Answer
$${y}=c_{1}\exp^{1+\sqrt{3}x}+c_{2}\exp^{1-\sqrt{3}x}$$
Work Step by Step
$$y''-2y'+2y=0$$Let $y-e^{\lambda{x}}$ so that $(\ln{y})'=\lambda$.
$${\lambda}^2-2{\lambda}+2=0$$
$$\lambda_{1,2}=\frac{2\pm{\sqrt{4+8}}}{2}=\frac{2\pm{2\sqrt3}}{2}=1\pm\sqrt3$$
$$\therefore{y}=c_{1}\exp^{1+\sqrt{3}x}+c_{2}\exp^{1-\sqrt{3}x}$$