= Solution
{c}
For the homogeneous <linear recurrence relation>, try $u_n=r^n$. Its <characteristic equation of a linear recurrence> is
$$
r^2+3r-4=(r-1)(r+4)=0.
$$
The distinct <roots of a polynomial> $1,-4$ give the complementary terms $A+B(-4)^n$. A constant cannot absorb the forcing because $1$ is already a <root of a polynomial> of the characteristic equation; try a <particular solution> $cn$. Substitution gives $c=-4c+1$, hence $c=1/5$. The <general solution> is $u_n=A+B(-4)^n+n/5$.
The <initial conditions> give $A+B=0$ and $A-4B=4/5$. Hence $A=4/25$, $B=-4/25$, and
$$
\boxed{u_n=\frac n5+\frac4{25}\bigl(1-(-4)^n\bigr),\qquad n\ge0.}
$$
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