Non-homogenous recurrence relation. How to find the particular solution?

Question

I have enclosed one image of two textbook pages.
There is a system of equation (see frame) on page 2.
I do not understand why both terms can be set equal to 0 (zero)to solve it.
Thank you for the assistance.

Answer 1

The argument they're making is a fairly common one. When you have two polynomials $f(n)$ and $g(n)$, you can say they're equal if each corresponding coefficient is equal. As an example, $f(x)=(4a)x^2+(2b)x+1$ is equal to $g(x)=3x+1$ precisely when $4a=0$, $2b=3$, and $1=1$. Hopefully that concept makes sense, now to get into their argument.

They started with $An+B=3A(n-1)+3B+n$. This is actually two polynomials in $n$. To see this, first I'm going to simplify the right-hand side: \begin{align} 3A(n-1)+3B+n&=3An-3A+3B+n \\ &=\underbrace{(3A+1)}_{a_1}n+\underbrace{(3B-3A)}_{a_0} \\ &=a_1n+a_0 \end{align} It should be clear from the above that the right-hand side is a degree 1 polynomial in $n$, with certain coefficients. So, I have some polynomial (for the left-hand side) $f(n)=An+B$, and on the right-hand side I have $g(n)=(3A+1)n+(3B-3A)$. These will be equal precisely when the constant terms are equal, and when the coefficients in front of the $n$ term are equal. So, we would say that:

$$3A+1=A\quad\text{ and}$$ $$B=3B-3A$$ These two equations can be rearranged into $2A+1=0$ and $2B-3A=0$, which is the system of equations the book gets to.