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!
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.