Can I change $$ x_{n+2} = 14x_{n+1} - 49x_n + n7^n\\ n>=0\\ x_0 = 1\\ x_2=14 $$ to $$ x_{n} = 14x_{n-1} - 49x_{n-2} + n7^n\\ n>=2\\ x_0 = 1\\ x_2=14 $$
And it's same? I need to find solutions of recursive equations, but have no idea how do it when have $$ x_{n+2} =... $$ not $$ a_n =... $$ like in others exercise.
Yes, you can do it but notice that when you changes $x_n$ to $x_{n-2}$ in the second term you should have done the same to $n\cdot 7^n$ so the recursion looks like $$x_n=14x_{n-1}-49x_{n-2}+(n-2)\cdot 7^{n-2}.$$