Superposition of right-hand sides

Homogeneous. As before, $y_h = (Ax+B)e^x$.

First particular solution. For $(x^2+1)e^x$, since $1$ is a double root, try $y_{p1} = P(x)e^x$ with $\deg P \le 4$. Substituting:
$$y_{p1}'' - 2y_{p1}' + y_{p1} = P''(x)e^x.$$
We need $P'' = x^2+1$. Integrating twice:
$$P(x) = \frac{x^4}{12} + \frac{x^2}{2} + Cx + D.$$
Taking $C=D=0$ for a particular solution:
$$y_{p1}(x) = \left(\frac{x^4}{12} + \frac{x^2}{2}\right)e^x.$$

Second particular solution. For $e^{3x}$, try $\alpha e^{3x}$. We found $\alpha = \frac14$.

General solution.
$$y(x) = \left(\frac{x^4}{12} + \frac{x^2}{2} + Ax + B\right)e^x + \frac14 e^{3x}.$$