<s>[INST] Solve $ x_{n+1} - x_n = 2n + 3$ Solve $$  x_{n+1} - x_n = 2n + 3, x_0 = 1, n \ge 0$$
I would try to find a homogen solution and used $$ r^2 - r = 0$$ and got $$x^h_n = A1^n$$ but this seems wrong and I'm stuck on how to continue. 
=== Edit ===
Homogen solution:
$r^2 - r = 0 $ has the solutions $r=0, r=1 $ So we have $$x_n^h = C1^n$$
Particulair solution:
$x_n^p = B(2n+3)$. Inserting this in the original equation gives:
$$ B(2(n+1)+3) - B(2n + 3) = 2n + 3$$
$$ B = \frac{2n + 3}{8}$$
Solution:
$$ x_n = C 1^n + \frac{2n + 3}{8}(2n + 3)$$
Using $x_0 = 1$ to find $C$ gives $C = - \frac{1}{8}$
So the answer is $$ x_n = x_n^h + x_n^p =  - \frac{1}{8}1^n + \frac{2n + 3}{8}(2n + 3)$$
However I know that the answer should be $x_n = (n+1)^2$ so my answer seems to be really off.
=== Edit 2 ===
Okay, so I realise that I should guess a polynom of the same size as my right hand part of the original equation. Since this doesn't work for this particulair problem, I'll try with size + 1. So:
$$ x_n^p = an^2 + bn + c $$
Insert this in the original equation:
$$a(n+1)^2 + b(n+1) + c - an^2 -bn -c = 2n +3$$ And this gives $a = 1, b = 2$.
Then I got $$ x_n = 1^n + n^2 +2n$$ Still not the answer I'm looking for. Specially not the $1^n$ part which will still be there no matter what particulair solution I find.
 [/INST] $$x_{n+1} - x_n = 2n + 3 \tag A$$
$$n = \frac 12 x_{n+1} - \frac 12 x_n - \frac 32 \tag B$$
$$n-1 = \frac 12 x_{n} - \frac 12 x_{n-1} - \frac 32 \tag C$$
Subtract previous two:
$$1 = \frac 12 x_{n+1} + \left(-\frac 12 - \frac 12\right)x_n + \frac 12 x_{n-1} \tag D$$
$$x_{n+1} = x_{n} - x_{n-1} + 2 \tag E$$
Now it is a regular affine equation.
$$\begin{bmatrix} x_{n+1} \\ x_{n} \\ 1 \end{bmatrix}
=
\begin{bmatrix}
1 & -1 & 2 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{bmatrix}^n
\begin{bmatrix} x_1 \\ x_0 \\ 1 \end{bmatrix} \tag F$$
</s>