Solution
= Solution
Rearrange as $x''-4x'+3x=10t$. The <characteristic polynomial> factors as $(r-1)(r-3)$, giving the <homogeneous solutions> $e^t,e^{3t}$. A particular <polynomial> solution $At+B$ obeys $3A=10$ and $3B-4A=0$. Hence
$$
\boxed{x(t)=C_1e^t+C_2e^{3t}+\frac{10}{3}t+\frac{40}{9}.}
$$
This is the full two-constant solution of the <second-order linear differential equation>.