When we are solving initial value problems, we usually need to write these as sets of first order equations, because most of the program packages require this.
Example 1: $$ y''(x)+y(x)=0,\ y(0)=a_0,\ y'(0)=b_0 $$
We may for instance write this equation in a system as follows, $$ \begin{align*} y'(x)=&g(x)\\ g'(x)=&-y(x)\\ y(0)=&a_0,\ g(0)=b_0 \end{align*} $$
Example 2:
Another of a third order ODE is: $$ \begin{align} &y'''(x)+2y''(x)-(y'(x))^2+2y(x)=x^2 \tag{2.21}\\ &y(0)=a_0,\ y'(0)=b_0,\ y''(0)=c_0 \nonumber \end{align} $$
We set \( y'(x)=g(x) \) and \( y''(x)=g'(x)=f(x) \), and the system may be written as $$ \begin{align*} y'(x)=&g(x)\\ g'(x)=&f(x)\\ f'(x)=&-2f(x)+(g(x))^2-2y(x)+x^2 \end{align*} $$
with initial values \( y(0)=a_0,\ g(0)=b_0,\ f(0)=c_0 \).
This is fair enough for hand calculations, but when we use program packages a more systematic procedure is needed. Let's use the equation above as an example.
We start by renaming \( y \) to \( y_0 \). We then get the following procedure: $$ \begin{align*} y'&=y'_0=y_1\\ y''&=y''_0=y'_1=y_2 \end{align*} $$
Finally, the third order ODE in (2.21) may be represented as a system of first order ODEs: $$ \begin{align*} y'_0(x)=&y_1(x)\\ y'_1(x)=&y_2(x)\\ y'_2(x)=&-2y_2(x)+(y_1(x))^2-2y_0(x)+x^2 \end{align*} $$
with initial conditions \( y_0(0)=a_0,\ y_1(0)=b_0,\ y_2(0)=c_0 \).
The general procedure to reduce a higher order ODE to a system of first order ODEs becomes the following:
Given the equation $$ \begin{align} y^{(m)} &=f(x,y,y',y'',\dots,y^{(m-1)}) \tag{2.22}\\ y(x_0) &=a_0, y'(x_0)=a_1, \dots,y^{(m-1)}(x_0)=a_{m-1} \nonumber \end{align} $$ where $$ \begin{align*} y^{(m)}\equiv \frac{d^my}{dx^m} \end{align*} $$ with \( y=y_0 \), we get the following system of ODEs: $$ \begin{align} y'_0 &=y_1 \nonumber \\ y'_1 &=y_2 \nonumber \\ &. \tag{2.23} \\ &.\nonumber \\ y'_{m-2}&=y_{m-1} \nonumber\\ y'_{m-1}&= f(x,y_0,y_1,y_2,\dots,y_{m-1}) \nonumber \end{align} $$ with the following boundary conditions: $$ \begin{align} y_0(x_0) &=a_0, y_1(x_0)=a_1, \dots,y_{m-1}(x_0)=a_{m-1} \nonumber \end{align} $$
First we write \( y'''=y'y''+(y')^2-2y+x^3 \).
By use of (2.23) we get $$ \begin{align*} &y_0'=y_1\\ &y_1'=y_2\\ &y_2'=y_1y_2+(y_1)^2-2y_0+x^3\\ &y_0(0)=a,\ y_1(0)=b,\ y_2=c \end{align*} $$