot \:82}}{2\cdot \:9}\quad
$$
Thus the possible values of $r$ are
$$r_{1}=-\frac{1}{3}+3i ,\quad r_{2}=-\frac{1}{3}-3i .$$
Therefore two solutions of Eq. (1) are
$$
y_{1}(t) = e^{(-\frac{1}{3}+3i)t}=e^{-\frac{1}{3}t}(\cos3t + i \sin 3t)
$$
and
$$
y_{2}(t) = e^{(-\frac{1}{3}-3i)t}=e^{-\frac{1}{3}t}(\cos3t - i \sin 3t)
$$
Thus the general solution of the differential equation is
$$
y(t) =e^{-\frac{1}{3}t}( c_{1} \cos3t +c_{2} \sin 3t) \quad (2)
$$
where $ c_{1} $ and $c_{2}$ are arbitrary constants.
To apply the first initial condition, we set $t = 0$ in Eq. (2); this gives
$ y(0) = c_{1} = -1. $
For the second initial condition we must differentiate Eq. (2) as follows
$$
y^{\prime}(t) =e^{-\frac{1}{3}t} [(-\frac{1}{3}c_{1}+3c_{2}) \cos t-\frac{1}{3}( c_{2}+3c_{1}) \sin t]
$$
and then set $t = 0 $. In this way we find that
$$
y^{\prime}(0) = (-\frac{1}{3}c_{1}+3c_{2}) =2 , \quad\quad c_{1} = -1.
$$
it follows that $ c_{2} =\frac{5}{9} $
Using these values of $ c_{1} = -1 , \quad c_{2} =\frac{5}{9} $, in Eq. (2), we obtain
$$
y(t) =e^{-\frac{1}{3}t}( - \cos3t +\frac{5}{9} \sin 3t)
$$
as the solution of the initial value problem (1).
Since $ \sin 3t $ and $ \cos3t $ are bounded and $ e^{-\frac{1}{3}t} $ goes to $0$ as $t → ∞ $, $y(t)$ goes to $0$ as $ t → ∞ $. From the graph we can tell that the function $ y(t)$ has decaying oscillation.
