Transform ODE system of first order into ODE of $n$-th order I know that you can always transform a higher order ODE into a system of first order ODEs.
In our lecture the professor made the statement that the reverse, namely transforming a given system of first order ODEs into one ODE of a higher order, is not always possible.
I was a bit confused by this statement because as far as I have understood the reverse is only possible in the very rare case where
\begin{align}
y_0'&=y_1\\
y_1'&=y_2\\
&\;\vdots\\
y_{n-1}'&=y_n\\
y_n'&=f(x,y_0,y_1, \cdots, y_n).
\end{align}
If the given system doesn't have this form you can't transform it into an ODE of $n$-th order.
Is this correct? Did I miss something?
 A: An autonomous system of differential equations
$$
\dot x= f(x),\quad x\in\Omega\subseteq\mathbb R^n,\quad f\in C^n(\Omega)
$$
can be transformed to the form of a n-th order differential equation iff there is a $C^{n}(\Omega)$ function
$\varphi:\; \mathbb R^n\to\mathbb R$ such that the system of functions
$$
\Phi(x)=(\varphi(x),\mathbf F \varphi(x),\ldots,\mathbf F^{n-1} \varphi(x))
$$
is functionally independent on $\Omega$. Here
$$
\mathbf F \varphi(x)=\sum_{i=1}^{n} f_i(x)\frac{\partial \varphi}{\partial x_i},
\;\mathbf F^2 \varphi(x)= \mathbf F( \mathbf F\varphi(x)),\;\ldots,\;
\mathbf F^{n-1} \varphi(x)= \underbrace{\mathbf F(\ldots \mathbf F(\mathbf F}_{n-1}\varphi(x))\ldots).
$$
The mapping $\mathbf F \varphi(x)$ (or $L_f \varphi(x)$) is sometimes called the Lie derivative and sometimes the derivative along the trajectories of the system (depending on the field of science involved).
If such a function $\varphi(x)$ is known, then the equivalent differential equation is
$$\tag{1}
y^{(n)}=\mathbf F^n \varphi(x){\Large|}_{x=\Phi^{-1}(\bar y)},\quad 
\bar y= (y,\dot y,\ldots,y^{(n-1)}).
$$
It can be obtained by using the change of variables
$$
y= \varphi(x),\;\dot y= \mathbf F \varphi(x),\;\ddot y= \mathbf F^2 \varphi(x),\;\ldots\; y^{(n-1)}=\mathbf F^{n-1} \varphi(x).
$$
For example, consider the Rossler system
$$
\left\{
\begin{array}{rcl}
\dot x_1&=&-x_2-x_3,\\
\dot x_2&=&x_1+ax_2,\\
\dot x_3&=&x_1x_3-bx_3+c.
\end{array}
\right.
$$
If we guess that we can choose $\varphi(x)=x_2$, then we can get the differential equation. The change of variables is
$$
y=x_2,
$$
$$
\dot y= \mathbf F y= \mathbf F x_2=  
(-x_2-x_3)\cdot 0+ (x_1+ax_2)\cdot1+ (x_1x_3-bx_3+c)\cdot0=
x_1+ax_2,
$$
$$
\ddot y= \mathbf F^2 x_2= \mathbf F (x_1+ax_2)= ax_1+(a^2-1)x_2-x_3;
$$
we can also express $x_1,x_2,x_3$ in terms of $y,\dot y,\ddot y$:
$$\tag{2}
x_1=\dot y-ay,\quad x_2=y,\quad x_3=-\ddot y+a\dot y-y.
$$
Finally, we can obtain the right part of the differential equation (1):
$$
\mathbf F^3 x_2= \mathbf F(ax_1+(a^2-1)x_2-x_3)
$$
$$
=a(-x_2-x_3)+(a^2-1)(x_1+ax_2)-(x_1x_3-bx_3+c).
$$
Applying (2) to express  $x_1,x_2,x_3$ in terms of $y,\dot y,\ddot y$ , we obtain the equation
$$
\dddot y=a\ddot y-\dot y+(\ddot y-a\dot y+y)(\dot y-ay-b)-c.
$$
A: It is also possible to transform slight modification of your example into a higher order ODE. For instance:
\begin{align}
y_1' &= 2y_2, \\
y_2' &= 6y_2 y_1.
\end{align}
Just define new state variables $z_1 = y_1$ and $z_2 = 2 y_2$. Then in this new variables you have
\begin{align}
z_1' &= z_2, \\
z_2' &= 3z_2 z_1,
\end{align}
which gives
$$
z_1'' = 3z_1'z_1 \quad \text{and hence}\quad y_1'' = 3y_1' y_1.
$$
Clearly you can do less obvious state variable transformations. I hope this solves your confusion.
