Answer
$$
y^{\prime \prime}+y^{\prime}+1.25 y=0, \quad y(0)=3, \quad y^{\prime}(0)=1
$$
The general solution of the given initial value problem is
$$ y(t) =y(t) =e^{-\frac{1}{2}t}(3 \cos t+\frac{5}{2} \sin t) $$
$ y(t) $ has decaying oscillation.
Work Step by Step
$$
y^{\prime \prime}+y^{\prime}+1.25 y=0, \quad y(0)=3, \quad y^{\prime}(0)=1 \quad (i)
$$
We assume that $ y = e^{rt}$, and it then follows that $r$ must be a root of the characteristic equation
$$
r^{2}+r+1.25 =100r^2+100r+125=0, $$
so its roots are
$$
\:r_{1,\:2}=\frac{-100\pm \sqrt{100^2-4\cdot \:100\cdot \:125}}{2\cdot \:100}
$$
Thus the possible values of $r$ are
$$r_{1}=-\frac{1}{2}+i ,\quad r_{2}=-\frac{1}{2}-i .$$
Therefore two solutions of Eq. (1) are
$$
y_{1}(t) = e^{(-\frac{1}{2}+i )t}=e^{-\frac{1}{2}t}(\cos t + i \sin t)
$$
and
$$
y_{2}(t) = e^{(-\frac{1}{2}-i )t}=e^{-\frac{1}{2}t}(\cos t - i \sin t)
$$
Thus the general solution of the differential equation is
$$
y(t) =e^{-\frac{1}{2}t}(c_{1} \cos t+c_{2} \sin t) \quad\quad\quad (ii)
$$
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} = 3. $
For the second initial condition we must differentiate Eq. (ii) as follows
$$
y^{\prime}(t) =e^{-\frac{1}{2}t}[(c_{2}-\frac{1}{2}c_{1}) \cos t-(\frac{1}{2} c_{2}+c_{1}) \sin t]
$$
and then set $t =0 $. In this way we find that
$$
\begin{split}
y^{\prime}(0) & =(c_{2}-\frac{1}{2}c_{1}) \cos (0)-(\frac{1}{2} c_{2}+c_{1}) \sin (0)=1 \\
& = (c_{2}-\frac{1}{2}c_{1})=1
\end{split}
$$
Substituting $c_{1} = 3. $ It follows that $ c_{2} = \frac{5}{2}. $
Using these values of $c_{1} = 3. $ and $ c_{2} = \frac{5}{2}. $ in Eq. (ii), we obtain
$$
y(t) =e^{-\frac{1}{2}t}(3 \cos t+\frac{5}{2} \sin t)
$$
as the solution of the initial value problem (i).
Since $ \sin t $ and $ \sin t $ are bounded and $ e^{−t/2 }$ 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.
