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I would like to get your help to understand if I am solving this correctly. Especially, I am not sure if I actually got the general solution, and what is the role of $\alpha =2,3$ that also solve the equivalent equation I got (and are true for $n=0$). Thank you. This is the Solution:

We'll find simplified equivalent equation: $$F_{n+1}-5F_{n}+6F_{n-1}=n-1$$By subtraction, leads to:$$F_{n+2}-6F_{n+1}+11F_{n}-6F_{n-1}=1$$And by shifting the series to:$$F_{n+1}-6F_{n}+11F_{n-1}-6F_{n-2}=1$$Which gives us, by subtraction again, and then shifting the series two indexes forward:$$F_{n+4}-7F_{n+3}+17F_{n+2}-17F_{n+1}+6F_{n}=0$$Now by guessing solution of the form $$F_{n}=\alpha^{n}$$, the characteristic polynomial is$$\alpha^{4}-7\alpha^{3}+17\alpha^{2}-17\alpha+6=\left(\alpha-1\right)^{2}\left(\alpha-2\right)\left(\alpha-3\right)=0$$Then, we can extract two solutions out of the root $\alpha=1$ of multiplicity $2$, which are: $1,n$. This means we guess that$$F_{n}=a\cdot n+b$$By pluging it into the first equation:$$a\left(n+2\right)+b-5\left(a\left(n+1\right)+b\right)+6\left(an+b\right)=n$$we get: $$2an-3a+2b=n$$ which implies:$$ a=\frac{1}{2},b=\frac{3}{4}.$$ Hence: $F_{n}=\frac{1}{2}n+\frac{3}{4}$

Martin Sleziak
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e.ad
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    Instead of shifting and subtracting, it is better to find constants $c,d$ such that $G_n := F_n + cn + d$ satisfies the same recurrence as $F_n$ but with the right hand side equal to zero. That way, your three term recurrence will stay a three term recurrence, instead of blowing up to five terms. And of course, your $c$ and $d$ will turn out to be just the $a$ and $b$ you already found. In fact, this is an instance where "the general solution to an inhomogeneous problem is equal to a particular solution to the inhomogeneous problem plus the general solution of the homogeneous problem". – R.P. Apr 08 '21 at 18:22
  • @RP_ Hey , I cant see from where comes the intuition to do that in the first place? – e.ad Apr 08 '21 at 19:24
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    Well the intuition is that homogeneous problems are easier than inhomogeneous problems. So naturally you would want to try a substitution that turns the inhomogeneous problem into a homogeneous one. In the case of linear recurrences (and linear differential equations) this is often possible, namely by subtracting a "particular" solution (meaning just any old solution, not a general one). – R.P. Apr 08 '21 at 19:27
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    So I guess that when I said $c$ and $d$ were equal to $a$ and $b$, I guess I should have said they differ by a sign. But anyway, I hope the method is clear. – R.P. Apr 08 '21 at 19:30
  • Yes, it is. thank you – e.ad Apr 09 '21 at 07:23

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The most general solution is $F_n=A2^n+B3^n+an+b$ with $a,\,b$ as per your calculations.

J.G.
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  • When I tried to find something with $A2^n+B3^n$ I just discovered that for any $A,B$ then if I plug it in $F_{n+2}-5F_{n+1}+6F_{n}$ I get $0$, So I guessed they are true just for $n=0$. So I guessed I should ignored them. Why am I wrong in that? – e.ad Apr 08 '21 at 18:16
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    Because, if you get $0$, doesn't that show that they satisfy the equation? – R.P. Apr 08 '21 at 18:25
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    @e.ad That's not how it works. Would you claim $u_n=0$ is the only solution of $u_{n+1}=2u_n$? – J.G. Apr 08 '21 at 18:50
  • @J.G. umm in the case you presented, I can choose any $u_1 = c$ and then multiply by $2$. But I think I miss the point. I mean, after you showed in your answer the most general solution, I understand I cant take solution of the form $A2^n+B3^n$ because there must be the $3/4$ factor. But for some reason this is not trivial for me – e.ad Apr 08 '21 at 19:16
  • @RP_ (This actually continues the point from my last comment) It does that just for $n=0$ no? I mean if you take $n=1$ then you get $ 0 = F_{n+2}-5F_{n+1}+6F_{n} = 1$ which is a contradiction as far as I see – e.ad Apr 08 '21 at 19:18
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    @e.ad I'm sorry, I was being stupid. Yes, the function $A \cdot 2^n + B \cdot 3^n$ by itself is not a solution of course, you need the $\frac{1}{2}n + \frac{3}{4}$ bit. But to find the general solution, you really need to pass to the homogeneous equation (by the substitution I described), and this homogeneous equation will have $A \cdot 2^n + B \cdot 3^n$ as its general solution. – R.P. Apr 08 '21 at 19:38
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    @e.ad So in my example the general solution is $u_n=A2^n$. If you susbtitute this in to find $A$, it simplifies to $0=0$. This implies any $A$ works, not that $A=0$, or that something claimed so far is "true just for $n=0$". – J.G. Apr 08 '21 at 20:15
  • Yes, I see. thank you – e.ad Apr 09 '21 at 07:23