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Is there a way to solve for example:

Where $D$ is the differential operator.

$$(D-4)x + D^2y = 0$$

$$(D+1)x + Dy = 0$$

Without the operator

$$x' - 4x + y'' = 0$$ $$x' + 1x + y' = 0$$

Using a matrix and then row reducing into echelon form?

I'm not sure how to set this up and solve it, to get the coefficients of the form $AD^2+BD+C$.

$(0D^2+1D-4)x + (D^2+0D+0)y = 0$

$(0D^2+1D+1)x + (0D^2+1D+0)y = 0$

rref([$0D^2+1D-4,D^2+0D+0,0][0D^2+1D+1,0D^2+1D+0,0$]) ?

I know this is a complex problem and appreciate any help!

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request for clarification: does $D = \frac{d}{dt}$, that is, differentiation with respect to some not-explicitly-presented independent variable? – Robert Lewis Jul 19 '13 at 22:54
Welcome to MSE! It is difficult to read some of your math. You can learn how to format math on this website here. – Daryl Jul 19 '13 at 23:02
@BobR: Can you please write your two sets of equations without the using the differential operator ($x', y''...$)? If I interpret your equations correctly, there is a way to solve them. Regards – Amzoti Jul 19 '13 at 23:40
@Amzoti I think that's what it is supposed to be without the operator. – Bob R Jul 20 '13 at 6:28
@RobertLewis Yes, sorry Im not good at formatting $$D = d/dt$$ $$D^2 = d^2/dt^2$$ – Bob R Jul 20 '13 at 6:28
up vote 0 down vote accepted

To do it the way OP wants to, I think we have to introduce a new variable, call it $v$, defined by $v=y'$. Then we have $$\eqalign{x'+v'&=4x\cr x'&=-x-v\cr y'&=v\cr}$$ which we can write as $$\eqalign{v'&=5x+v\cr x'&=-x-v\cr y'&=v\cr}$$ or in matrix form $z'=Az$ where $z=(v,x,y)$ and $$A=\pmatrix{1&5&0\cr-1&-1&0\cr1&0&0\cr}$$ and now you're on your way.

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