# Thread: solve the system through systematic elimination...

1. ## solve the system through systematic elimination...

ok im not sure if im doing this right, my numbers are getting a little funky i was wondering if someone could check my work and let me know if im doing it right. this is the problem

$\large \begin{Bmatrix} Dx-5x+7y=0 & \\ Dy-2x-9y=0 \end{Bmatrix}$

i solved it for y by eliminating the x

i got the following...

$\large y(14+D^{2}-14D+45)=0$

im not done with the problem, i jus wanted to know if i was right up to the point of where I am. thanks in advance.

2. D is operator? If the problem is like that you shouldn't separate from variable!

3. Yes, "D" is the derivative operator. But it is common notation in linear differential equations with constant coefficients to write, say, $\frac{d^2y}{dt^2}+ 5\frac{dy}{dt}+ 7y= 0$ as $(D^2+ 5D+ 7)y= 0$. As long as the coefficients are constant, the " $D^2+ 5D+ 7$" 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, $(D^2- 9D)y- 2Dx= 0$. From the first equation, $Dx= 5x- 7y$ so $(D^2- 9D)y- 2(5x- 7y)= (D^2- 9D+ 14)y- 10x= 0$. From the second equation, [tex]2x= (D- 9)y[tex] so [tex]-10x= (-5D+ 45)y. Then $(D^2- 9D+ 14)y- 10x= (D^2- 9D+ 14)y+ (-5D+ 45)y= (D^2- 14D+ 59)y= 0$. You appear to have dropped the "14y".

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

4. I didn't see. You are right in your world. But in my, completely wrong. You didn't finish a problem (in your world). I will get solutions always before you in my world.

5. ok i will rework the problem thanks alot for your help!

6. 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, $\frac{d^2y}{dt^2}+ 5\frac{dy}{dt}+ 7y= 0$ as $(D^2+ 5D+ 7)y= 0$. As long as the coefficients are constant, the " $D^2+ 5D+ 7$" 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, $(D^2- 9D)y- 2Dx= 0$. From the first equation, $Dx= 5x- 7y$ so $(D^2- 9D)y- 2(5x- 7y)= (D^2- 9D+ 14)y- 10x= 0$. From the second equation, [tex]2x= (D- 9)y[tex] so [tex]-10x= (-5D+ 45)y. Then $(D^2- 9D+ 14)y- 10x= (D^2- 9D+ 14)y+ (-5D+ 45)y= (D^2- 14D+ 59)y= 0$. You appear to have dropped the "14y".

$D^2- 14D+ 59$ is not going to factor easily but the equation $D^2- 14D+ 59= 0$ can be solved by completing the square or the quadratic formula.
thanks so much for your help, i really appreciate it, i will rework the problem with the suggested correction.

7. Regards