# Linear systems question

1. Oct 21, 2007

### gaganaut

Hi all,
I have a system of linear DE 's as under.

$$\dot{x}_1(t)=-tx_1+x_2$$

and

$$\dot{x}_2(t)=-tx_2+x_1$$

Now how do we find the solution

$$x(t)^{T}=[x_1, x_2]$$

I tried to find a similar post but could not. Any help would be highly appreciated. I am in the midst of writing a code wherein I need to solve this system of linear DE's.

Gaganaut.

2. Oct 21, 2007

### HallsofIvy

Staff Emeritus
I can think of about a dozen different ways to do that problem. Unfortunately, since you have not shown how you would attempt such a probem, I have no idea which of them is appropriate for you. Do you see my dilemma?

3. Oct 21, 2007

### Manchot

The most straightforward way seems to be that you could subtract the first from the second to obtain one (separable) equation for their difference.

4. Oct 21, 2007

### gaganaut

Linear differential equations: Solution so far

First of all thanks for getting back. So far I have brought the system in a matrix form as under.

$$$\left[ \begin{array}{c} \dot{x}_1\\ \dot{x}_2 \end{array} \right]=\left[ \begin{array}{cc} -t & 1\\ 1 & -t \end{array} \right] \[ \left[ \begin{array}{c} x_1\\ x_2 \end{array} \right]$$$

I am a bit skeptical about the further steps that I did and that's when I decided to get help on this. I have written a formula for integrating the system as under.

$$\underline{x}(t) = exp\left(\int_{t_0}^t A(\tau)\,d\tau \right)\underline{x}_0(t)$$

where $$A(\tau)=\left[ \begin{array}{cc} -\tau & 1\\ 1 & -\tau \end{array} \right]$$

I cannot get any further. I definitely want to take the matrix approach as it is easier to code in for me. Also, I might be wrong with the integral method, so I would appreciate a better method preserving the matrix and vector form.

Thank you.

Gaganaut

Last edited: Oct 21, 2007