Answer
$$
y^{\prime \prime}+4 y=0, \quad y(0)=0, \quad y^{\prime}(0)=1
$$
The general solution of the given initial value problem is
$$ y(t) =\frac{1}{2} \sin 2t $$
For increasing $t$,
$ y $ will be steadily oscillating
Work Step by Step
$$
y^{\prime \prime}+4 y=0, \quad y(0)=0, \quad y^{\prime}(0)=1 \quad (i)
$$
We assume that $ y = e^{rt}$, and it then follows that $r$ must be a root of the characteristic equation
$$
r^{2}+4=0,$$
so its roots are
$$
\quad \:r_{1,\:2}=\frac{\pm \sqrt{-4\cdot \:1\cdot \:4}}{2\cdot \:1}
$$
Thus the possible values of $r$ are
$$r_{1}=2i ,\quad r_{2}=-2i .$$
Therefore two solutions of Eq. (1) are
$$
y_{1}(t) = e^{2i}=(\cos 2t + i \sin 2t)
$$
and
$$
y_{2}(t) = e^{-2i}=(\cos 2t - i \sin 2t)
$$
Thus the general solution of the differential equation is
$$
y(t) =( c_{1} \cos 2t+c_{2} \sin 2t) \quad (ii)
$$
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} = 0. $
For the second initial condition we must differentiate Eq. (ii) as follows
$$
y^{\prime}(t) =( -2c_{1} \sin 2t+2c_{2} \cos 2t)
$$
and then set $t = 0 $. In this way we find that
$$
y^{\prime}(0) =2c_{2} =1
$$
from which $ c_{2} = \frac{1}{2}$.
Using these values of $c_{1}=0$ and $c_{2}=\frac{1}{2}$ in Eq. (ii), we obtain
$$
y(t) =\frac{1}{2} \sin 2t
$$
as the solution of the initial value problem (i).
For increasing $t$,
$ y $ will be steadily oscillating
