= Solution
For the homogeneous <linear recurrence relation>, a trial $u_k=r^k$ gives the <characteristic equation of a linear recurrence> $r^2+5r+6=(r+2)(r+3)=0$. Thus the two homogeneous sequences are $(-2)^k$ and $(-3)^k$. Seek a particular solution $u_k=ak+b$; substitution gives $12ak+12b-5a=12k+1$, so $a=1$ and $b=1/2$. Hence
$$
\boxed{u_k=A(-2)^k+B(-3)^k+k+\frac12.}
$$
The two independent constants supply arbitrary initial values for this second-order <linear recurrence relation>.
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