what happens if you have both repeated and non-repeated roots? i know there are different forms for both, so if given roots say $5, -3, -3, -3$ would it then be $A(5)^n + B(-3)^n + Cn(-3)^n + Dn^2 (-3)^n$?
since $a(n) = b(n) + p(n)$ for non homogenous recurrence relations, if this was your RR (e.g. $a(n) = 5(n-2) + (n-3) + 3(n-4) + 4*5^{n-2}$ how would p(n) look? would it look like $p(n) = 4Bn(5)^{n-2}+4C(5)^{n-2}$ (since 5 along with 3, 3, 3 is a root of the equation)?
i get quite confused about the p(n) part of a(n) = b(n) + p(n) especially with the whole guessing part
sorry in advance for the formatting (i don't know how to use latex and this is my first post)
Your answer to the first question is correct.
I’m guessing that in the second question you meant to write
$$a(n)=5a(n-2)+a(n-3)+3a(n-4)+4\cdot5^{n-2}\;,$$
so that the non-homogeneous part is just $4\cdot5^{n-2}$. However, in that case the roots of the characteristic equation of the homogeneous part are not $5,3,3$, and $3$. If you did have a homogeneous part whose characteristic equation had those roots, however, then $p(n)$ would have the form $B\cdot5^nn^1=Bn5^n$.
The general rule here is that if the non-homogeneous part has the form $r^nQ(n)$, where $r$ is a root of multiplicity $m$ of the characteristic equation of the homogeneous part, and $Q(n)$ is a polynomial of degree $d$, then
$$p(n)=n^mr^n\left(p_0+p_1n+\ldots+p_dn^d\right)\;.\tag{1}$$
In your problem $r=5$, $m=1$, and $4\cdot5^{n-2}=\frac4{5^2}\cdot5^n=\frac4{25}\cdot5^n$, so $Q(n)=\frac4{25}$, and $d=0$.
If $r$ is not a root of the characteristic equation of the homogeneous part, leave out the factor of $n^m$ in $(1)$. (Equivalently, view the multiplicity of $r$ as $0$, and note that $n^0\equiv 1$.)