Find the general solution of the given differential equation.
$$
2 y^{\prime \prime}-3 y^{\prime}+y=0
$$
$$
\begin{array}{l}{\text { Assume } y=e^{r t}} \\ {\text { then } 2 r^{2}-3 r+1=0} \\ {\qquad(2 r-1)(r-1)=0} \\ {\qquad r_{1}=\frac{1}{2}, r_{2}=1}\end{array}
$$
$\because$ The general solution follows the form
$$
y=c_{1} e^{r_{1} t}+c_{2} e^{r_2 t}
$$
$$
\therefore \quad y=c_{1} e^{\frac{t}{2}}+c_{2} e^{t}
$$
