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