Particular solution to difference equation

Question

Given $$x_{n+2}-2x_{n+1}+x_n=1,\text{ }x_0=3,\text{ }x_1=6$$ I found the comp solution to be $x_{nc}=3+3n$. When approaching the particular solution I tried $x_{np}=a$ and $x_{np}=an+b$ but in both cases, when I substitute them into the original equation, all the terms cancel. How am I supposed to approach this?

Answer 1

Observe that if you put $y_n = x_n - x_{n-1}, y_1 = x_1 - x_0 = 6 - 3 = 3$. Then you get a new equation in $y's$: $y_{n+2} - y_{n+1} = 1, y_1 = 3, n \ge 0$. Thus $y_n = y_1 + (n-1)d= 3 + 1(n-1) = 2+n$. Hence $x_n - x_{n-1} = 2+n$. So $x_n = (x_n - x_{n-1})+(x_{n-1} - x_{n-2})+\cdots + (x_2 - x_1) + (x_1-x_0)+x_0= (2+n)+(2+(n-1))+\cdots +(2+2)+(2+1)+3=2n + \dfrac{n(n+1)}{2}+3$.