Operational determinants in ODE

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The discussion focuses on solving a system of ordinary differential equations (ODEs) represented in operator notation. The equations x'' - 3y' - 2x = 0 and y'' + 3x' - 2y = 0 are transformed into operator form as (D^2 - 2)x - 3Dy = 0 and 3Dx + (D^2 - 2)y = 0. The variable D denotes the derivative operator, where D^2 represents the second derivative. Participants clarify the transition from the original equations to the operator notation and explore the implications of the operator D.

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  • Understanding of ordinary differential equations (ODEs)
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  • Knowledge of derivative concepts, specifically first and second derivatives
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cue928
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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.
 
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D represents the first derivative (with what ever the independent variable is in this problem) and [itex]D^2[/itex] 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?)
 
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.
 

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