Operational determinants in ODE

[cue928]

I am to find the general solution of the following two equations, using operator notation:
x''-3y'-2x=0
y''+3x'-2y=0
The book suggests starting out with:
(D^2 -2)x - 3Dy = 0
3Dx+(D^2 - 2)y = 0 but for the life of me, I do not see how they got this from the first two equations.

 

[HallsofIvy]

D represents the first derivative (with what ever the independent variable is in this problem) and $D^2$ is the second derivative.
The first equation is x''- 3y'- 2x= D^2x- 3Dy- 2x= D^2x- 2x- 3y= (D^2-2)x- 3y= 0.

The second equation is y''+ 3X'- 2y= D^2y+ 3Dx- 2y= D^2y- 2y+ 3Dx= (D^2- 2)y+ 3Dx= 0.

(You titled this "Operational determinants in ODE. Were you thinking that "D" was the determinant of something?)

 

[cue928]

No, not really. To be honest, I wasn't sure what it meant. Let me try running through the problem with that piece of insight, which I appreciate, and see if it makes any sense.