Mary: You may consider your equation for $n+2$ instead of $n+1$; this gives us $$ a(n+2)=2a(n+1)-a(n)-1.$$ Subtract the first equation from this one. We obtain $$\begin{array}{rl}a(n+2)-a(n+1)&=(2a(n+1)-a(n)-1)-(2a(n)-a(n-1)-1)\\&=2a(n+1)-3a(n)+a(n-1),\end{array}$$ or $$a(n+2)=3a(n+1)-3a(n)+a(n-1).$$
In general, if your equation is "almost homogeneous" but for a constant, the same trick of subtracting consecutive instances of the recurrence gives us a homogeneous equation.
The associated characteristic equation is now $x^3-3x^2+3x-1=0$, or $(x-1)^3=0$. This means that $a(n)=A+Bn+Cn^2$ for some constants $A,B,C$.
In general, if the equation has the form $(x-r)^k(x-b)^s\dots=0$, then the general solution will have the form $$a(n)=(A+Bn+\dots+C n^{k-1})r^n+(D+En+\dots+F n^{s-1})b^n+\dots$$ (in the most studied case, the characteristic equation has no repeated roots, but as in your example here, repeated roots may occur, so we need this more general version).
To find $A,B,C$ in our equation for $a(n)=A+Bn+Cn^2$, we use the given initial conditions. Are you sure the condition $a(10)=0$ is correct, rather than $a(1)=0$?
With $a(1)=0$, we have $0=A+B+C$. Also, $a(0)=0$, so $A=0$. Finally, $a(2)=2a(1)-a(0)-1=-1$, using the original recurrence, so $A+2B+4C=-1$. This easily gives us $A=0$, $B=1/2$, and $C=-1/2$, or $$ a(n)=\frac{n-n^2}2. $$
Note that we needed to compute $a(2)$ before we could find $A,B,C$. This is because the original equation was not homogeneous, so even though it is of second order, we needed an additional initial condition to the two given to us. (Note the homogeneous equation we obtained is of third order, needing 3 initial conditions.)
Unfortunately, I do not know of a decent reference on recurrence relations at an elementary level (I know of one, in Spanish, a translation of a small Russian booklet published by Mir eds. years ago). There are several standard approaches which might not be as elementary as you may want, but you may enjoy looking at them anyway: Using linear algebra or generating functions. For the latter, a good reference is "generatingfunctionology",
http://www.math.upenn.edu/~wilf/DownldGF.html
and for the former a good reference is Volume II of Apostol's Calculus book.
Hmm... With $a(10)=0$, we can proceed as follows: We have $$a(n)=A+Bn+Cn^2$$ and so $$A=0$$ (since $a(0)=0$) and $$10B+100C=0$$ (since $a(10)=0$ and $A=0$). We also have $$a(n+2)=2a(n+1)-a(n)-1,$$ or
$$ B(n+2)+C(n+2)^2=2B(n+1)+2C(n+1)^2-Bn-Cn^2-1, $$ which simplifies to $2C=-1$, or $$C=-1/2.$$ Then $$B=5,$$ from the equation for $a(10)=0$, and $$ a(n)=5n-\frac{n^2}2. $$
Let
$$
b_n=a_n+2^{n+2} \quad \forall n\ge 1.
$$
Then
$$
b_1=a_1+2^3=9,\quad b_2=a_2+2^4=27,
$$
and for every $n\ge 3$ we have:
\begin{eqnarray}
b_n&=&4(b_{n-1}-2^{n+1})-3(b_{n-2}-2^n)+2^{n+2}+2^n\\
&=&4b_{n-1}-3b_{n-2}+(-8+3+4+1)2^n\\
&=&4b_{n-1}-3b_{n-2}.
\end{eqnarray}
The characteristic equation associated to the recurrence relation
$$
b_n-4b_{n-1}+3b_{n-2}=0
$$
is
$$
r^2-4r+3=0,
$$
and its solutions are
$$
r_1=1, \quad r_2=3.
$$
Therefore
$$
b_n=c_1+c_2\cdot3^n,
$$
where $c_1$ and $c_2$ satisfy:
$$
c_1+3c_2=9,\quad c_1+9c_2=27,
$$
i.e.
$$
c_1=0,\quad c_2=3.
$$
Hence
$$
a_n=b_n-2^{n+2}=3^{n+1}-2^{n+2}
$$
Best Answer
Let $x_{np}$ a particular solution for the recurrence relation. Since the recurrence relations is of the form $g(n)=a^n$, then $x_{np}=qa^n$ except if $a$ is solution for the characteristic equation with multiplicity $s$, in which case $x_{np}=qn^sa^n$.
It was your error: 2 is solution for the caracteristic equation, which is $x-2=0$, with multiplicity $s=1$. Then, the particular solution is of the form $x_{np}=qn2^n$.
Now you can continue.