# Systems of DEs

1. Jul 26, 2005

### EvLer

I have this system of DEs, but after getting through all the simplifications I can't factor the auxiliary equation:

x1' = -2x1 + x2 + x3
x2' = x1 - x2 + 3x3
x3' = -x2 - 3x3

Then I eliminated all variables except for x2, using Differentiation operator.
my AE from x2 is this:

r3 + 6r2 + 12r + 12 = 0

And I'm stuck!

2. Jul 26, 2005

### George Jones

Staff Emeritus
For what it's worth, this factors into

$$\left( r + a \right) \left(r^2 + \left( 6 - a \right) r + \frac{12}{a} \right) = 0,$$

with

$$a = 2 + 2^{\frac{2}{3}}$$

Regards,
George

3. Jul 26, 2005

### qbert

I get a different equation for x2.

I get: -(D+2)(D^2 + 4D + 5)x2 = 0
or (D^3 + 6D^2 + 13D + 10)x2 = 0
the aux has roots: -2, -2 +i, -2 - i.

We can check that the eigenvalues of
$$\left( \begin{array}{ccc} -2 & 1 & 1 \\ 1 & -1 & 3 \\ 0 & -1 & -3 \end{array} \right)$$
are -2, -2+i, -2-i

so the solutions should be linear combinations
of exp(-2t), exp(-2t)*sin(t), exp(-2t)*cos(t).

4. Jul 27, 2005

### EvLer

How did you get that equation?
I tried eliminating x1 first, after that I have 2 equations:

x2 + (D + 3)x3 = 0
(D + 1)(D + 2)x2 - x2 - x3 - 3(D + 2)x3 = 0

so from here

x2 + (D + 3)x3 = 0
[(D + 1)(D + 2) - 1]x2 - [3(D + 2) + 1]x3 = 0

thanks.

Last edited: Jul 27, 2005
5. Jul 27, 2005

### qbert

starting here
x2 + (D + 3)x3 = 0
[(D + 1)(D + 2) - 1]x2 - [3(D + 2) + 1]x3 = 0

therefore
[3(D+2)+1]x2 + (D+3)[3(D+2)+1]x3 = 0
(D+3)[(D + 1)(D + 2) - 1]x2 - (D+3)[3(D + 2) + 1]x3 = 0

and some mindless algebra gives:
[3(D+2)+1 + (D+3)[(D+1)(D+2) -1]] x2 = 0;
[3D + 6+1 + (D+3)[D^2 + 3D + 1]]x2 = 0
[3D +7 + D^3 + 3D^2 + D + 3D^2 + 9D + 3]x2 = 0
[D^3 + 6D^2 + 13D + 10]x2 = 0

Last edited: Jul 27, 2005
6. Jul 27, 2005

### EvLer

argh.... Ok, thanks much! I guess I just expected a nice and simple solution...