The equations are
[tex]\frac{dx}{dt}= -x+ z[/tex]
[tex]\frac{dy}{dt}= -y+ z[/tex]
[tex]\frac{dz}{dt}= -x+ y[/tex]
or, letting "D" represent differentiation with respect to t,
[tex]Dx= -x+ z[/tex]
[tex]Dy= -y+ z[/tex]
[tex]Dz= -x+ y[/tex]
Since those are three first order equations in three dependent variables, they will reduce to a single third order equation in one dependent variable. Differentiating the first equation again, [itex]D^2x= -Dx+ Dz[/itex].
From the first and third equation, that is the same as [itex]D^2x= -x+ z- x+ y= -2x+z+ y[/itex]. But, also from the first equation, z= Dx+ x so we can write that as [itex]D^2x= -2x+ Dx+ x+ y[/itex] or [itex]D^2x- Dx= -x+ y[/itex]. Differentiating again, [itex]D^3x- D^2x= -Dx+ Dy[/itex].
From the second equation, [itex]Dy= -y+ z[/itex] so we can write that as [itex]D^3x- D^2x+ Dx= -y+ z[/itex]. Again, [itex]z= Dx+ x[/itex] so that gives [itex]D^3x- D^2x+ Dx= -y+ Dx+ x[/itex] or [itex]D^3x- D^2x- x= y[/itex].
Finally, from the third equation, [itex]y= Dz- x[/itex] and, since [itex]z= Dx+ x[/itex], [itex]Dz= D^2x+ Dx[/itex], [itex]y= D^2x+ Dx- x[/itex] so that [itex]D^3x- D^2x- x= D^2x+ Dx- x[/itex] or [itex]D^3x- 2D^2x- Dx= (D^3+ 2D^2- D)x= 0[/itex].
Note that solving that equation will give you x(t) already involving 3 undetermined constants. In order not to introduce any other constants, get y(t) and z(t) from
[itex]z= Dx+ x[/itex] and then [itex]y= Dz+ x[/itex].
Frankly, that is not the way I would have done this problem. I would have written it as the matrix equation
[tex]\frac{\begin{pmatrix}x(t) \\ y(t)\\ z(t)\end{pmatrix}}{dt}= \begin{pmatrix}-1 & 0 & 1 \\ 0 & -1 & 1 \\ -1 & 1 & 0 \end{pmatrix}\begin{pmatrix}x(t) \\ y(t) \\ z(t)\end{pmatrix}[/tex]
and found the eigenvalues and eigenvectors of the coefficient matrix.
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