Convert vector differential equation's order

Here is a 2nd order vector differential equation: $$\overrightarrow{Y}''= \begin{pmatrix}a & b \\c & d \end{pmatrix} \overrightarrow{Y}$$ Don't work it out, but write it as a vector differential eqn in $1$st order in higher dimensions.

I am not sure where to begin.

How does one convert from 2nd to 1st order? Hints are appreciated.

Here's a simple example. If you have the second order differential equation $$y'' =y$$ you can write it as the pair of first order differential equations $$u' = y \\ y' = u$$ which is the vector diff-eq $$\vec w' = A\vec w$$ where $\vec w=(u,y)^T$ and $$A = \begin{pmatrix}0&1\\1&0\end{pmatrix}$$
• I am still unable to proceed. I dont see how $$u' = y$$ and $$y' = u$$, that doesnt seems to make sense? – Aditya Kalra Jul 7 '17 at 3:25
• @AdityaKalra Differentiate the second equation and use the result to eliminate $u'$ from the first. It makes sense... $u$ is just a new function that only exists to couple to the first one. It's like saying the equation $x=31$ can be rewritten as the system $\{x =y+1, y = 30\}$ – spaceisdarkgreen Jul 7 '17 at 3:28
• What you are saying is $$y'' = y$$ ?? how is that true? Ie if I differnetiate then substitute? – Aditya Kalra Jul 7 '17 at 23:24
• Well... yes, as I said, differentiate and substitute. I'm claiming the equation $y''=y$ is implied by the system of equations (1) $u'=y$ and (2) y'=u. The derivative of eq (2) is $y'' = u'.$ And then plugging that in for $u'$ in the LHS of eq (1) gives $y'' = y.$ – spaceisdarkgreen Jul 7 '17 at 23:31
Let $$\vec{x} = \begin{pmatrix} x_1 \\ x_2 \\ \end{pmatrix} = \begin{pmatrix} y_1' \\ y_2' \\ \end{pmatrix} = \vec{y}'.$$ Then $$\vec{x}' = \begin{pmatrix} x_1' \\ x_2' \\ \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \begin{pmatrix} y_1\\ y_2 \\ \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \vec{y}.$$ In coordinates, $$\begin{pmatrix} \vec{x}' \\ \vec{y}' \\ \end{pmatrix} = \begin{pmatrix} x_1' \\ x_2' \\ y_1' \\ y_2' \\ \end{pmatrix} = \begin{pmatrix} 0 & 0 & a & b \\ 0 & 0 & c & d \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ y_1 \\ y_2 \\ \end{pmatrix} = \begin{pmatrix} \vec{x} \\ \vec{y} \\ \end{pmatrix},$$ or $$\begin{pmatrix} \vec{x}' \\ \vec{y}' \\ \end{pmatrix} = \begin{pmatrix} 0 & A \\ I & 0 \\ \end{pmatrix} \begin{pmatrix} \vec{x} \\ \vec{y} \\ \end{pmatrix},$$ where $$A= \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \mbox{ and } I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}.$$