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I've been asked to solve the following differential equation:


I know how to solve it taking the trail solution $y=e^{mx}.$ But is the following approach using $D= \dfrac{d}{dx}$ valid?:


So the solution is $y=Ae^{2x}+Be^{-3x}$

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Not correct. You set $y = e^{rx}$ and then, solve for $r$, which gives the solutions for this problem. – NasuSama Jan 13 '14 at 3:15
This is sometimes used, though there is a bit more to this method, for a simple illustration, check… and – Macavity Jan 13 '14 at 3:19
You solution is correct. See a related technique. – Mhenni Benghorbal Jan 13 '14 at 3:49

Your approach as @Mhenni and @Nasu noted is correct. If we set $D=\frac{dy}{dx}$ so $D^2=\frac{d^2y}{dx^2}$ and so we have the following $D$ operation:

$$P(D)y=(D^2+D-6)y=D^2y+Dy-6y=y''+y'-6y=0$$ Moreover $P(D)=(D-2)(D+3)$. Setting $(D+3)y=u$, we get $(D-2)u=0$ or $u'=2u$ or $u=Ae^{2x}$. We already set $u=y'+3y$, so we have $$y'+3y=Ae^{2x}$$ Try to solve this simple one by yourself.

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