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.
 A: Do the sum on both sides, which gives you $x_n-x_0 = \sum_{k=0}^{n-1} (2k +3)$
That should be easy to finish it from there.
A: $$x_{n+1} - x_n = 2n + 3 \tag A$$
$$n = \frac 12 x_{n+1} - \frac 12 x_n - \frac 32 \tag B$$
$$n-1 = \frac 12 x_{n} - \frac 12 x_{n-1} - \frac 32 \tag C$$
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.
$$\begin{bmatrix} x_{n+1} \\ x_{n} \\ 1 \end{bmatrix}
=
\begin{bmatrix}
1 & -1 & 2 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{bmatrix}^n
\begin{bmatrix} x_1 \\ x_0 \\ 1 \end{bmatrix} \tag F$$
A: Since you're interested in general methods of solving linear recurrences, here is the fact you need:
Consider the recurrence relation $$a_n = c_1 a_{n-1} + c_2 a_{n-2} + \dots + c_k a_{n-k} + F(n),$$ where $F(n)$ is of the form $f(n)s^n$ for some polynomial $f(x)$ of degree $t$ and some constant $s$. Then there is a particular solution of the form $a_n = n^m g(n) s^n$, where $g(x)$ is a polynomial of degree at most $t$ and $m$ is the multiplicity of the root $s$ in the characteristic equation $x^k - c_1 x^{k-1} - c_2 x^{k-2} - \dots - c_{k-1}x - c_k = 0$. We agree that $m$ is zero when $s$ is not a root.
In your situation, you have $F(n) = (2n + 3)1^n$, so you apply the foregoing with $f(x) = 2x + 3$ and $s = 1$. $1$ is a root of multiplicity one of the characteristic equation $x - 1 = 0$, so here $m = 1$. Furthermore, $f$ has degree $1$. Thus there is a particular solution of the form $a_n = n(cn + d)$.
This should be enough to get you started, but if you need more help, please edit your question to say what you've tried with this. 
EDIT: No, the particular solution won't be $B(2n + 3)$. $c$ and $d$ are not necessarily $2$ and $3$. They are constants that you'll have to solve equations to find. Also, why are you using $B$? As I said, $m=1$ in this case, so you need to multiply by $n$. Set $x_n = cn^2 + dn$.
EDIT: The fact that you'd have to have degree $2$ was predictable from the rule I stated. $g$ can have degree as large as $1$, and $m=1$.
Once you've found the particular solution $x_n = n^2 + 2n$, then to get the general solution, you add the general homogeneous solution, which is $k \cdot 1^n$ = $k$, so is constant. Thus the general solution to the recurrence $x_{n+1} - x_n = 2n + 3$ is $x_n = n^2 + 2n + k$.
This doesn't take into account that $x_0 = 1$. Pick $k$ so that the initial condition is satisfied.
A: Substitute $n=n+1$ we have system of equations $$x_{n+1}-x_n=2n+3\\x_{n+2}-x_{n+1}=2n+5\\x_{n+2}-2x_{n+1}+x_n=2$$
Than doing the same again
$$x_{n+2}-2x_{n+1}+x_n=2\\x_{n+3}-2x_{n+2}+x_{n+1}=2\\x_{n+3}-3x_{n+2}+3x_{n+1}-x_n=0$$
So now you can solve the homogeneous equation which gives $a_n=n^2+n+1$
A: This telescopes conveniently. 
Summing from $n=0$ to $n=m-1$ gives 
$$\begin{align}\require{cancel}
\sum_{n=0}^{m-1}x_{n+1}-x_n&=\sum_{n=0}^{m-1}\left(2n+3\right)\\
x_m-\cancelto{1}{x_0}&=2\sum_{n=0}^{m-1}n+3m\\
x_m-1&=m(m-1)+3m\\
x_m&=m^2+2m+1\\
x_m&=(m+1)^2\\
x_n&=(n+1)^2\qquad \blacksquare
\end{align}$$
Check:
$$x_{n+1}-x_n=(n+2)^2-(n+1)^2=(2n+3)(1)=2n+3$$
