Determine values of the constants $A$ and $B$ such that $a_n=An+B$ is a solution of the recurrence relation $a_n=2a_{n-1}+n+5$.
I know that the characteristic equation is $r-2 = 0$ which has the root $r = 2$. Usually I find constants $A$ and $B$ by $a_n=Ar^n+Br^n$. It looks like I cannot apply this here because $a_n=An+B$ is given. The solution is $A = -1$ and $B = -7$. I don't know how to get this answer. Can anyone help me please.
Consider the homogeneous recurrence relation $a_n=2a_{n-1}$ where $n\geq1$. The characteristic equation is $x-2=0$, where $x=2$ is the root of the characteristic root. Thus the general solution is $a_n=a2^n$ where $a$ is a constant. Let $a_n=bn+c$ where $b$ and $c$ are constants. We do this because $b_n=n+5$, so a linear equation is a proper guess. Substituting this into our original recurrence relation we see that $bn+c=2(b(n-1)+c)+(n-1)+5$ implies that $bn+c=(2b+1)n+(-2b+2c+5)$. Equating the coefficients of these polynomials we see that $b=-1$ and $c=-7$. Thus the particular solution is $a_n=-n-7$. Combining the general solution and the particular solution we have $a_n=a2^n-n-7$.