What's wrong in this method of solving a difference equation?

Question

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?

Answer 1

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.

Answer 2

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 $n$).

Answer 3

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.}

Answer 4

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 $-a$.

Answer 5

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 $\,y_n=\ldots\,$