# Non-autonomous system of ODE's

• Nov 3rd 2012, 01:07 PM
wsldam
Non-autonomous system of ODE's
$\displaystyle \frac{dx}{dt} = ay - bx + f(t)$

$\displaystyle \frac{dy}{dt} = cx - dy + g(t)$

$\displaystyle a,b,c,d \in \mathbb{R}$ and $\displaystyle f(t),g(t)$ are continuous functions of $\displaystyle t$

Is anyone familiar with a possible method to solve such a system (or some kind of perturbation to approx the solution)?
• Nov 3rd 2012, 02:07 PM
TheEmptySet
Re: Non-autonomous system of ODE's
Quote:

Originally Posted by wsldam
$\displaystyle \frac{dx}{dt} = ay - bx + f(t)$

$\displaystyle \frac{dy}{dt} = cx - dy + g(t)$

$\displaystyle a,b,c,d \in \mathbb{R}$ and $\displaystyle f(t),g(t)$ are continuous functions of $\displaystyle t$

Is anyone familiar with a possible method to solve such a system (or some kind of perturbation to approx the solution)?

The system can be written in Matrix form as

$\displaystyle \frac{d}{dt}\begin{bmatrix} x \\ y \\ \end{bmatrix}=\begin{bmatrix} -b & a \\ c & -d \\\end{bmatrix}\begin{bmatrix} x \\ y \\ \end{bmatrix}+\begin{bmatrix} f(t) \\ g(t) \\\end{bmatrix}$

So as a matrix equation this gives

$\displaystyle \dot{Y}=AY+\vec{F}(t)$

From here try to Diagonalize the matrix, if that is not possible use the Jordan canonical form. So now we have a matrix that gives

$\displaystyle PAP^{-1}=J$

So make the substitution $\displaystyle Y=P^{-1}X$

This gives

$\displaystyle P^{-1}\dot{X}=AP^{-1}X+\vec{F}(t)$ and now multiply by P to get

$\displaystyle \dot{X}=PAP^{-1}X+P\vec{F}(t)$

If the matrix was diagonalizable, this will uncouple the linear system into two ODEs. If you could only find JCF then the system will look like

$\displaystyle \frac{d}{dt}\begin{bmatrix} \hat{x} \\ \hat{y} \\ \end{bmatrix}=\begin{bmatrix} \lambda & 1 \\ 0 & \lambda \\\end{bmatrix}\begin{bmatrix} \hat{x} \\ \hat{y} \\ \end{bmatrix}+P\vec{F}(t)$

So you have solve for $\displaystyle \hat{y}$ and sub that into the first equation