Find solutions of recursive equations

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Stuck... Find solutions of recursive equations using generating functions. $$ x_{n+2} = 14x_{n+1} - 49x_n + n7^n, n\ge 0\\ x_0 = 1\\ x_1=14 $$

What I tried: $$ x_{n} = 14x_{n-1} - 49x_{n-2} + (n-2)7^{n-2}, n\ge 2\\ a_{n} = 14a_{n-1} - 49a_{n-2} + (n-2)7^{n-2}\\ F(x) = \sum_{n=0}a_nx^n = 1 + \sum_{n=1}a_nx^n\\ F(x) - 1 = \sum_{n=1}a_nx^n$$ \begin{align} F(x) &= 1 + 14x + \sum_{n=2}^\infty (14a_{n-1} - 49a_{n-2} + (n-2)7^{n-2})x^n\\ &= 1 + 14x + 14\sum_{n=2}^\infty a_{n-1}x^n -49\sum_{n=2}^\infty a_{n-2}x^n + \sum_{n=2}^\infty (n-2)7^{n-2}x^n \\ &= 1 + 14x + 14x\sum_{n=2}^\infty a_{n-1}x^{n-1} -49x^2\sum_{n=2}^\infty a_{n-2}x^{n-2} + x^2\sum_{n=2}^\infty (n-2)7^{n-2}x^{n-2} \\ &= 1 + 14x + 14x\sum_{n=1}^\infty a_{n}x^{n} -49x^2\sum_{n=0}^\infty a_{n}x^{n} + x^2\sum_{n=0}^\infty n7^{n}x^{n} \\ &= 1 + 14x + 14x(F(x) - 1) - 49x^2F(x) + x^2\sum_{n=0}^\infty n7^{n}x^{n} \end{align}

And there is a problem. Is this correct? If yes, how to transform $ x^2\sum_{n=0}^\infty n7^{n}x^{n} $ into something like $x^2F(x)$?

It's a method which our teacher showed us to get generating function, but he didn't show us how make it with inhomogeneous.

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This looks correct so far. To proceed, note that $$\sum_{n \ge 0} n z^n = \sum_{n \ge 1} n z^n = z \sum_{n \ge 1} n z^{n-1} = z \frac{d}{dz}\sum_{n \ge 1} z^n = z \frac{d}{dz}\frac{z}{1-z} = \frac{z}{(1-z)^2}$$