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I was looking here and it said $$\frac{d^2y}{dx^2}-2\frac{dy}{dx}-3y=0$$ $$D^2-2D-3=0$$ $$(D-3)(D+1)=0$$ $$\therefore D=3 \text{ or } D=-1$$What does $D = 3$ mean?

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    Try $y=e^{Dx}$. – 2017-01-25
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    Where in your reference does it say "$D=3$ or $D=-1$"? – 2017-01-25
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    @Hurkyl when you go to the place that says "The Use Of The D Operator To Find The Complementary Function For Linear Equations" and click "2". – 2017-01-26

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$D$ and $3$ are both operators that act on the solution space to this equation. $D$ acts by sending $u \mapsto \frac{\mathrm{d}u}{\mathrm{d}x}$. $3$ acts by sending $u \mapsto 3u$.

$D=3$ is the assertion that they are the same operator. The truth set of this equation is precisely those $u$ such that $3u = \frac{\mathrm{d}u}{\mathrm{d}x}$.

The disjunction "$D=3$ or $D=-1$" means1 that the truth sets of $D=3$ and $D=-1$ jointly cover the entire solution space to the original equation. Or equivalently, for each particular solution $u$, at least one of $Du=3u$ and $Du = (-1)u$ is true.

The conclusion "$D=3$ or $D=-1$", as it turns out, is wrong. The space of operators has zero divisors; one cannot conclude from $ST=0$ that $S=0$ or $T=0$. And in truth, very few of the solutions to the original equation satisfy that statement.

It turns out, however, that a basis for the solution space can be obtained by taking one solution from the truth set of $D=3$ and one solution from the truth set of $D=-1$.

1: It could also be the claim that at least one of the statements "$D=3$ is true everywhere" and "$D=-1$ is true everywhere". You have to infer from context which interpretation is meant.

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    To the OP: We can conclude from $(D-3)(D+1)f=0$ that if $g=(D+1)f$ then $(D-3)g=0.$ That is we have $g'=3g.$ So $f'(x)+f(x)=g(x)=Ae^{3x}$ for some constant $A.$ – 2017-01-25
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$D=\frac{d}{dx}$

You have $(D-3)(D+1)y=0$

Either $\frac{dy}{dx}=3y$ or $\frac{dy}{dx}=-y$

You can go from there...

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    That is not completely true, these two equations provide solutions, but not all solutions. Use it to form a first order system $(D-3)v=0$, $v=(D-1)y$. – 2017-01-25
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    Or add the step $y=\frac12((D-1)y-(D-3)y=u+v$ to get a decomposition into solutions of $(D-3)u=0$ and $(D-1)v=0$. – 2017-01-25
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    @LutzL You are right. This is not complete. – 2017-01-25
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If you don't eliminate $y$ too fast in the second equation, you may have seen $D^2y-2Dy-3y = 0$. Eventually you'll get Dy = 3y which means y' = 3y.

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By an operator we mean a transformation that transforms a function into another function. Let $D $ denote the differentiation with respect to $x $, that is, $D \equiv \frac {d}{dx} $, and we write $$Df (x) =Df = f' =\frac {df}{dx} $$ Thus $D $ transforms $f (x)$ into its derivative $f'(x) $. For example $D (x^3)=3x^2, D (\sin x)=\cos x $ and so on.

So, the homogenous linear DE $a_0\frac {d^2y}{dx^2}+ a_1\frac{dy}{dx} + a_2$ can be transformed to $a_0D^2y + a_1Dy + a_2=0 =F (D)y $. When all $a_i$'s $i $ from $0$ to $n $ are constants, $F (D) $ can be factorised, that is why, we have $D^2-2D-3 =(D+1)(D-3) $. Hope it helps.