Solution
= Solution
The homogeneous <characteristic equation> is $r^2-4r+3=(r-1)(r-3)=0$, giving $e^x$ and $e^{3x}$. Since the forcing $e^{3x}$ already appears among these solutions, try a <particular solution> $y_p=Ax e^{3x}$. Differentiation and substitution give
$$
y_p''-4y_p'+3y_p=2Ae^{3x},
$$
so $A=1/2$. The <general solution> is
$$
\boxed{y=C_1e^x+C_2e^{3x}+\frac{x}{2}e^{3x}.}
$$