Systems of linear differential equations by elimination?

[zack7]

The original form:
http://img190.imageshack.us/i/79624585.png/

http://img685.imageshack.us/i/70092098.png/

I have manged to get to

(D + 1)x + (D² +D +1)z = 0

but after that I am kind of lost.

 

[tiny-tim]

hi zack7!

(try using the X² icon just above the Reply box )

what is D ? is it d/dt ?

can't you separate the variables, and solve for (D + 1)x = 0 and (D² +D +1)z = 0 separately?

 

[zack7]

hi zack7!

(try using the X² icon just above the Reply box )

what is D ? is it d/dt ?

can't you separate the variables, and solve for (D + 1)x = 0 and (D² +D +1)z = 0 separately?

I can't seem to find a way to eliminate z ?

 

[HallsofIvy]

The equations are
\frac{dx}{dt}= -x+ z
\frac{dy}{dt}= -y+ z
\frac{dz}{dt}= -x+ y
or, letting "D" represent differentiation with respect to t,
Dx= -x+ z
Dy= -y+ z
Dz= -x+ y

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, D^2x= -Dx+ Dz.

From the first and third equation, that is the same as D^2x= -x+ z- x+ y= -2x+z+ y. But, also from the first equation, z= Dx+ x so we can write that as D^2x= -2x+ Dx+ x+ y or D^2x- Dx= -x+ y. Differentiating again, D^3x- D^2x= -Dx+ Dy.

From the second equation, Dy= -y+ z so we can write that as D^3x- D^2x+ Dx= -y+ z. Again, z= Dx+ x so that gives D^3x- D^2x+ Dx= -y+ Dx+ x or D^3x- D^2x- x= y.

Finally, from the third equation, y= Dz- x and, since z= Dx+ x, Dz= D^2x+ Dx, y= D^2x+ Dx- x so that D^3x- D^2x- x= D^2x+ Dx- x or D^3x- 2D^2x- Dx= (D^3+ 2D^2- D)x= 0.

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
z= Dx+ x and then y= Dz+ x.

Frankly, that is not the way I would have done this problem. I would have written it as the matrix equation
\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}
and found the eigenvalues and eigenvectors of the coefficient matrix.