Originally Posted by

**HallsofIvy** Yes, "D" is the derivative operator. But it is common notation in linear differential equations with constant coefficients to write, say, $\displaystyle \frac{d^2y}{dt^2}+ 5\frac{dy}{dt}+ 7y= 0$ as $\displaystyle (D^2+ 5D+ 7)y= 0$. As long as the coefficients are constant, the "$\displaystyle D^2+ 5D+ 7$" can be treated like a polynomial (if the coefficients are not constant, you run into problems with commutativity).

Slapmaxwell1, that is not what I get. Differentiating the second equation, $\displaystyle (D^2- 9D)y- 2Dx= 0$. From the first equation, $\displaystyle Dx= 5x- 7y$ so $\displaystyle (D^2- 9D)y- 2(5x- 7y)= (D^2- 9D+ 14)y- 10x= 0$. From the second equation, [tex]2x= (D- 9)y[tex] so [tex]-10x= (-5D+ 45)y. Then $\displaystyle (D^2- 9D+ 14)y- 10x= (D^2- 9D+ 14)y+ (-5D+ 45)y= (D^2- 14D+ 59)y= 0$. You appear to have dropped the "14y".

$\displaystyle D^2- 14D+ 59$ is not going to factor easily but the equation $\displaystyle D^2- 14D+ 59= 0$ can be solved by completing the square or the quadratic formula.