How to transform a differential equation to a system of differential equations Lets say I have a differential equation like $$y''+y+4=0$$ and I have to convert it to a system of first order equations? How is that done. 
I am interested in the method (and an explanation of it) rather than the answer.
 A: The general procedure is to replace the derivatives of $y$ with new symbols - so in this case, we might set
$$u = y, v = y'$$
Note that these are automatically related via $u' = v$. Now consider a differential equation such as $y'' - 12 y= 0$. Rewriting this as
$$(y')' - 12y = 0$$
and expressing this in the new symbols, we have
$$v' - 12 u = 0 \implies v' = 12 u$$
So for this ODE, the system could be
$$\left\{\begin{array}{l} u' &= v \\ v' &= 12 u \end{array}\right.$$

For a higher order equation, like $y^{(n)} = ...$, make a total of $n + 1$ substitutions, e.g. $u = y, u_1 = y', u_2 = y''$, and so on.
A: The offending term (that makes it higher than first order), is $y''$, so convert it to first order. How? Let $z = y'$, then $z' = y''$. It's as simple as that. So now the single second order equation $y'' + y + 4 = 0$ becomes the system of two first-order equations
$z' + y + 4 = 0\\
z = y'$
A: $\newcommand{\+}{^{\dagger}}
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$\ds{y'' + y + 4 = 0:\ {\large ?}}$.

\begin{align}
{y' \choose y}' &=\pars{\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}}
{y' \choose y} + {4 \choose 0}\quad\imp\quad
\vec{r}' + \ic\sigma_{y}\vec{r} = \vec{R}
\end{align}

where
$$
\vec{r} \equiv {y' \choose y}\,,\qquad
\sigma_{y} \equiv \pars{\begin{array}{rr}0 & -\ic \\ \ic & 0\end{array}}\,,
\qquad
\vec{R} \equiv {4 \choose 0}
$$

$$
\totald{\pars{\expo{\ic\sigma_{y}x}\vec{r}}}{x} =\expo{\ic\sigma_{y}x}\vec{R}
\qquad\imp\qquad\vec{r} =\expo{-\ic\sigma_{y}x}
\int\expo{\ic\sigma_{y}x}\vec{R}\,\dd x + \expo{-\ic\sigma_{y}x}\vec{C}\,,\quad
\vec{C}:\ \mbox{constant vector}
$$

$$
\vec{r} = -\ic\sigma_{y}\vec{R} + \expo{-\ic\sigma_{y}x}\vec{C}\,,
\qquad \expo{-\ic\sigma_{y}x} =\cos\pars{x} - \ic\sigma_{y}\sin\pars{x}
=\pars{\begin{array}{rr}\cos\pars{x} & -\sin\pars{x} \\ \sin\pars{x} & \cos\pars{x}\end{array}} 
$$

$$
\color{#88f}{\large y = -4 + C_{x}\sin\pars{x} + C_{y}\cos\pars{x}}
$$

