Answer
$${y}=c_{1}\exp^{\frac{9+3\sqrt5}{2}x}+c_{2}\exp^{\frac{9-3\sqrt5}{2}x}$$
Work Step by Step
$$y''-9y'+9y=0$$Let $y-e^{\lambda{x}}$ so that $(\ln{y})'=\lambda$.
$${\lambda}^2-9{\lambda}+9=0$$
$$\lambda_{1,2}=\frac{9\pm{\sqrt{81-36}}}{2}=\frac{9\pm{\sqrt45}}{2}=\frac{9\pm{3\sqrt5}}{2}$$
$$\therefore{y}=c_{1}\exp^{\frac{9+3\sqrt5}{2}x}+c_{2}\exp^{\frac{9-3\sqrt5}{2}x}$$
