Answer
$${y}=c_{1}\exp^{\frac{3}{2}x}+c_{2}\exp^{\frac{-3}{2}x}$$
Work Step by Step
$$4y''-9y=0$$Let $y-e^{\lambda{x}}$ so that $(\ln{y})'=\lambda$.
$$4{\lambda}^2-9=0$$
$$\lambda_{1,2}=\pm\frac{3}{2}$$
$$\therefore{y}=c_{1}\exp^{\frac{3}{2}x}+c_{2}\exp^{\frac{-3}{2}x}$$