Answer
$$
y^{\prime \prime}+2 y^{\prime}+2 y=0, \quad y(\pi / 4)=2, \quad y^{\prime}(\pi / 4)=-2
$$
The general solution of the given initial value problem is
$$ y(t) = \sqrt{2}e^{-(t -\pi / 4)} ( \cos t+ \sin t) $$
$ y(t) $ has decaying oscillation.
Work Step by Step
$$
y^{\prime \prime}+2 y^{\prime}+2 y=0, \quad y(\pi / 4)=2, \quad y^{\prime}(\pi / 4)=-2 \quad (1)
$$
We assume that $ y = e^{rt}$, and it then follows that $r$ must be a root of the characteristic equation
$$
r^{2}+2r+2 = 0, $$
so its roots are
$$
\:r_{1,\:2}=\frac{-2\pm \sqrt{2^2-4\cdot \:1\cdot \:2}}{2\cdot \:1}
$$
Thus the possible values of $r$ are
$$r_{1}=-1+i ,\quad r_{2}= -1-i .$$
Therefore two solutions of Eq. (1) are
$$
y_{1}(t) = e^{(-1+i )t}=e^{-t}(\cos t + i \sin t)
$$
and
$$
y_{2}(t) = e^{(-1-i )t}=e^{-t}(\cos t - i \sin t)
$$
Thus the general solution of the differential equation is
$$
y(t) =e^{-t} (c_{1} \cos t+c_{2} \sin t) \quad\quad\quad (2)
$$
where $ c_{1} $ and $c_{2}$ are arbitrary constants.
To apply the first initial condition, we set $t = \pi / 4 $ in Eq. (2); this gives
$$
\begin{split}
y(\pi / 4) &=e^{-\pi / 4} (c_{1} \cos (\pi / 4)+c_{2} \sin (\pi / 4)=2 \\
& = e^{-\pi / 4} (\frac{\sqrt{2}}{2}.c_{1} +\frac{\sqrt{2}}{2} .c_{2} ) =2 \quad (3)
\end{split}
$$
For the second initial condition we must differentiate Eq. (2) as follows
$$
y^{\prime}(t) =e^{-t}[(c_{2}-c_{1}) \cos t-( c_{2}+c_{1}) \sin t]
$$
$$
\begin{split}
y^{\prime}(\pi / 4) &= e^{-\pi / 4}[(c_{2}-c_{1}) \cos (\pi / 4)-( c_{2}+c_{1}) \sin (\pi / 4)]=-2 \\
& = e^{-\pi / 4}[\frac{\sqrt{2}}{2} (c_{2}-c_{1}) - \frac{\sqrt{2}}{2}( c_{2}+c_{1}) ]=-2 \quad (4)
\end{split}
$$
from eq.(3) and eq.(4), it follows that
$$
\left\{\begin{array}{ll}{e^{-\pi / 4} (\frac{\sqrt{2}}{2}.c_{1} +\frac{\sqrt{2}}{2} . c_{2}} ) & =2
\\ {e^{-\pi / 4}[\frac{\sqrt{2}}{2} (c_{2}-c_{1}) - \frac{\sqrt{2}}{2}( c_{2}+c_{1}) ] } & {=-2}\end{array}\right.
$$
This system of linear equations evaluates to
from which $ c_{1} = \sqrt{2}e^{\pi / 4} ,\quad c_{2} = \sqrt{2}e^{\pi / 4} $.
Using these values of $ c_{1} = \sqrt{2}e^{\pi / 4} , \quad c_{2} = \sqrt{2}e^{\pi / 4} $, in Eq. (2), we obtain
$$
y(t) =e^{-t} (\sqrt{2}e^{\pi / 4} \cos t+\sqrt{2}e^{\pi / 4} \sin t)
$$
$$
\begin{split}
y(t) &= e^{-t} (\sqrt{2}e^{\pi / 4} \cos t+\sqrt{2}e^{\pi / 4} \sin t) \\
& = \sqrt{2}e^{\pi / 4 -t} ( \cos t+ \sin t)
\\
& = \sqrt{2}e^{-(t -\pi / 4)} ( \cos t+ \sin t)
\end{split}
$$
as the solution of the initial value problem (1).
Since $ \sin t $ and $ \cos t $ are bounded and $ e^{-(t -\pi / 4)} $ 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.
