Resonance with a double root

Homogeneous equation. $r^2 - 2r + 1 = (r-1)^2 = 0$, so $r=1$ is a double root and
$$y_h(x) = (Ax+B)e^x.$$

Particular solution. Try $y_p = \alpha e^{3x}$. Then $y_p' = 3\alpha e^{3x}$, $y_p'' = 9\alpha e^{3x}$, and
$$y_p'' - 2y_p' + y_p = (9 - 6 + 1)\alpha e^{3x} = 4\alpha e^{3x}.$$
We need $4\alpha = 1$, so $\alpha = \frac14$.

General solution.
$$y(x) = (Ax+B)e^x + \frac14 e^{3x}.$$