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EinsteinCross

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##

A = \begin{bmatrix}

a & b\\

c & d

\end{bmatrix}

## and ##

\mathbf{x} = \begin{bmatrix}

x(t)\\ y(t))

\end{bmatrix}

##

Now to solve this equation we transform it into a 4x4 first order system ##\mathbf{X'} = M\mathbf{X} ## where

##

M = \begin{bmatrix}

0 & 1 & 0 & 0 \\

a & 0 & b & 0\\

0 & 0 & 0 & 1 \\

c & 0 & d & 0

\end{bmatrix}## and ##

\mathbf{X} = \begin{bmatrix}

x(t)\\ x'(t)

\\ y(t)

\\ y'(t)

\end{bmatrix}

##

Now solving the first order system by calculating the eigenvalues and corresponding eigenvectors gives us the fundamental matrix for the 4x4 system

##

\Phi (t) = [[\xi_{1}],[\xi_{2}],[\xi_{3}],[\xi_{4}]]

\begin{bmatrix}

e^{\lambda_{1}t }

\\ e^{\lambda_{2}t }

\\ e^{\lambda_{3}t }

\\ e^{\lambda_{4}t }

\end{bmatrix}^{T} ## where the xi brackets are the eigenvectors in column form.

But my question is this: What I am looking to do is to find the fundamental matrix of the original second order 2x2 system which is represented as ##

\Psi (t) = \begin{bmatrix}

\psi_{1}(t) & \psi_{2}(t)

\\ \psi_{3}(t)

& \psi_{4}(t)

\end{bmatrix} ## such that ##\Psi ''(t) = A\Psi (t)##. So how does one

*extract*the solution(s) to the original 2x2 system from the fundamental matrix of the first order 4x4 system?