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Math Help - Non-autonomous system of ODE's

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    Non-autonomous system of ODE's

    \frac{dx}{dt} = ay - bx + f(t)

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

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

    Is anyone familiar with a possible method to solve such a system (or some kind of perturbation to approx the solution)?
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    Re: Non-autonomous system of ODE's

    Quote Originally Posted by wsldam View Post
    \frac{dx}{dt} = ay - bx + f(t)

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

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

    \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

    \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

    PAP^{-1}=J

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

    This gives

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

    \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

    \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 \hat{y} and sub that into the first equation
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