In this website, http://www.codecogs.com/library/maths/calculus/differential/the-d-operator.php, the following problem and solution is given:
Problem:
Find the Particular Integral of $$\frac{d^2y}{dx^2} - 5\,\frac{dy}{dx} + 6y = \sin\,2x$$
Solution:
This can be re-written as $$\tag1 y = \frac{1}{D^2 - 5D + 6}\sin\,2x $$
Using equation 1 we can put $D^2 = -4$
Therefore: $$y = \frac{1}{-\,4\;-\,5D + 6}\sin\,2x =\frac{1}{2 - 5D}\sin\,2x$$
If we multiply the top and bottom of this equation by 2 + 5D
$$y =\frac{2 + 5D}{4 - 25D^2}\sin\,2x$$
But $D^2=-4$
Therefore $$ y =\frac{2 + 5D}{104}\sin\,2x = \frac{1}{104}\left(2\sin\,2x + 5D\sin\,2x \right)$$ But since $D\sin\,2x = 2\cos\,2x$ $$y = \frac{1}{104}\left(2\sin\,2x + 10\cos\,2x \right) $$
The biggest problem I have with this is the line, $D^2 = -4$.
So far as I understand, $D$ is an operator on functions, and $D^2$ represents the double application of that operator. So how, can it be equal to -4? How can an operator on functions be equal to an integer? They're two things of fundamentally different types, how can they be equal?
You can interpret $D^2=-4$ as $D^2=-4I$, where $I$ is the identity operator; that's no big deal. If anything, you should be more worried about "dividing" by an operator.
To expand a bit about what's going on in that solution, the author knows that the solution will be a linear combination of $\sin(2x)$ and $\cos(2x)$. For those two functions (and their linear combinations), $D^2=-4I$.
Note that guessing all those manipulations is harder than postulating that the particular solution is of the form $\alpha\sin(2x)+\beta\cos(2x)$ and substitute in the equation to find $\alpha$ and $\beta$ (what is usually known as Undetermined Coefficients).