Answer
$$
4 y^{\prime \prime}+9 y=0
$$
The general solution of that equation is given by
$$ y(t) =( c_{1} \cos \frac{3t}{2} +c_{2} \sin \frac{3t}{2}) $$
where $ c_{1} $and $c_{2}$ are arbitrary constants.
Work Step by Step
$$
4 y^{\prime \prime}+9 y=0 \quad\quad\quad \quad (1)
$$
We assume that $ y = e^{rt}$, and it then follows that $r$ must be a root of the characteristic equation
$$
4r^{2}+9=0,$$
so its roots are
$$
\quad \:r_{1,\:2}=\frac{\pm \sqrt{-4\cdot \:1\cdot \:9}}{4\cdot \:1}
$$
Thus the possible values of $r$ are
$$r_{1}=\frac{3}{2}i ,\quad r_{2}=-\frac{3}{2}i .$$
Therefore two solutions of Eq. (1) are
$$
y_{1}(t) = e^{(\frac{3t}{2}) i}=(\cos \frac{3t}{2} + i \sin \frac{3t}{2})
$$
and
$$
y_{2}(t) = e^{(-\frac{3t}{2}) i}=(\cos \frac{3t}{2} - i \sin \frac{3t}{2})
$$
Thus the general solution of the differential equation is
$$
y(t) =( c_{1} \cos \frac{3t}{2} +c_{2} \sin \frac{3t}{2})
$$
where $ c_{1} $and $c_{2}$ are arbitrary constants.
