Non-homogeneous recurrence equation problem, i dont understand how we reached this particular solution

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So I have the equation $y(x+3) - 7y(x+2) +16y(x+1) - 12y(x) = 8 \cdot 2^x$

First I find the general form

The roots of the characteristic polynomial are $2$ (double) and $3$

So the general form is $$c_12^x +c_2x2^x +c_33^x=y^{0}_x$$

Now to find the particular one

I tried to set $ψ_x=c \cdot 2^x$ and solve after, but apparently the correct setting is $ψ_x=c \cdot x^2 \cdot 2^x$

I dont get it??? How we reached $ψ_x=c \cdot x^2 \cdot 2^x$

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Suppose you have a non-homogenous recurrence relation given by

$$F_{n}+AF_{n-1}+BF_{n-2}=c x^{n}$$ and let $f_{h}$ be the solution to the homogenous recurrence relation and $f_{n}$ the solution to the associated non-homogenous recurrence relation. Let $x_{1}$ and $x_{2}$ be the roots of the characteristic equation $x^2+Ax+B=0.$

Then

$\bullet$ If $x\neq x_{1}$ and $x\neq x_{2}$, then $f_{n}=dx^{n}$ since $x^n$ will not be part of the homogenous solution $f_{h}$

$\bullet$ If $x= x_{1}$ and $x\neq x_{2}$, then $f_{n}=dnx^{n}$ since the term $x_{1}^n$ also appears in $f_{h}$, so it will not satisfy the non-homogenous recurrence relation.

$\bullet$ If $x=x_{1}= x_{2}$, then $f_{n}=dn^2x^{n}$, since both $x_{1}^n$ and $nx_1^n$ form $f_{h}$, and they will not contribute in the non-homogenous recurrence relation.

See also 1.