What's wrong in this method of solving a difference equation? Consider:
$$y_{n+1} = 2y_n + 1$$
To solve this I think I need to find "any" one particular solution and add it to a homogeneous solution.
A homogeneous solution is $2^ny_0$

For a particular solution, if I substitute $y_n = an$, $$a(n+1) = 2an + 1
> \implies a = \dfrac{1}{1-n}$$
This gives the complete solution as $$y_n = \color{red}{\dfrac{n}{1-n}}+2^ny_0$$

However for a particular solution, if I substitute $y_n=b$, I get
$$b=2b+1 \implies b=-1$$
This gives the complete solution as $$y_n=\color{red}{-1}+2^ny_0$$
These two solutions seem to be very different. I don't see where I've made an error.
Any particular solution will work in the complete solution, right?
If so, why the the two particular solutions above gave seemingly different general solutions?
 A: Note that $a=1/(1-n)$ is not constant so there is no particular solution of the form $y_n=an$. On the other hand there is one of the form $y_n=b$ with $b=-1$ (which does not depend on $n$).
A: I think where you really went wrong is this:
You wrote that, if $y_n = an$, then $a = \frac{1}{1 - n}$. But the assumption $y_n = an = \frac{n}{1-n}$ is false to begin with, so the fact that you derived something from a false assumption means nothing.
Now why is the assumption false?
Well1, according to your assumption $y_n = an$ we have $$y_n = \frac{n}{1-n}$$ right?
So plug that into the original equation. Does it work?
$$\frac{(n+1)}{1-(n+1)} = 2 \frac{n}{1-n} + 1$$
Remember, this is implied by your assumption. But this only holds for $n = -1$.
In other words, it won't work for any other $n$ than $-1$... neither $-2$, nor $0$, nor $1$, etc...  
So your assumption that $y_n=an$ holds for all $n$ contradicts itself, hence it cannot be true.
1 Someone else contended that you derived this incorrectly too, but that's a math error separate from what I'm trying to show, which is the mistake in your reasoning. I just assume you did the math right and show where the logic went wrong.
A: 
Homogeneous solution is $2^ny_0$  

No, the general homogeneous solution is $\,C \cdot 2^n\,$ for some constant $\,C\,$.

This gives the complete solution as $\quad y_n=\color{red}{-1}+2^ny_0$

Except this doesn't work if you try it for $\,n=0\,$ or $\,n=1\,$. See note above why.
Instead, you should get that $\,y_n = -1 + C \cdot 2^n\,$, where the constant $\,C\,$ is determined from the initial condition $\,y_0 = -1 + C \cdot 2^0 \iff C = y_0+1\,$

To solve this I think I need find "any" one particular soln and add it to homogeneous solution.

Alternatively, you can solve it directly by rewriting the recurrence as $\,y_{n+1}+1=2\left(y_n+1\right)\,$. It follows that $\,y_n+1\,$ is a geometric progression, and therefore $\,y_n+1 = 2^n(y_0+1)\,$, so $\,y_n=\ldots\,$
A: There's no particular solution of the form $y_n = an$, since, assuming $a$ is constant, you found that $a$ must satisfy $a=1/(1-n)$, contrary to the assumption that $a$ is constant.
A: The problem is that $a$ is a function of n:
$$y_n = a_nn$$
$$a_{n+1}(n+1) = 2a_nn + 1$$
$$a_{n+1}n+a_{n+1} = 2a_nn + 1$$
$$a_{n+1}n-2a_nn+a_{n+1}=1$$
$$(a_{n+1}-2a_n)n+a_{n+1}=1$$
From here, you can't combine $a_{n+1}-2a_n$ into just $-a$.
