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