Algebra of Linear differential operators, question on Commutativity and Association The following is a discussion on the following second differential equation
$$  \frac{dy^2}{dx} - y = 0 $$
So, let us introduce the following, convention and definition, represent the derivative operator by
$$ D = \frac{d}{dx} $$
So that our first equation can be represented as
$$ \left( D^2 - 1 \right) y(x) = 0 $$
My professor discussed on the idea, on "factoring" (The space on which the derivative acts?) the operators, his discussion lead to the apparent nonuniqueness of factoring as either
$$ (D - \tanh(x))(D + \tanh(x)) y = 0 \quad \textrm{and } \quad (D - 1)(D + 1) = 0 $$
My first question is that how is that  $ (D - \tanh(x))(D + \tanh(x))  ``=" (D^2 - 1) $ and how is unfoiled?, because I think is not commutative because
$$ f(x)\frac{d}{dx}y \neq \frac{d}{dx}f(x)y(x) $$
One is first derive something and multiply by $f$, and the other is derive the product.
Second, my professor move to the general
$$ y'' + a(x)y' + b(x) = 0 $$
or
$$ \big(D^2 + a(x)D + b(x)\big)y(x) = 0 $$
and he wants to factor the above as
$$ \big( D + A(x)\big)\big(D + B(x)\big)y(x) = 0 $$
So then he unfoils the above equation as:
$$ \big( D^2 + AD + AB + B' + BD \big)y = 0 $$
and he argues that the terms $B'$ and $BD$ comes from the fact that either we take the derivative of $B$ or take the derivative and multiply by B (Why that doesn't apply to the terms $ AD  $ and we should add $ A'$ ? 
Also, do you happen to know where to find reference on solving ODEs by this approach? Thanks
 A: You seem to have the right concerns, you just need to clarify what the definitions mean and work through them.
Operators do form an algebra, and indeed that algebra is non-commutative.  We can embed functions by making them into operators which multiply by that function.  For example, the function $f$ becomes the operator $g \mapsto (x \mapsto f(x)g(x))$.  The differential operator is $D = g \mapsto (x \mapsto g'(x))$.  The multiplication of the algebra of operators is composition of the operators.  In particular, $$Df = D \circ f = g \mapsto D(x \mapsto f(x)g(x)) = g \mapsto (x \mapsto f(x)g'(x) + f'(x)g(x))$$
meanwhile $$fD = f \circ D = g \mapsto (x \mapsto f(x)g'(x))$$
which leads to the operator equation $$Df = fD + f'$$ the most notable instance being when $f(x) = x$ and is usually notated $$Dx = xD + 1$$
Now let's work through that first example: $$
\begin{align}
(D + \tanh)(D - \tanh)  
&= D^2 + \tanh D - D\tanh - \tanh^2 \\
&= D^2 + \tanh D - \tanh D - \text{sech}^2 - \tanh^2 \\
&= D^2 - (\text{sech}^2 + \tanh^2) \\
&= D^2 - 1
\end{align}$$
You should be able to apply the above logic to the general case.  The rules we are using beyond the ones I've explicated are the normal rules of an algebra, i.e. addition is commutative and multiplication distributes but is not, in general, commutative.  The particular algebra being used is an operator algebra and specifically the operational calculus pioneered by Heaviside.  There are nice connections between this approach and the theory of formal power series leading to the umbral calculus which is applied in more discrete/combinatorial branches of mathematics.
