# Solution to Matrix Differential Equation

1. Oct 11, 2008

### WolfOfTheSteps

How do I write the form of the solution to this equation:

$$\dot{\vec{x}}(t) = \left [ \begin{array}{cc} a_{11}(t) & a_{12}(t) \\ a_{21}(t) & a_{22}(t) \end{array} \right ] \vec{x}(t)$$

I just need to be able to write x1(t) and x2(t) so I can do the rest of the problem I'm working on. Getting this would just be a small step in my solution, but I am very rusty with my differential equations!! :(

Initially, I thought to write:

$$x_1(t) = \int_{t_0}^{t}x_1(\tau)a_{11}(\tau) + x_2(\tau)a_{12}(\tau)d\tau$$

$$x_2(t) = \int_{t_0}^{t}x_1(\tau)a_{21}(\tau) + x_2(\tau)a_{22}(\tau)d\tau$$

But that has the solutions with dependence on x1(t) and x2(t). That's not the way to write it, is it?

Thanks.

2. Oct 11, 2008

### Staff: Mentor

Let's look at your system of DEs this way:
$$\dot{\vec{x}}(t) = A \vec{x}(t)$$

What you'd like is a system that looks like this:
$$\dot{\vec{x}}(t) = D \vec{x}(t)$$
where D is a diagonal matrix.

This will untangle things so that you have x1'(t) = d11 x1(t) and x2'(t) = d22 x2(t).

These are easy to solve, since each one involves only a single variable.

Getting the matrix D is the hard part, though, since doing this involves changing to a different basis (for R2 in your case). Without going into too many details, you'll want to find a matrix D that is similar to your original matrix A, which I'm assuming is invertible. Similarity is precisely defined this way: If A ~ D, then for some invertible matrix P, AP = PD.

Equivalently, P$$^{-1}$$AP = P$$^{-1}$$PD = D.

You will need to come up with a matrix P whose columns are the new basis, and a matrix P$$^{-1}$$, the inverse of P.

To wind this up, the columns of P are the eigenvectors of matrix A, and it turns out that the diagonal entries of D are the eigenvalues of A.

I hope I've given you enough to at least get you started searching for the things to learn more about. Solving a system of DEs, even the simplest possible system in two variables requires a significant amount of understanding in linear algebra.

Mark

3. Oct 12, 2008

### WolfOfTheSteps

Yes, my linear algebra is a few years out of service, but it is not non-existent. You have refreshed my memory. Thanks.